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Count pairs with given sum in C++
We are given an integer array and the task is to count the total number of pairs that can be formed using the given array values such that the sum of the pairs is equal to the given sum.
Input − int arr[] = {2, 8, 1, 5, 11}, sum = 10
Output − Count of pairs with given sum 13 is − 2
Explanation −
| a1 | a2 | a1 + a2 |
| 2 | 8 | 10 |
| 2 | 1 | 3 |
| 2 | 5 | 7 |
| 2 | 11 | 13 |
| 8 | 1 | 9 |
| 8 | 5 | 13 |
| 8 | 11 | 19 |
| 1 | 5 | 6 |
| 1 | 11 | 12 |
| 5 | 11 | 16 |
Input − int arr[] = {2, 8, -1, 5, -11}, sum = 6
Output − Count of pairs with given sum 6 is − 1
Explanation −
| a1 | a2 | a1 + a2 |
| 2 | 8 | 10 |
| 2 | -1 | 1 |
| 2 | 5 | 7 |
| 2 | -11 | -9 |
| 8 | -1 | 7 |
| 8 | 5 | 13 |
| 8 | -11 | -3 |
| 1 | 5 | 6 |
| 1 | -11 | -10 |
| 5 | -11 | -6 |
Approach used in the below program is as follows
Input an array of integer elements to form a pair and the integer value of sum.
Calculate the size of an array pass the data to the function for further processing
Create a temporary variable count to match the pairs with the given sum
Start loop FOR from i to 0 till the size of an array
Inside the loop, start another loop FOR from j to i+1 till the size of an array
Inside the loops, set the temporary variable total as arr[i] + arr[j]
Check IF total == sum then increment the count by 1
Return the count
Print the result
Example
#include <bits/stdc++.h>
using namespace std;
//Count pairs with given sum
int Pair_Sum(int arr[], int size, int sum){
int count = 0;
for (int i=0; i<size; i++){
for (int j=i+1; j<size; j++){
int total = arr[i] + arr[j];
if (total == sum){
count++;
}
}
}
return count;
}
int main(){
int arr[] = {2, 6, 1, 7, 9, 8} ;
int sum = 9;
int size = sizeof(arr)/sizeof(arr[0]);
cout<<"Count of pairs with given sum "<<sum<<" is: "<<Pair_Sum(arr, size, sum);
return 0;
}Output
If we run the above code it will generate the following output −
Count of pairs with given sum 9 is: 2
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