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2 Sum - Count Pairs with given Sum in Sorted Array

Last Updated : 12 Jul, 2025
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Given a sorted array arr[] and an integer target, the task is to find the number of pairs in the array whose sum is equal to target.

Examples: 

Input: arr[] = [-1, 1, 5, 5, 7], target = 6
Output: 3
Explanation: Pairs with sum 6 are (1, 5), (1, 5) and (-1, 7).         

Input: arr[] = [1, 1, 1, 1], target = 2
Output: 6
Explanation: Pairs with sum 2 are (1, 1), (1, 1), (1, 1), (1, 1), (1, 1) and (1, 1).

Input: arr[] = [-1, 10, 10, 12, 15], target = 125
Output:  0

In this post, we are counting pairs with given sum when the input array is sorted. To count the pairs when the input array is not sorted, refer to 2 Sum – Count pairs with given sum.

Table of Content

[Naive Approach] By Generating All Possible Pairs - O(n^2) time and O(1) space

The very basic approach is to generate all the possible pairs and check if any pair exists whose sum is equals to given target value, then increment the count variable.

To know more about the implementation, please refer 2 Sum – Count pairs with given sum.

[Expected Approach] Using Two Pointer Technique - O(n) and O(1) Space

The idea is to use the two-pointer technique by maintaining two pointers, say left and right and initialize them to the first and last element of the array respectively. According to the sum of left and right pointers, we can have three cases:

  • arr[left] + arr[right] < target: We need to increase the sum of pair, move the left pointer towards right.
  • arr[left] + arr[right] > target: We need to decrease the sum of pair, move the right pointer towards left.
  • arr[left] + arr[right] = target: We have found a pair whose sum is equal to target. We can find the product of the count of both the elements and add them to the result.
C++ 
// C++ program for counting pairs with given sum 
// using Two Pointer Technique

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int countPairs(vector<int> &arr, int target) {
    int res = 0;
    int n = arr.size();
    int left = 0, right = n - 1;

    while (left < right) {
      
        // If sum is greater 
        if (arr[left] + arr[right] < target)
            left++;

        // If sum is lesser
        else if (arr[left] + arr[right] > target)
            right--;

        // If sum is equal
        else {
          
            int cnt1 = 0, cnt2 = 0;
            int ele1 = arr[left], ele2 = arr[right];
          
            // Count frequency of first element of the pair
            while (left <= right and arr[left] == ele1) {
                left++;
                cnt1++;
            }
          
			// Count frequency of second element of the pair
            while(left <= right and arr[right] == ele2) {
                right--;
                cnt2++;
            }
           
            // If both the elements are same, then count of
            // pairs = the number of ways to choose 2 
            // elements among cnt1 elements
            if(ele1 == ele2) 
            	res += (cnt1 * (cnt1 - 1))/2;
          
            // If the elements are different, then count of
            // pairs = product of the count of both elements
            else 
            	res += (cnt1 * cnt2);
        }
    }
    return res;
}

int main() {
    vector<int> arr = {-1, 1, 5, 5, 7};
    int target = 6;

    cout << countPairs(arr, target);

    return 0;
}
C 
// C program for counting pairs with given sum 
// using Two Pointer Technique

#include <stdio.h>

// Function to return the count of pairs
int countPairs(int arr[], int n, int target) {
    int res = 0;
    int left = 0, right = n - 1;

    while (left < right) {
      
        // If sum is greater 
        if (arr[left] + arr[right] < target)
            left++;

        // If sum is lesser
        else if (arr[left] + arr[right] > target)
            right--;

        // If sum is equal
        else {
            int cnt1 = 0, cnt2 = 0;
            int ele1 = arr[left], ele2 = arr[right];
          
            // Count frequency of first element of the pair
            while (left <= right && arr[left] == ele1) {
                left++;
                cnt1++;
            }
          
            // Count frequency of second element of the pair
            while(left <= right && arr[right] == ele2) {
                right--;
                cnt2++;
            }
           
            // If both the elements are same, then count of
            // pairs = the number of ways to choose 2 
            // elements among cnt1 elements
            if(ele1 == ele2)
                res += (cnt1 * (cnt1 - 1)) / 2;
          
