COMPANY SPECIFIC SERIES

ACCENTURE – CODING - TRAINER HANDOUT

PROGRAMS ASKED IN RECENT ACCENTURE DRIVES:

1. PROGRAM 1 - EDWARD’S BIRTHDAY

It is Edward's birthday today. His friends have bought him a huge circular cake.
Edward wants to find out the maximum number of pieces he can get by making exactly N
straight vertical cuts on the cake.

Your task is to write a function that returns the maximum number of pieces that can be
obtained by making N number of cuts.
Note: Since the answer can be quite large, modulo it by 1000000007

Input Specification:
input1: An integer N denoting the number of cuts
Output Specification:
Return the maximum number of pieces that can be obtained by making N cuts on the cake

Example:
Input Output Explanation
1 2 Given the above scenario, if we make 1 cut on the cake,
the maximum number of pieces of cake we obtain is 2 and
2%1000000007=2. Hence, 2 is returned as output.

5 16 Given the above scenario, if we make 5 cuts on the cake,

the maximum number of pieces of cake we obtain is 16
and 16% 1000000007=16. Hence, 16 is returned as output.

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Code Solution in C
#include<stdio.h>
int maximumCuts(int n){
const unsigned int M = 1000000007;
int sum=1;
while(n!=0)
sum+=n--;
return sum%M;
}
int main(){
int n;
scanf("%d",&n);
int ans = maximumCuts(n);
printf("%d\n",ans);
return 0;
}

Code Solution in C++

#include<bits/stdc++.h>
using namespace std;
int maximumCuts(int n){
const unsigned int M = 1000000007;
int sum=1;
while(n!=0)
sum+=n--;
return sum%M;
}
int main(){
int n;
cin >> n;
int ans = maximumCuts(n);
cout << ans <<"\n";
return 0;
}

Code Solution in JAVA

import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{

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public static void main (String[] args) throws IOException
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
Solution ob = new Solution();
System.out.println(ob.maximumCuts(n));
}
}
class Solution
{
static long maximumCuts(int n)
{
long M = 1000000007;
if(n==1 )
{
return 2;
}
if(n==0)
{
return 1;
}
return (n + maximumCuts(n-1))%M;
}
}

Code Solution in Python

class Solution:
def maximumCuts(self, n):
M = 1000000007
num=((n*(n+1))//2) +1
return num % M
if __name__ == '__main__':
n = int(input())
ob = Solution()
ans = ob.maximumCuts(n)
print(ans)

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2. PROGRAM 2 - REGIONS ON A PLANE

Problem statement
Mr. Professor is a great scientist, but he is not able to find a solution to one problem. There
are N straight lines that are not parallel, and no three lines go through the same point. The
lines divide the plane into M regions. Write a function to find out the maximum number of
such regions he can get on the plane.

Input Specification:
Input1: An integer N representing the number of straight lines (0 <=N<= 100)

Output Specification:
Return the maximum number of regions

Example:
Input Output Explanation
2 4 Given the above scenario, 2 lines divide the plane into 4
regions. Therefore, 4 is returned as the output.

3 7 Given the above scenario, 3 lines divide the plane into 7

regions. Therefore, 7 is returned as the output

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Code Solution in C
#include<stdio.h>
void maxRegions(int n)
{
int num;
num = n * (n + 1) / 2 + 1;
printf("%d",num);
}
int main()
{
int n;
scanf("%d",&n);
maxRegions(n);
return 0;
}

Code Solution in C++

#include<bits/stdc++.h>
using namespace std;
void maxRegions(int n)
{
int num;
num = n * (n + 1) / 2 + 1;
cout << num;
}
int main(){
int n;
cin >> n;
maxRegions(n);
return 0;
}

Code Solution in JAVA

import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws IOException
{

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Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
Solution ob = new Solution();
ob.maxRegions(n);
}
}
class Solution
{
public static void maxRegions(int n)
{
int num;
num = n * (n + 1) / 2 + 1;
System.out.println(num);
}
}

Code Solution in Python

class Solution:
def maxRegions(self, n):
num = n * (n + 1) // 2 + 1
return num
if __name__ == '__main__':
n = int(input())
ob = Solution()
ans = ob.maxRegions(n)
print(ans)

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3. PROGRAM 3 - SUM OF ODD INTEGERS IN ARRAY

Problem statement
An odd number is an integer which is not a multiple of 2.
You are required to implement the following function:
Int SumOddIntegers(int arr[], int n);
The function accepts an integer array ‘arr’ of length ‘n’ as its argument. You are required to
calculate the sum of all odd integers in an array ‘an’ and return the same
Note:
Array can have negative integers
n>0
Computed values lie within integer range

Example:
Input:
arr: 1 4 6 7 10 12 11 5
n: 8
Output:
24
Explanation:
The odd integers in array {1, 4, 6, 7, 10, 12, 11, 5} are {1, 7,11, 5} and their sum is 24, hence
24 is returned.
The custom input format for the above case:
8
1 4 6 7 10 12 11 5
(The first line represents ‘n’ the second line represents the elements of the array)

Sample Input
arr: 2 4 9 7 11 13 25 31 6 8 10 24
n: 12
Sample Output
96
The custom input format for the above case:
12
2 4 9 7 11 13 25 31 6 8 10 24
(The first line represents ‘n’ the second line represents the elements of the array)

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Code Solution in C++

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4. PROGRAM 4 - INVERSION COUNT IN ARRAY

Problem statement
Let j and k be two indices in an array A.
If j<K and A[j] > A[k], then the pair (j,k) is known as an “Inversion pair”.
You are required to implement the following function:
int InversionCount(int *A, int n);
The function accepts an array ‘A’ of ‘n’ unique integers as its argument. You are required to
Calculate the number of ‘Inversion pair’ in an array A, and return.
Note:
If ‘A’ is NULL (None in case of python). return – 1
I_f_ _‘n_’ _<_2_,_ _r_e_t_u_r_n_ _0_ _
Example:
Input:
A: 1 20 6 4 5
n: 5
Output:
5
Explanation:
The inversion pair in array A are (20,6),(20,4),(20,5),(6,4) and (6,5), the count of the
inversions are 5, hence 5 is returned.
The custom input format for the above case:
5
1 20 6 4 5
(The first line represents the size of the array, the second line represents the elements of the
array)
Sample Input
A: 13 10 9 6 21 15 14
n: 7
Sample Output
9
The custom input format for the above case:
7
13 10 9 6 21 15 14

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5. PROGRAM 5 - SUPERIOR ARRAY ELEMENT

Problem statement
In an array a superior element is one which is greater than all elements to its right.
The rightmost element will always be considered as a superior element.

You are given a function,

Int FindNumberOfSuperiorElements(int* arr, int n);
The function accepts an integer array ‘arr’ and its length ‘n’. Implement the function to find
and return the number of superior elements in array ‘arr’.

Assumptions:
1. N > 0.
2. Array index starts from 0.

Example:
Input Output Explanation
Arr: 7 9 5 2 8 7 3 9 is greater than all the elements to its right, 8 is
greater than element to its right and 7 is the rightmost
element. Hence total 3 superior elements.

The Custom input format for the above case:

6
795287
(The first line represents ‘n’ the second line represents the elements of the array ‘arr’)

Sample Input Output

Arr: 2 8 9 7 4 2 4

The custom input format for the above case:

Input:
6
289742
(The first line represents ‘n’, the second line represents the elements of the array ‘arr’)
Output:
4
Explanation: 9 is greater than all the elements to its right, 7 is greater than element to its right,
4 is greater than element to its right and 2 is the rightmost element. Hence total 4 superior
elements.

