Count pairs with given sum | Set 2

Given an array arr[] and an integer sum, the task is to find the number of pairs of integers in the array whose sum is equal to sum.

Examples:

Input: arr[] = {1, 5, 7, -1}, sum = 6
Output: 2
Pairs with sum 6 are (1, 5) and (7, -1)

Input: arr[] = {1, 5, 7, -1, 5}, sum = 6
Output: 3
Pairs with sum 6 are (1, 5), (7, -1) & (1, 5)

Input: arr[] = {1, 1, 1, 1}, sum = 2
Output: 6

Approach: Two Different methods have already been discussed here. Here, a method based on sorting will be discussed.

  1. Sort the array and take two pointers i and j, one pointer pointing to the start of the array i.e. i = 0 and another pointer pointing to the end of the array i.e. j = n – 1.
    • If arr[i] + arr[j] is

    • Greater than the sum then decrement j.
    • Lesser than the sum then increment i.
    • Equals to the sum then count such pairs.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the count of pairs
// from arr[] with the given sum
int pairs_count(int arr[], int n, int sum)
{
    // To store the count of pairs
    int ans = 0;
  
    // Sort the given array
    sort(arr, arr + n);
  
    // Take two pointers
    int i = 0, j = n - 1;
  
    while (i < j) {
        // If sum is greater
        if (arr[i] + arr[j] < sum)
            i++;
  
        // If sum is lesser
        else if (arr[i] + arr[j] > sum)
            j--;
  
        // If sum is equal
        else {
            // Find the frequency of arr[i]
            int x = arr[i], xx = i;
            while (i < j and arr[i] == x)
                i++;
  
            // Find the frequency of arr[j]
            int y = arr[j], yy = j;
            while (j >= i and arr[j] == y)
                j--;
  
            // If arr[i] and arr[j] are same
            // then remove the extra number counted
            if (x == y) {
                int temp = i - xx + yy - j - 1;
                ans += (temp * (temp + 1)) / 2;
            }
            else
                ans += (i - xx) * (yy - j);
        }
    }
  
    // Return the required answer
    return ans;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 5, 7, 5, -1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int sum = 6;
  
    cout << pairs_count(arr, n, sum);
  
    return 0;
}

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Python3

# Python3 implementation of the approach

# Function to return the count of pairs
# from arr with the given sum
def pairs_count(arr, n, sum):

# To store the count of pairs
ans = 0

# Sort the given array
arr = sorted(arr)

# Take two pointers
i, j = 0, n – 1

while (i < j): # If sum is greater if (arr[i] + arr[j] < sum): i += 1 # If sum is lesser elif (arr[i] + arr[j] > sum):
j -= 1

# If sum is equal
else:

# Find the frequency of arr[i]
x = arr[i]
xx = i
while (i < j and arr[i] == x): i += 1 # Find the frequency of arr[j] y = arr[j] yy = j while (j >= i and arr[j] == y):
j -= 1

# If arr[i] and arr[j] are same
# then remove the extra number counted
if (x == y):
temp = i – xx + yy – j – 1
ans += (temp * (temp + 1)) // 2
else:
ans += (i – xx) * (yy – j)

# Return the required answer
return ans

# Driver code
arr = [1, 5, 7, 5, -1]
n = len(arr)
sum = 6

print(pairs_count(arr, n, sum))

# This code is contributed by Mohit Kumar

Output:

3


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Improved By : mohit kumar 29