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6. PROGRAM 6 - RAT COUNT HOUSE

Problem Description:
The function accepts two positive integers ‘r’ and ‘unit’ and a positive integer array ‘arr’ of
size ‘n’ as its argument ‘r’ represents the number of rats present in an area, ‘unit’ is the
amount of food each rat consumes and each ith element of array ‘arr’ represents the amount
of food present in ‘i+1’ house number, where 0 <= i

Note:
● Return -1 if the array is null
● Return 0 if the total amount of food from all houses is not sufficient for all the rats.
● Computed values lie within the integer range.

Example:

Input:

● r: 7
● unit: 2
● n: 8
● arr: 2 8 3 5 7 4 1 2

Output: 4

Explanation:

Total amount of food required for all rats = r * unit

= 7 * 2 = 14

The amount of food in 1st houses = 2+8+3+5 = 18. Since, the amount of food in 1st 4 houses
is sufficient for all the rats. Thus, output is 4.

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7. PROGRAM 7 - BINARY STRING

Problem Description:
The Binary number system only uses two digits, 0 and 1 and the number system can be called
binary string. You are required to implement the following function:

int OperationsBinaryString(char* str);

The function accepts a string str as its argument. The string str consists of binary digits
eparated with an alphabet as follows:
● – A denotes AND operation
● – B denotes OR operation
● – C denotes XOR Operation

You are required to calculate the result of the string str, scanning the string to right taking one
operation at a time, and return the same.
Note:
● No order of priorities of operations is required
● Length of str is odd
● If str is NULL or None (in case of Python), return -1

Input:
str: 1C0C1C1A0B1
Output:
1

Explanation:
The alphabets in str when expanded becomes “1 XOR 0 XOR 1 XOR 1 AND 0 OR 1”, result
of the expression becomes 1, hence 1 is returned.

Sample Input:
0C1A1B1C1C1B0A0
Output:
0

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8. PROGRAM 8 - PASSWORD CHECKER

You are given a function.

int CheckPassword(char str[], int n);

The function accepts string str of size n as an argument. Implement the function which

returns 1 if given string str is valid password else 0.

str is a valid password if it satisfies the below conditions.

● At least 4 characters
● At least one numeric digit
● At Least one Capital Letter
● Must not have space or slash (/)
● Starting character must not be a number

Assumption:

Input string will not be empty.

Example:

Input 1:

aA1_67

Input 2:

a987 abC012

Output 1:

Output 2:

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9. PROGRAM 9

You are given a function,

int findCount(int arr[], int length, int num, int diff);

The function accepts an integer array ‘arr’, its length and two integer variables ‘num’ and
‘diff’. Implement this function to find and return the number of elements of ‘arr’ having an
absolute difference of less than or equal to ‘diff’ with ‘num’.

Note: In case there is no element in ‘arr’ whose absolute difference with ‘num’ is less than or
equal to ‘diff’, return -1.

Example:
Input:
● arr: 12 3 14 56 77 13
● num: 13
● diff: 2
Output:
3

Explanation:
Elements of ‘arr’ having absolute difference of less than or equal to ‘diff’ i.e. 2 with ‘num’
i.e. 13 are 12, 13 and 14.

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10. PROGRAM 10

Implement the following Function

def differenceofSum(n. m)

The function accepts two integers n, m as arguments Find the sum of all numbers in range
from 1 to m(both inclusive) that are not divisible by n. Return difference between sum of
integers not divisible by n with sum of numbers divisible by n.

Assumption:
● n>0 and m>0
● Sum lies between integral range

Example
Input
n:4
m:20
Output
90

Explanation
● Sum of numbers divisible by 4 are 4 + 8 + 12 + 16 + 20 = 60
● Sum of numbers not divisible by 4 are 1 +2 + 3 + 5 + 6 + 7 + 9 + 10 + 11 + 13 + 14 +
15 + 17 + 18 + 19 = 150
● Difference 150 – 60 = 90

Sample Input
n:3
m:10

Sample Output
19

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11. PROGRAM 11

You are required to implement the following Function

def LargeSmallSum(arr)

The function accepts an integers arr of size ’length’ as its arguments you are required to
return the sum of second largest element from the even positions and second smallest from
the odd position of given ‘arr’
Assumption:
● All array elements are unique
● Treat the 0th position as even
NOTE
● Return 0 if array is empty
● Return 0, if array length is 3 or less than 3
Example
Input
arr:3 2 1 7 5 4
Output
7

Explanation
● Second largest among even position elements(1 3 5) is 3
● Second smallest among odd position element is 4
● Thus output is 3+4 = 7

Sample Input
arr:1 8 0 2 3 5 6
Sample Output
8

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12. PROGRAM 12

Implement the following Function

def ProductSmallestPair(sum, arr)

The function accepts an integer sum and an integer array arr of size n. Implement the function
to find the pair, (arr[j], arr[k]) where j!=k, Such that arr[j] and arr[k] are the least two
elements of array (arr[j] + arr[k] <= sum) and return the product of element of this pair
NOTE
● Return -1 if array is empty or if n<2
● Return 0, if no such pairs found
● All computed values lie within integer range

Example
Input
sum:9
size of Arr = 7
Arr:5 2 4 3 9 7 1
Output
2

Explanation
Pair of least two element is (2, 1) 2 + 1 = 3 < 9, Product of (2, 1) 2*1 = 2. Thus, output is 2

Sample Input
sum:4
size of Arr = 6
Arr:9 8 3 -7 3 9

Sample Output
-21

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13. PROGRAM 13

N-base notation is a system for writing numbers that uses only n different symbols, This
symbols are the first n symbols from the given notation list(Including the symbol for o)
Decimal to n base notation are (0:0, 1:1, 2:2, 3:3, 4:4, 5:5, 6:6, 7:7, 8:8, 9:9, 10:A,11:B and so
on up to 35:Z)

Implement the following function

Char* DectoNBase (int n, int num):

The function accept positive integer n and num Implement the function to calculate the
n-base equivalent of num and return the same as a string

Steps:
1. Divide the decimal number by n, Treat the division as the integer division
2. Write the the remainder (in n-base notation)
3. Divide the quotient again by n, Treat the division as integer division
4. Repeat step 2 and 3 until the quotient is 0
5. The n-base value is the sequence of the remainders from last to first

Assumption:
1 < n < = 36

Example
Input
n: 12
num: 718
Output
4BA

Explanation
num Divisor quotient remainder
718 12 59 10(A)
59 12 4 11(B)
4 12 0 4(4)

Sample Input
n: 21
num: 5678

Sample Output
CI8

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14. PROGRAM 14

Implement the following functions

char*MoveHyphen(char str[],int n);

The function accepts a string “str” of length ‘n’, that contains alphabets and hyphens (-).
Implement the function to move all hyphens(-) in the string to the front of the given string.

NOTE: Return null if str is null.

Example:
Input:
str.Move-Hyphens-to-Front
Output:
—MoveHyphenstoFront

Explanation:
The string “Move-Hyphens -to-front” has 3 hyphens (-), which are moved to the front of the
string, this output is “— MoveHyphen”

Sample Input
Str: String-Compare

Sample Output
StringCompare

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15. PROGRAM 15

Problem Statement
A carry is a digit that is transferred to left if sum of digits exceeds 9 while adding two
numbers from right-to-left one digit at a time

You are required to implement the following function.

Int NumberOfCarries(int num1 , int num2);

The function accepts two numbers ‘num1’ and ‘num2’ as its arguments. You are required to
calculate and return the total number of carries generated while adding digits of two numbers
‘num1’ and ‘ num2’.

Assumption: num1, num2>=0

Example:
● Input
○ Num 1: 451
○ Num 2: 349
● Output
○ 2

Explanation:
Adding ‘num 1’ and ‘num 2’ right-to-left results in 2 carries since ( 1+9) is 10. 1 is carried
and (5+4=1) is 10, again 1 is carried. Hence 2 is returned.

Sample Input
Num 1: 23
Num 2: 563

Sample Output
0

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16. PROGRAM 16

Problem Statement
You are given a function,
Void *ReplaceCharacter(Char str[], int n, char ch1, char ch2);

The function accepts a string ‘ str’ of length n and two characters ‘ch1’ and ‘ch2’ as its
arguments . Implement the function to modify and return the string ‘ str’ in such a way that
all occurrences of ‘ch1’ in original string are replaced by ‘ch2’ and all occurrences of ‘ch2’
in original string are replaced by ‘ch1’.

Assumption: String Contains only lower-case alphabetical letters.

Note:
● Return null if string is null.
● If both characters are not present in string or both of them are same , then return the
string unchanged.
Example:
● Input:
○ Str: apples
○ ch1:a
○ ch2:p
● Output:
○ paales
Explanation:
‘A’ in the original string is replaced with ‘p’ and ‘p’ in original string is replaced with ‘a’,
thus output is paales.

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17. PROGRAM 17 - OPERATION CHOICES

Problem Statement
You are required to implement the following function.
Int OperationChoices(int c, int n, int a , int b)

The function accepts 3 positive integers ‘a’ , ‘b’ and ‘c ‘ as its arguments. Implement the
function to return.
● ( a + b ) , if c=1
● ( a – b ) , if c=2
● ( a * b ) , if c=3
● (a / b) , if c =4

Assumption: All operations will result in integer output.

Example:
● Input
○ c :1
○ a:12
○ b:16
● Output:
○ Since ‘c’=1 , (12+16) is performed which is equal to 28 , hence 28 is returned.
Sample Input
c:2
a : 16
b : 20
Sample Output
-4

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18. PROGRAM 18

Problem Statement

You are given a function,

Int MaxExponents (int a , int b);

You have to find and return the number between ‘a’ and ‘b’ ( range inclusive on both ends)

which has the maximum exponent of 2.

The algorithm to find the number with maximum exponent of 2 between the given range is

1. Loop between ‘a’ and ‘b’. Let the looping variable be ‘i’.
2. Find the exponent (power) of 2 for each ‘i’ and store the number with maximum
exponent of 2 so faqrd in a variable , let say ‘max’. Set ‘max’ to ‘i’ only if ‘i’ has
more exponent of 2 than ‘max’.
3. Return ‘max’.

Assumption: a <b

Note: If two or more numbers in the range have the same exponents of 2 , return the small

number.

Example

● Input:
○ 7
○ 12
● Output:
○ 8

Explanation:

Exponents of 2 in:
7-0
8-3
9-0
10-1
11-0
12-2
Hence maximum exponent if two is of 8.

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19. PROGRAM 19

Problem Statement
You are required to implement the following function:
Int Calculate(int m, int n);

The function accepts 2 positive integer ‘m’ and ‘n’ as its arguments.You are required to
calculate the sum of numbers divisible both by 3 and 5, between ‘m’ and ‘n’ both inclusive
and return the same.

Note
0 < m <= n

Example
Input:
m : 12
n : 50
Output
90

Explanation:
The numbers divisible by both 3 and 5, between 12 and 50 both inclusive are {15, 30, 45}
and their sum is 90.

Sample Input
m : 100
n : 160

Sample Output
510

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20. PROGRAM 20

Problem Statement
You are required to input the size of the matrix then the elements of matrix, then you have to
divide the main matrix in two sub matrices (even and odd) in such a way that element at 0
index will be considered as even and element at 1st index will be considered as odd and so
on. Then, you have to sort the even and odd matrices in ascending order and then print the
sum of the second largest number from both the matrices.

Example
● enter the size of array : 5
● enter element at 0 index : 3
● enter element at 1 index : 4
● enter element at 2 index : 1
● enter element at 3 index : 7
● enter element at 4 index : 9

Sorted even array: 1 3 9

Sorted odd array: 4 7

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21. PROGRAM 21

Instructions: You are required to write the code. You can click on compile and run anytime
to check compilation/execution status. The code should be logically/syntactically correct.

Problem: Write a program in C to display the table of a number and print the sum of all the
multiples in it.

Test Cases:
Test Case 1:
Input:
5
Expected Result Value:
5, 10, 15, 20, 25, 30, 35, 40, 45, 50
275

Test Case 2:
Input:
12
Expected Result Value:
12, 24, 36, 48, 60, 72, 84, 96, 108, 120
660

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22. PROGRAM 22 - INTERMEDIATE PALINDROME NUMBERS

Instructions: You are required to write the code. You can click on compile and run anytime
to check compilation/execution status. The code should be logically/syntactically correct.

Question: Write a program such that it takes a lower limit and upper limit as inputs and print
all the intermediate palindrome numbers.

Test Cases:

TestCase 1:
Input :
10 , 80
Expected Result:
11 , 22 , 33 , 44 , 55 , 66 , 77.

Test Case 2:
Input:
100, 200
Expected Result:
101 , 111 , 121 , 131 , 141 , 151 , 161 , 171 , 181 , 191.

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23. PROGRAM 23

Instructions: You are required to write the code. You can click on compile & run anytime to
check the compilation/ execution status of the program. The submitted code should be
logically/syntactically correct and pass all the test cases.

Question: The program is supposed to calculate the sum of distance between three points
from each other.

For
x1 = 1 y1 = 1
x2 = 2 y2 = 4
x3 = 3 y3 = 6
Distance is calculated as : sqrt(x2-x1)^2 + (y2-y1)^2

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24. PROGRAM 24 - FIND THE MAXIMUM VALUE AND ITS INDEX IN THE
ARRAY

Problem Statement:
You are given a function, void MaxInArray(int arr[], int length); The function accepts an
integer array ‘arr’ of size ‘length’ as its argument. Implement the function to find the
maximum element of the array and print the maximum element and its index to the standard
output
(STDOUT). The maximum element and its index should be printed in separate lines.

Note:
● Array index starts with 0
● Maximum element and its index should be separated by a line in the output
● Assume there is only 1 maximum element in the array
● Print exactly what is asked, do not print any additional greeting messages

Example:
Input:
23 45 82 27 66 12 78 13 71 86
Output:
86
9

Explanation:
86 is the maximum element of the array at index 9.

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25. PROGRAM 25 - AUTOBIOGRAPHICAL NUMBER

Problem Statement:
An Autobiographical Number is a number N such that the first digit of N represents the count
of how many zeroes are there in N, the second digit represents the count of how many ones
are there in N and so on.

You are given a function, def FindAutoCount(n):

The function accepts string “n” which is a number and checks whether the number is an
autobiographical number or not. If it is, an integer is returned, i.e. the count of distinct
numbers in ‘n’. If not, it returns 0.

Assumption:
● The input string will not be longer than 10 characters.
● Input string will consist of numeric characters.

Note:
If string is None return 0.

Example:
Input:
n: “1210”
Output:
3

Explanation:
0th position in the input contains the number of 0 present in input, i.e. 1, in 1st position the
count of number of 1s in input i.e. 2, in 2nd position the count of 2s in input i.e. 1, and in 3rd
position the count of 3s i.e. 0, so the number is an autobiographical number.
Now unique numbers in the input are 0, 1, 2, so the count of unique numbers is 3. So 3 is
returned.

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26. PROGRAM 26 - BALANCE FRUITS

Implement the following function:

int BalanceFruits(int a, int m, int rs);

You have a basket full of apples and mangoes, your job is to make the number of apples and
given a function that accepts three integers 'a', 'm' and 'rs' as its argument where 'a' and a
basket respectively and 'rs' is the rupees that you have. Implement the function to balance the
basket.
• If 'a' > 'm', then buy (a - m) mangoes at the rate of Rs 1 per mango.
• If 'a' < 'm', then sell (m - a) mangoes at the rate of Rs 1 per mango.
Return the total rupees left with you after balancing the fruits.

Assumption:
• a > = 0, m > = 0 and rs > = 0
• rs > = (a - m)
Note: If a = m, return rs unchanged

Example
Sample Input
a:8
m:4
Rs : 6

Sample Output
2

Explanation: Since a > m, (a - m) mangoes are bought at Rs 1 per mango, so rs = 6 - 4 = 2.

Thus, output is 2.

Case 1
Input
846
Output
2

Case 2
Input
746
Output
3

56
Code Solution:

#include<iostream>
using namespace std;
int main()
{
int a, m, rs;
cin >> a >> m >> rs;
if(a > m)
{
cout << rs - (a - m);
}
else if( a == m)
{
cout << rs;
}
else if( a < m)
{
cout << rs + (m - a);
}
}

57
27. PROGRAM 27 - TALLEST TREE

A TreeHeight is represented by the following structure:

struct TreeHeight
{
int feets;
int inches;
};

You are given a function,

int TallestTree(struct TreeHeight trees[], int n);

The function accepts an array 'trees' of type 'TreeHeight' consisting of 'n' elements as its
argument. 'TreeHeight' consists of two integers 'feets' and 'inches' which represents the height
of a tree. Implement the function to find the tallest tree among the 'trees' and return its height
(in inches).

Height of a tree = (12 * feets) + inches

Assumption:
• feets > 0
• 0 > = inches > = 11
Note:
• Computed value lies within integer range.
• Return -1 if trees is null (None, in case of Python).
• trees(0) is the first element.

Example:
Sample Input:
Feets Inches
10 4
23 5
21 2
27 7
Sample Output:
33 1

58
Explanation:
Height of trees (in inches):
• trees[0] = (10 * 12) + 4 = 124
• trees[1] = (23 * 12) + 5 = 281
• trees[2] = (21 * 12) + 2 = 254
• trees[3] = (27 * 12) + 7 = 331
Maximum of {124, 281, 254, 331} = 331, thus output is 331.

Case 1
Input
4
10 4
23 5
21 2
27 7
Output
331

Case 2
2
46 10
55 6
Output
666

59
Code Solution:

#include<iostream>
using namespace std;
struct TreeHeight
{
int feets;
int inches;
};
int main()
{
int n;
cin >> n;
struct TreeHeight arr[n];
for(int i = 0; i < n; i++)
{
cin >> arr[i].feets;
cin >> arr[i].inches;
}
int max = 0;
for(int i = 0; i < n; i++)
{
int temp = arr[i].feets * 12 + arr[i].inches;
if(max < temp)
max = temp;
}
cout << max;
}

60
28. PROGRAM 28 - FRAGMENTS DIVISIBLE BY ELEVEN

You are given a function,

def DivisibilityByEleven(num):

The function accepts an integer 'num' as input. You have to implement the function such that
it returns the number of contiguous integer fragments of 'num' that are divisible by 11.
Contiguous integer fragments of a number, say 1273, are:

Example:
Input:
1215598
Output:
4

Explanation: 55, 121, 12155 and 15598 are contiguous fragments of the number 1215598 that
are divisible by 11.

Case 1:
Input
1215598
Output
5

Case 2:
Input
55
Output
1

61
Code Solution:

#include<bits/stdc++.h>
using namespace std;
int main()
{
int n;
cin >> n;
int temp = n;
int sum, count = 0;
vector <int> v;
int len = 0;
while(n != 0)
{
v.push_back(n % 10);
n = n / 10;
}
reverse(v.begin(), v.end());
for(int i = 0; i < v.size(); i++)
{
sum = v[i];
for(int j = i+1; j < v.size(); j++)
{
sum = sum * 10 + v[j];
if(sum % 11 == 0)
count++;
} }
cout << count; }

62
29. PROGRAM 29 - CONVERT TO PALINDROME

A palindrome is a sequence of character that has the property of reading the same way either
in direction. You are given a function,
char *ConvertPalindrome(char* str)

The function accepts the string str. Implement the function to find and return the minimum
characters required to append at the end of string str to make it palindrome

Assumption
1.String will contain only lower case English alphabets.
2.Length of string is greater than or equal to 1

Note:
1. If string is already palindrome return “NULL”.
2. You have to find the minimum characters required to append at the end of string to make it
palindrome.

Example:
Sample Input
abcdc
Sample Output
ba
Explanation: If we append ‘ba’ at the end of the string ‘abcdc’ it becomes abcdcba i.e. a
palindrome.

Case 1:
Input
abcdc
Output
ba

Case 2:
Input
ab
Output
a

63
Code Solution:

#include<bits/stdc++.h>
using namespace std;
bool checkPalindrome(string s,int i,int j)
{
while(i<j)
{
if(s[i] != s[j])
return false;
i++;
j--;
}
return true;
}
int main()
{
string inp;
cin>>inp;
int n=inp.length();
int i;
for(i=0;i<n;i++)
{
if(checkPalindrome(inp,i,n-1))
break;
}
for(int j = i-1; j >= 0; j--)
cout << inp[j];
return 0;
}

64
30. PROGRAM 30 - COUNT SPECIFIC NUMBERS

You are required to implement the following function:

int CountSpecificNumbers(int m, int n);

The function accept two arguments m and n which are integers. You are required to calculate
the count of numbers having only 1, 4 and 9 as their digits between the numbers lying in the
range m and n both inclusive, and return the same. Return -1 if m>n.

Example:
Sample Input:
100
200
Sample Output:
9
Explanation: The numbers between 100 and 200, both inclusive having only 1, 4 and 9 as
their digits are 111, 114, 119, 141, 144, 149, 191, 194, 199. The count is 9 hence 9 is
returned.

Case 1:
Input
100
200
Output
9

Case 2:
Input
1
100
Output
12

65
Code Solution:

#include<bits/stdc++.h>
using namespace std;
int CountSpecificNumber(int m,int n);
int main()
{
int m,n;
cin>>m>>n;
cout<<CountSpecificNumber(m,n);
return 0;
}
int CountSpecificNumber(int m,int n)
{
if(m<=n)
{
int i,count=0;
for(i=m;i<=n;i++)
{
int num=i,flag=1;
while(num)
{
int n = num%10;
num=num/10;
if(n == 1 || n == 4 || n == 9)
continue;
else
{
flag=0;
break;
}
}
if(flag==1)
count++;
}
return count;
}
return -1;
}

66
31. PROGRAM 31 - PICKING TICKETS

Consider an array of n tickets prices, tickets. A number, m, is defined as the size of some
subsequence, s, of tickets where each element covers an unbroken range of integers. That is
to say if you were to sort the elements in s, the absolute difference between any elements j
and j+1 would be either 0 or 1. Determine the maximum length of a subsequence chosen
from the ticket array.

Example:
Tickets = [8,5,4,8,4]
Valid subsequences, sorted, are{4,4,5} and {8,8}.These subsequences have m, values of 3
and 2, respectively. Return 3.
Return:
Int: an integer that denotes the maximum possible value of m.

Constraints:
1<= n <=105
1<= tickets[i] <= 109

Example:
Sample Input:
STDIN FUNCTION
--------- ------------------
4 -> tickets [] size n=4
4 -> tickets = [4,13,2,3]
13
2
3
Sample Output:
3

Explanation: There are two subsequences of tickets that contain consecutive integers when
sorted: {2,3,4} and {13}. These subsequences have m values of 3 and 1, respectively. Return
the maximum values of m, which is 3.

Case 1:
Input
4
4
13
2

67
3
Output
3

Case 2:
Input
5
8 5 4 8 4
Output
3

68
Code Solution:

#include<bits/stdc++.h>
using namespace std;
int maxSubsequence(int a[],int n)
{
int max=0,i,count=0,flag=0;
for(i=0;i<n-1;i++)
{
int dif = a[i+1]-a[i];
if(dif==1||dif==0)
{
count++;
flag=1;
}
else
{
if(count>max)
max=count;
count=0;
}
}
if(flag)
return count>max?count+1:max+1;
else
return count;
}
int main()
{
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++)
cin>>a[i];
sort(a,a+n);
cout<<maxSubsequence(a,n);
}

69
32. PROGRAM 32 - COUNT OCCURRENCES OF DIGIT

You are required to implement the following function:

int CountDigiOoccurrences(int l, int u, int x);

The function accepts 3 positive integers ‘l’, 'u' and 'x' as its arguments. You are required to
calculate the number of occurrences of a digit 'x' in the digits of numbers lying in the range
between ‘I’ and 'u' both inclusive, and return the same.

Note:
• I<u
• 0<x<9

Example:
Sample Input:
I: 2
u: 13
x: 3
Sample Output:
2

Explanation: The number of occurrences of digit 3 in the digits of numbers lying in the
range [2, 13] both inclusive is 2, ie (3, 13), hence 2 is returned.

Case 1:
Input
2 13 3
Output
2
Case 2:
Input
1 30 2
Output
13

70
Code Solution:

#include<iostream>
using namespace std;
int countDigitOccurrences(int l, int u, int x)
{
int rem, count = 0;
for(int i = l; i <= u; i++)
{
int temp = i;
while(temp != 0 )
{
if(temp % 10 == x)
count++;
temp = temp / 10;
}
}
return count;
}
int main()
{
int l, u, x;
cin >> l >> u >> x;
cout << countDigitOccurrences(l, u, x);
return 0;
}

71
33. PROGRAM 33 - FIND THE BMI

A person’s body mass index is a simple calculation based on height and weight that classifies
the person as underweight, overweight, or normal. The formula for metric unit is,
BMI = weight in kilograms / (height in mesters)2

You are given a function,

Int GetBMICategory(int weight, float height);

The function accepts the weight (an integer number) and height (a floating point number) of a
person as its arguments. Implement the function such that it calculations the BMI of the
person and returns an integer, the person’s BMI category as per the given rules:
1. If BMI < 18, return 0.
2. If 18 >= BMI < 25, return 1.
3. If 25 >= BMI <30, return 2.
4. If 30 >= BMI < 40, return 3.
5. If BMI >= 40, return 4.

Note:
Weight > 0 and its unit is kilogram.
Height > 0 and its unit is metre.
Compute BMI as a floating-point.

Example:
Input:
42
1.54
Output:
0

Explanation: The person’s weight is 42Kg and height is 1.54 metres BMI = weight in
kilograms / (height in meters)2 = 42/(1.54 * 1.54) = 17.70
Since, BMI < 18, as per given rules, the output is 0.

Sample Input:
62
1.52
Sample Output:
2

72
Case 1:
Input
42
1.54
Output
0

Case 2:
Input
62
1.52
Output
2

73
Code Solution:

#include <iostream>
using namespace std;
int findBMI(int weight,float height){
float ans = (float) (weight / (height * height));
if(ans < 18)
return 0;
else if(ans >= 18 && ans < 25)
return 1;
else if(ans >= 25 && ans < 30)
return 2;
else if(ans >= 30 && ans < 40)
return 3;
else
return 4;
}

int main() {
int weight;
float height;
cin>>weight>>height;
cout<<findBMI(weight,height);

return 0;
}

74
34. PROGRAM 34 - NUMEROLOGICAL REDUCTION

Numerological reduction of a number is the sum of digits calculated recursively until the
resulting value is a single digit.

You are given a function, int FindNumerologicalReduction(int n);

The function accepts a positive integer ‘n’. Implement the function to find and return the
numerological reduction of ‘n’.

Assumption: n> = 0

Example:
Sample Input:
3245
Sample Output:
5

Explanation:
3245 = 3+2+4+5 = 14
14 = 1+4 = 5
Hence 5 is the numerological reduction of 3245.

Case 1:
Input
3245
Output
5

Case 2:
Input
12345
Output
6

75
Code Solution:

#include <iostream>
using namespace std;
int main()
{
int n;
cin >> n;
if(n % 9 == 0)
cout << "9";
else
cout << n % 9;

return 0;
}

76
35. PROGRAM 35 - EMAIL ID OBFUSCATION

Personal information like email id and other contact details need to be handled in a way so
that privacy of the user is safeguarded. One way to do this is to hide or partially hide such
information, also called obfuscation of the information.

Given the email id as input, the program should obfuscate the id as follows:
1) For the given mail id, the part that comes before @ is referred to as the first part. If there is
no first @ character, the mail id is invalid.
2) If the first part of the email id is less than or equal to 5 characters in length, replace all
characters in the first part with *
3) If the first of the email id is greater than 5 characters in length, print the first 3 characters
as it is and replace the remaining characters of the first part with *
4) If the email id is invalid, print Invalid Input.

Example 1:
Input :
abc@gmail.com
Output :
***@gmail.com

Explanation: Here the first part is 'abc'. This is less than 5 characters in length, so replace all
characters in the first part with *.

Example 2:
Input :
abcdefghi@gmail.com
Output :
abc******@gmail.com
Explanation: Here the first part is 'abcdefghi'. This is more than 5 characters in length, so
replace all characters in the first part except the first with *.

Complete the function obfuscateMailId in the editor. The function must print the obfuscate
mail id as per the rules mentioned. If the mail id is invalid, it should print Invalid Input.
obfuscateMailId has the following parameter(s) :
mailid : a string representing the mail id
Constraints :
1) A valid mail id should contain the character @ ( e.g. myid_at_gmail is invalid)
2) A valid mail id should have first part with at least one or more characters ( e.g. @myid is
invalid)

77
36. PROGRAM 36 - SUM OF NUMBERS DIVISIBLE BY BOTH 3 AND 5

You are required to implement the following function:

Int Calculate (int m, int n);

The function accepts 2 positive integer ‘m’ and ‘n’ as its arguments.You are required to
calculate the sum of numbers divisible both by 3 and 5, between ‘m’ and ‘n’ both inclusive
and return the same.

Note: 0 < m <= n

Example
Input:
m : 12
n : 50
Output
90

Explanation: The numbers divisible by both 3 and 5, between 12 and 50 both inclusive are
{15, 30, 45} and their sum is 90.

Sample Input
m : 100
n : 160
Sample Output
405

Case 1
Input (stdin)
12
50
Output (stdout)
90

Case 2
Input (stdin)
20
30
Output (stdout)
30

78
Code Solution:

#include<iostream>
using namespace std;

int Calculate(int, int);

int main()
{
int m, n, result;
cin>>m;
cin>>n;
result = Calculate(n,m);
cout<<result;
return 0;
}

int Calculate(int n, int m)

{
int i, sum = 0;
for(i=m;i<=n;i++)
{
if((i%3==0)&&(i%5==0))
{
sum = sum + i;
}
}
return sum;
}

79
37. PROGRAM 37 - TABLE OF NUMBERS

Write a program to display the table of a number and print the sum of all the multiples in it.

Case 1
Input (stdin)
12
Output (stdout)
12 24 36 48 60 72 84 96 108 120
660

Case 2
Input (stdin)
5
Output (stdout)
5 10 15 20 25 30 35 40 45 50
275

80
Code Solution:

#include <iostream>
using namespace std;

int main()
{
int n, i, value=0, sum=0;
cin>>n;
for(i=1; i<=10; ++i)
{
value = n * i;
cout<<value<<" ";
sum=sum+value;
}
cout<<endl<<sum ;
return 0;
}

81
38. PROGRAM 38

Robert is expert in strings where he challenges everyone to write a program for the below
implementation.
Two strings and comprising of lower case English letters are compatible if they are equal or
can be made equal by following this step any number of times:
Select a prefix from the string (possibly empty), and increase the alphabetical value of all the
characters in the prefix by the same valid amount.
For example if the string is abc and we select the prefix ab then we can convert it to bcc by
increasing the alphabetical value by 1. But if we select the prefix abc then we cannot increase
the alphabetical value.
Your task is to determine if given strings are compatible.

Input format
First line: String A
Next line: String B

Output format:
For each test case, print YES if string can be converted to string , otherwise print NO.

Constrain: 1<=(len of A,B)<1000005

Sample Input:
abaca
cdbda
Sample Output:
YES
Explanation: The string abaca can be converted to bcbda in one move and to cdbda in the
next move.

Test Case 1
Input
abaca cdbda
Output
YES
Test Case 2
Input
abcda abcda
Output
NO

82
Code Solution:

#include<iostream>
#include<string.h>
using namespace std;
int main()
{
char str1[1000005];
char str2[1000005];
cin>>str1>>str2;
int max = str2[0] - str1[0];
if(max > 0)
{
for(int i =1;i<strlen(str1);i++)
{
int x = str2[i] - str1[i];
if(x > max)
{
cout<<"NO";
return 0;
}
}
cout<<"YES";
}
else
cout<<"NO";
return 0;
}

83
39. PROGRAM 39 - SECOND GREATEST ELEMENT

Write a program to print the second greatest element in the given array. In case, if the second
largest element is not present in the array, print "There is no second largest element" and if
the size of the array is less than 2, print "Invalid Input".

Sample Input:
5
12 5 7 3 90
Sample Output:
12

Case 1
Input (stdin)
5
12 5 7 3 90
Output (stdout)
12

Case 2
Input (stdin)
10
10 20 50 43 60 51 70 82 93 55
Output (stdout)
82

84
Code Solution:
#include<iostream>
using namespace std;
#include <limits.h>

void print2largest(int arr[], int arr_size)

{
int i, first, second;
if (arr_size < 2)
{
cout<<"Invalid Input";
return;
}
first = second = INT_MIN;
for (i = 0; i < arr_size ; i ++)
{
if (arr[i] > first)
{
second = first;
first = arr[i];
}
else if (arr[i] > second && arr[i] != first)
second = arr[i];
}

if (second == INT_MIN)
cout<<"There is no second largest element";
else
cout<<second;
}

int main()
{
int n;
cin>>n;
int arr[n];
for(int i=0;i<n;i++)
cin>>arr[i];
print2largest(arr, n);
return 0;
}

85
40. PROGRAM 40 - MOST OCCURRING CHARACTER

Write a program to find the most occurring character in the string.

Sample Input:
Happy coding
Sample Output:
p

Case 1
Input (stdin)
Happy coding
Output (stdout)
p

Case 2
Input (stdin)
programming is good
Output (stdout)
p

86
Code Solution:

#include<iostream>
using namespace std;
#include <string.h>

int main()
{
char str[100], result;
int i, len;
int max = -1;
int freq[256] = {0};
scanf("%[^\n]s",str);
len = strlen(str);
for(i = 0; i < len; i++)
{
freq[str[i]]++;
}
for(i = 0; i < len; i++)
{
if(max < freq[str[i]])
{
max = freq[str[i]];
result = str[i];
}
}

cout<<result;
return 0;
}

87
41. PROGRAM 41 - INTEGER DIFFERENCE

Given an array arr and two integer value n and m as input, take the first element and find the
difference between the first element and the integer value n. If the difference is less than m
then increment that particular array element. Do this for all the element in an array and print
the final modified array as output.

Sample Input:
5
21457
32
Sample Output:
32557

Case 1
Input (stdin)
5
21457
32
Output (stdout)
32557

Case 2
Input (stdin)
5
12345
35
Output (stdout)
23456

88
Code Solution:

#include<iostream>
using namespace std;

int result (int arr[], int x, int n,int m);

int main ()
{
int x, i, n,m;
cin>>x;
int arr[x];
for (i = 0; i < x; i++)
{
cin>>arr[i];
}
cin>>n;
cin>>m;
result (arr, x, n, m);
for(i=0;i<x;i++)
{
cout<<arr[i]<<" ";
}
return 0;
}

int result (int arr[], int x, int n,int m)

{
int i;
for (i = 0; i < x; i++)
{
if ( (arr[i]-n ) < m)
{
arr[i]=arr[i]+1;
}
}
}

89
42. PROGRAM 42 - SEARCH AN ELEMENT

Write a program to search an element in an array and print its index value. If the element is
not present, then print -1.

Sample Input:
5
46728
7
Sample Output:
index = 2

Case 1
Input (stdin)
5
46728
7
Output (stdout)
index = 2

Case 2
Input (stdin)
5
46328
5
Output (stdout)
-1

90
Code Solution:

#include<iostream>
using namespace std;

int search (int arr[], int ele, int n);

int main ()
{
int n, i, ele;
cin>>n;
int arr[n];
for (i = 0; i < n; i++)
{
cin>>arr[i];
}
cin>>ele;
search (arr, ele, n);
return 0;
}

int search (int arr[], int ele, int n)

{
int i, flag = 0;

for (i = 0; i < n; i++)

{
if (ele == arr[i])
{
cout<<"index = "<< i;
flag = 1;
break;
}
}

if (flag == 0)
{
cout<<"-1";
}
}

91
43. PROGRAM 43 - DECIMAL TO BINARY EQUIVALENT

Write a program to convert a decimal number to a binary equivalent number.

Sample Input:
10
Sample Output:
1010

Case 1
Input (stdin)
10
Output (stdout)
1010

Case 2
Input (stdin)
2
Output (stdout)
10

92
Code Solution:

#include<iostream>
using namespace std;

int main()
{
int a[50],n,i=0;
cin>>n;
while(n>0)
{
a[i++]=n%2;
n=n/2;
}
for(i=i-1;i>=0;i--)
cout<<a[i];
return 0;
}

93
44. PROGRAM 44 - BINARY TO DECIMAL EQUIVALENT

Write a program to convert a binary number to an equivalent decimal value.

Sample Input:
1010
Sample Output:
10

Case 1
Input (stdin)
1010
Output (stdout)
10

Case 2
Input (stdin)
11111
Output (stdout)
31

94
Code Solution:

#include<iostream>
using namespace std;

int main()
{
int num, binary, decimal = 0, base = 1, rem;
cin>>num;
binary = num;
while (num > 0)
{
rem = num % 10;
decimal = decimal + rem * base;
num = num / 10 ;
base = base * 2;
}
cout<<decimal;
return 0;
}

95
45. PROGRAM 45 - INSERT AN ELEMENT

Write a program to create a dynamic array and insert an element in it, in the specified
position.

Input Format
First line consists of integer value N
Second line consists of an array elements
Third line consists of integer value as position and value
Output Format
Output consists of an array element

Refer the sample output for formatting

Sample Input:
5
12345
2 10
Sample Output:
1 10 2 3 4 5

Case 1
Input (stdin)
5
12345
2 10
Output (stdout)
1 10 2 3 4 5

Case 2
Input (stdin)
5
17845
2 20
Output (stdout)
1 20 7 8 4 5

96
Code Solution:

#include <iostream>
#include<cstdlib>
using namespace std;

int main()
{
int *array, position, i, n, value;
cin>>n;
array = (int*) malloc(n * sizeof(int));
if(array == NULL)
{
cout<<"Error! memory not allocated.";
exit(0);
}
for(i = 0; i < n; ++i)
{
cin>>array[i];
}
cin>>position>>value;

for (i = n - 1; i >= position - 1; i--)

array[i+1] = array[i];

array[position-1] = value;

for (i = 0; i <= n; i++)

cout<<array[i]<<" ";
return 0;
}

97
46. PROGRAM 46 - DELETE AT SPECIFIC POSITION

Write a program to create a dynamic array and delete an element from an array from the
specified position. If deletion is not possible, print "Deletion not possible."

Sample Input:
5
12345
2
Sample Output:
1345

Case 1
Input (stdin)
5
12345
2
Output (stdout)
1345

Case 2
Input (stdin)
5
12345
5
Output (stdout)
1234

98
Code Solution:

#include <stdio.h>
#include<stdlib.h>
int main()
{
int *array, position, i,c, n, value;
scanf("%d", &n);
array = (int*) malloc(n * sizeof(int));
if(array == NULL)
{
printf("Error! memory not allocated.");
exit(0);
}
for(i = 0; i < n; ++i)
{
scanf("%d", &array[i]);
}
scanf("%d", &position);

if ( position >= n+1 )

printf("Deletion not possible.\n");

else
{
for ( c = position - 1 ; c < n - 1 ; c++ )
array[c] = array[c+1];
for( c = 0 ; c < n - 1 ; c++ )
printf("%d ", array[c]);
}
return 0;
}

99
47. PROGRAM 47 - SEGREGATE 0’s and 1’s

Write a program to segregate all the 0’s in the left side and 1’s in the right side in the same
array.

Sample Input:
5
01010
Sample Output:
00011

Case 1
Input (stdin)
5
01010
Output (stdout)
00011

Case 2
Input (stdin)
6
110000
Output (stdout)
000011

100
Code Solution:

#include<iostream>
using namespace std;

void segregate0and1(int arr[],int n)

{
int left = 0, right = n-1;
while(1)
{
if(left >= right)
break;
if(arr[left] == 0)
left++;
else if(arr[right] == 1)
right--;
else
{
arr[left] = 0;
arr[right] = 1;
left++;
right--;
}
}
}

int main()
{
int n;
cin>>n;
int arr[n];
for(int i=0;i<n;i++)
cin>>arr[i];
segregate0and1(arr, n);
for(int i=0;i<n;i++)
cout<<arr[i];
return 0;
}

101
48. PROGRAM 48 - INSERTION SORT

Write a program to create a dynamic array and sort the elements in it using Insertion sort.

Input Format
First line consists of integer value N
Second line consists of an array element
Output Format
Output consists of an array element

Refer the sample output for formatting

Sample Input:
6
123433
Sample Output:
123334

Case 1
Input (stdin)
6
123433
Output (stdout)
123334

Case 2
Input (stdin)
7
8125642
Output (stdout)
1224568

102
Code Solution:

#include <iostream>
#include <cstdlib>
using namespace std;
int main()
{
int n, i, j, *ptr, temp;
cin>>n;
ptr = (int*) malloc(n * sizeof(int));
if(ptr == NULL)
{
cout<<"Error! memory not allocated.";
exit(0);
}
for(i = 0; i < n; ++i)
{
cin>>ptr[i];
}
for(i=1;i<n;i++)
{
temp = ptr[i];
j=i-1;
while(temp < ptr[j] && j>=0)
{
ptr[j+1] = ptr[j];
--j;
}
ptr[j+1]=temp;
}
for(i=0; i<n; i++)
cout<<ptr[i]<<" ";
free(ptr);
return 0;
}

103
49. PROGRAM 49 - SORTING THE FIRST HALF OF THE ARRAY IN
ASCENDING ORDER AND THE SECOND HALF IN DESCENDING ORDER

Write a program to sort the first half of an array in ascending order and the second half in
descending order.

Input Format:
The first line contains an integer 'N', denoting the size of the array.
The next line contains space-separated integers denoting the elements of the array.

Output Format:
Print sorted array

Sample Input:
6
123456
Sample Output:
123654

Case 1
Input (stdin)
9
524793168
Output (stdout)
245798631

Case 2
Input (stdin)
6
10 30 20 40 60 50
Output (stdout)
10 20 30 60 50 40

104
Code Solution:

#include<iostream>
using namespace std;
int insertion_sort(int n,int list[]);
int main()
{
int n;
cin>>n;
int list[n];
for(int index = 0;index < n; index++)
{
cin>>list[index];
}
insertion_sort(n,list);
return 0;
}
int insertion_sort(int n,int list[])
{
for (int idx1 = 1; idx1 < n/2; idx1++)
{
int key = list[idx1];
int idx2 = idx1 - 1;
while((idx2 >= 0) && (list[idx2] > key))
{
list[idx2 + 1] = list[idx2];
idx2--;
}
list[idx2 + 1] = key;
}
for (int idx1 = n/2; idx1 < n; idx1++)
{
int key = list[idx1];
int idx2 = idx1 - 1;
while((idx2 >= n/2) && (list[idx2] < key))
{
list[idx2 + 1] = list[idx2];
idx2--;
}
list[idx2 + 1] = key;
}
for(int i = 0; i < n; i++)
{
cout<< list[i]<<" ";
}
return 0;
}

105
50. PROGRAM 50 - REGULAR EXPRESSION

Write a program to check the regular expression matching. Hint : ‘+’ indicates consecutive
multiple occurrences of that particular character.

Input Format
First line consists of a string s
Second line consists of string s
Output Format
Output consists of string whether it is matched or not

Refer the sample output for formatting

Sample Input:
a+b+c
aaabbc
Sample Output:
Matched

Case 1
Input (stdin)
a+b+c
aaabbc
Output (stdout)
Matched

Case 2
Input (stdin)
b+c+x
bbcxxz
Output (stdout)
Not Matched

106
Code Solution:

#include<iostream>
using namespace std;
int main()
{
char arr[20],ans[20];
int i=0,j=0,len;
cin>>arr>>ans;
for(len=0;arr[len]!='\0';len++);
while(arr[i]==ans[j] && arr[i]!='\0')
{
if(arr[i+1]=='+')
{
for( ; ans[j]==arr[i] ; j++);
i=i+2;
}
else
{
i++;
j++;
}
}

if(i>=len && ans[j]=='\0')

{
cout<<"Matched";
}
else
{
cout<<"Not Matched";
}
return 0;
}

107
51. PROGRAM 51

In this lockdown a family of N members decided to play a game the rules of which are :-
● All N members are made to sit uniformly in a circle (ie. from 1 to N in clockwise
direction).
● The game start with the person sitting at first position.
● A song is played in the background. The lyrics of the song are denoted by a string
which consists of only letters 'x' and 'y'. Assume that each lyric of the song is a single
letter.
● If the lyric 'x' occurs in the song, the member who is currently holding the Parcel
passes it on to the next member. This passing takes place in clockwise direction.
● If the lyric 'y' occurs in the song, the member who is currently holding the Parcel loses
his/her chances of winning the game. He/she hands over the parcel to the next
member (in clockwise direction) and moves out.
● The game continues until a single member survives in the end. He/she will be the
winner of the game.
● Note that the song repeats continuously ie. while the game is going on, if at all the
song ends, the stereo system will automatically start playing the song from the start
without any delay.
You have to find out the member who wins the game.

Input: The input consists of 2 lines. The first line consists of N, the member of family in the
class. The next line consists of a string denoting the lyrics of the song the teacher plays.
Output: Print a single integer denoting the roll number of the student who wins the game.

Constraints:
2≤N≤100000
1≤|S|≤10000, where |S| denotes the length of the input string. It is guaranteed that at least 1
lyric in the song will be a 'y'

Sample Input:
3
xyx
Sample Output:
1

Explanation:
Starting from 1 lyrics : 'x' therefore he passes the ballto 2nd
2nd turn lyrics : 'y' therefore 2nd member gets out of game and passes to 3rd
3rd turn lyrics : 'x' therefore 3rd passes ball to first.
4th turn lyrics : 'x' passes to 3rd

108
5th turn lyrics: 'y' therefore gets eliminated.

Hence the person sitting at position 1 won this game.

Test Case 1
Input
3 xyx
Output
1

Test Case 2
Input
6 xxyyxy
Output
2

109
Code Solution:

#include<bits/stdc++.h>
#include<string>
using namespace std;
#define ll long long
ll n,slen;

void fun (ll size,ll k,string s, ll l, ll *ingame)

{
if(size==1)
{
ll i=1;
while(ingame[i]!=1)
{i++;

}
cout<<i<<endl;
return;
}
if(l==slen) l=0;
if(k==n+1) k=1;
if(ingame[k]==0)
{

fun(size,k+1,s,l,ingame);
}
else
{
if(s[l]=='x')
fun(size,k+1,s,l+1,ingame);
else{
ingame[k]=0;
fun(size-1,k+1,s,l+1,ingame);
}
}
return;
}

int main()
{

110
 cin>>n;
string s;
cin>>s;
slen=s.length();
ll ingame[n+1];
for(int i =1;i<=n;i++)
{
ingame[i]=1;
}
fun(n,1,s,0,ingame);
}

111
52. PROGRAM 52

Write a program to find the equilibrium element in the given array.

Hint : A[i] is equilibrium element if A[0] + A[1] + … A[i-1] == A[i+1] + A[i+2] + … +
A[n-1]

Sample Input 1:
6
123433
Sample Output 1:
4

Sample Input 2:
6
564231
Sample Output 2:
No Equilibrium element found

Test Case 1
Input
6123433
Output
4

Test Case 2
Input
6564231
Output
No Equilibrium element found

112
Code Solution:

#include<iostream>
using namespace std;

int findElement(int arr[], int n)

{
int prefixSum[n];
prefixSum[0] = arr[0];
for (int i = 1; i < n; i++)
prefixSum[i] = prefixSum[i - 1] + arr[i];
int suffixSum[n];
suffixSum[n - 1] = arr[n - 1];
for (int i = n - 2; i >= 0; i--)
suffixSum[i] = suffixSum[i + 1] + arr[i];
for (int i = 1; i < n - 1; i++)
if (prefixSum[i] == suffixSum[i])
return arr[i];
return -1;
}

int main()
{
int n;
cin>>n;
int arr[n];
for(int i=0;i<n;i++)
cin>>arr[i];
int result = findElement(arr, n);
if(result == -1)
cout<<"No Equilibrium element found";
else
cout<<result;
return 0;
}

113
53. PROGRAM 53 - NON REPEATING ELEMENTS

Write a program to eliminate the common elements in the given 2 arrays and print only the
non repeating elements and the total.

Sample Input:
54
12865
2 6 8 10
Sample Output:
1 5 10
3

Case 1
Input (stdin)
54
12865
2 6 8 10
Output (stdout)
1 5 10
3

Case 2
Input (stdin)
55
10 20 30 40 50
60 70 80 90 50
Output (stdout)
10 20 30 40 60 70 80 90
8

114
54. PROGRAM 54 - FIND THE Nth ELEMENT

Write a program to create a Singly Linked List and find the Nth element from the end of the
list.

Hint: Insertion at the beginning.

Output: output consists of single line, print the element in the given position in a list, else
print "No node found"

Sample Input:
5
10 8 6 4 0
2
Output:
8

Case 1
Input (stdin)
5
10 8 6 4 0
2
Output (stdout)
8

Case 2
Input (stdin)
10
100 200 300 400 500 600 700 800 900 101
11
Output (stdout)
No node found

115
Code Solution:

#include<iostream>
#include<stdlib.h>
using namespace std;
struct Node
{ int num;
struct Node* link;
};
void insertAtBeginning(struct Node** head_ref, int new_data)
{
struct Node* new_node = (struct Node*) malloc(sizeof(struct Node));
new_node->num = new_data;
new_node->link = (*head_ref);
(*head_ref) = new_node;
}
int printNthFromLast(struct Node* head, int n)
{
int len = 0, i;
struct Node *temp = head;
while (temp != NULL)
{
temp = temp->link;
len++;
}
if (len < n)
return 0;
temp = head;
for (i = 1; i < len-n+1; i++)
temp = temp->link;
cout<<temp->num;
return 1;
}
int main()
{
struct Node* head = NULL;
int total_elem_to_insert, tmp, i,n;
cin>>total_elem_to_insert;

for(i=0;i<total_elem_to_insert;i++)
{

116
 cin>>tmp;
insertAtBeginning(&head, tmp);
}
scanf("%d",&n);

if(0==(printNthFromLast(head, n)))
cout<<"No node found"<<endl;
return 0;
}

117
55. PROGRAM 55 - ANAGRAM

Write a program to find whether the given string is the anagram of the first string.

Sample Input 1:
eat
ate
Sample Output 1:
Anagram

Sample Input 2:
eat
abc
Sample Output 2:
Not Anagram

Case 1
Input (stdin)
eat
ate
Output (stdout)
Anagram

Case 2
Input (stdin)
eat
abc
Output (stdout)
Not Anagram

118
Code Solution:

#include<iostream>
using namespace std;

int find_anagram(char [], char []);

int main()
{
char array1[100], array2[100];
int flag;
cin>>array1;
cin>>array2;
flag = find_anagram(array1, array2);
if (flag == 1)
cout<<"Anagram";
else
cout<<"Not Anagram";
return 0;
}
int find_anagram(char array1[], char array2[])
{
int num1[26] = {0}, num2[26] = {0}, i = 0;
while (array1[i] != '\0')
{
num1[array1[i] - 'a']++;
i++;
}
i = 0;
while (array2[i] != '\0')
{
num2[array2[i] -'a']++;
i++;
}
for (i = 0; i < 26; i++)
{
if (num1[i] != num2[i])
return 0;
}
return 1;
}

119
56. PROGRAM 56 - MERGE TWO SORTED LINKED LIST

Write a program to merge two sorted linked lists. Given two sorted linked lists say list1 and
list2, merge list2 with list1. Insertion has to be done at the end of the list.

Input Format
The first line consists of the size of the first sorted linked list.
The second line consists of list of elements of first sorted linked list separated by space.
The third line consists of the size of the second sorted linked list.
The fourth line consists of list of elements of second sorted linked list separated by space.

Output Format: The output should be the merged list of two linked list.

Refer the sample input & output for formatting specifications.

Sample Input
3
135
3
246
Sample Output
1->2->3->4->5->6->NULL

Case 1
Input (stdin)
3
135
3
246

Output (stdout)
1->2->3->4->5->6->NULL

120
57. PROGRAM 57 - APPEND WITHOUT DUPLICATES

John assigned a roll number to each student in ascending order. Every time he assigns a
number to a student he wants to check whether the number is already assigned to any other
student to avoid the duplicates in the roll number. Add the number to the list only if the
number is not present already. Implement this concept using a Linked List. If the list is
empty, print "List is empty".

Input Format: The input consists of a list of integers, negative value indicates the end of the
linked list.

Output Format: The output should be numbers in the list in a separate line.

Refer the sample input & output for formatting specifications.

Sample Input
11
22
33
22
33
44
-77

Sample Output
11
22
33
44

Case 1
Input (stdin)
11
22
33
22
33
44
-77

121
Output (stdout)
11
22
33
44

Case 2
Input (stdin)
-5

Output (stdout)
List is empty

122
Code Solution:

#include<iostream>
using namespace std;
class Node
{
public:
int data;
Node *next;
};
void append(Node **headadd,int data)
{
Node *temp,*newnode,*temp1;
temp = *headadd;
temp1 = *headadd;
int flag = 1;
while(temp1 != NULL)
{
if(temp1->data == data)
flag = 0;
temp1 = temp1->next;
}
if(flag == 1)
{
newnode = new Node();
newnode->data = data;
newnode->next = NULL;
if(*headadd == NULL)
*headadd = newnode;
else
{
while(temp->next != NULL)
temp = temp->next;
temp->next = newnode;
}
}
}
void display(Node *head)
{
if(head == NULL)
{

123
 cout<< "List is empty";
return;
}
while(head != NULL)
{
cout<<head->data<<endl;
head = head->next;
}
}
int main()
{
Node *head = NULL;
int data;
do
{
cin>>data;
if(data>0)
append(&head,data);
}while(data>0);
display(head);
return 0;
}

124