            // If the elements are different, then count of
            // pairs = product of the count of both elements
            else 
                res += (cnt1 * cnt2);
        }
    }
    return res;
}

int main() {
    int arr[] = {-1, 1, 5, 5, 7};
    int target = 6;
    int n = sizeof(arr) / sizeof(arr[0]);

    printf("%d\n", countPairs(arr, n, target));

    return 0;
}
Java 
// Java program for counting pairs with given sum 
// using Two Pointer Technique

import java.util.Arrays;

class GfG {

    // Function to return the count of pairs
    static int countPairs(int[] arr, int target) {
        int res = 0;
        int n = arr.length;
        int left = 0, right = n - 1;

        while (left < right) {
          
            // If sum is greater 
            if (arr[left] + arr[right] < target)
                left++;

            // If sum is lesser
            else if (arr[left] + arr[right] > target)
                right--;

            // If sum is equal
            else {
                int cnt1 = 0, cnt2 = 0;
                int ele1 = arr[left], ele2 = arr[right];
              
                // Count frequency of first element of the pair
                while (left <= right && arr[left] == ele1) {
                    left++;
                    cnt1++;
                }
              
                // Count frequency of second element of the pair
                while (left <= right && arr[right] == ele2) {
                    right--;
                    cnt2++;
                }
               
                // If both the elements are same, then count of
                // pairs = the number of ways to choose 2 
                // elements among cnt1 elements
                if (ele1 == ele2) 
                    res += (cnt1 * (cnt1 - 1)) / 2;
              
                // If the elements are different, then count of
                // pairs = product of the count of both elements
                else 
                    res += (cnt1 * cnt2);
            }
        }
        return res;
    }

    public static void main(String[] args) {
        int[] arr = {-1, 1, 5, 5, 7};
        int target = 6;

        System.out.println(countPairs(arr, target));
    }
}
Python 
# Python program for counting pairs with given sum
# using Two Pointer Technique

def countPairs(arr, target):
    res = 0
    n = len(arr)
    left = 0
    right = n - 1

    while left < right:

        # If sum is greater
        if arr[left] + arr[right] < target:
            left += 1

        # If sum is lesser
        elif arr[left] + arr[right] > target:
            right -= 1

        # If sum is equal
        else:
            cnt1 = 0
            cnt2 = 0
            ele1 = arr[left]
            ele2 = arr[right]

            # Count frequency of first element of the pair
            while left <= right and arr[left] == ele1:
                left += 1
                cnt1 += 1

            # Count frequency of second element of the pair
            while left <= right and arr[right] == ele2:
                right -= 1
                cnt2 += 1

            # If both the elements are same, then count of
            # pairs = the number of ways to choose 2
            # elements among cnt1 elements
            if ele1 == ele2:
                res += (cnt1 * (cnt1 - 1)) // 2

            # If the elements are different, then count of
            # pairs = product of the count of both elements
            else:
                res += (cnt1 * cnt2)

    return res


if __name__ == "__main__":
    arr = [-1, 1, 5, 5, 7]
    target = 6

    print(countPairs(arr, target))
C# 
// C# program for counting pairs with given sum using 
// Two Pointer Technique

using System;

class GfG {
  
    // Function to return the count of pairs
    static int countPairs(int[] arr, int target) {
        int res = 0;
        int n = arr.Length;
        int left = 0, right = n - 1;

        while (left < right) {
          
            // If sum is greater 
            if (arr[left] + arr[right] < target)
                left++;

            // If sum is lesser
            else if (arr[left] + arr[right] > target)
                right--;

            // If sum is equal
            else {
                int cnt1 = 0, cnt2 = 0;
                int ele1 = arr[left], ele2 = arr[right];

                // Count frequency of first element of the pair
                while (left <= right && arr[left] == ele1) {
                    left++;
                    cnt1++;
                }

                // Count frequency of second element of the pair
                while (left <= right && arr[right] == ele2) {
                    right--;
                    cnt2++;
                }

                // If both the elements are same, then count of
                // pairs = the number of ways to choose 2 
                // elements among cnt1 elements
                if (ele1 == ele2)
                    res += (cnt1 * (cnt1 - 1)) / 2;

                // If the elements are different, then count of
                // pairs = product of the count of both elements
                else
                    res += (cnt1 * cnt2);
            }
        }
        return res;
    }

    static void Main(string[] args) {
        int[] arr = { -1, 1, 5, 5, 7 };
        int target = 6;

        Console.WriteLine(countPairs(arr, target));
    }
}
JavaScript 
// JavaScript program for counting pairs with given sum 
// using Two Pointer Technique

function countPairs(arr, target) {
    let res = 0;
    const n = arr.length;
    let left = 0, right = n - 1;

    while (left < right) {
      
        // If sum is greater 
        if (arr[left] + arr[right] < target) {
            left++;
        }

        // If sum is lesser
        else if (arr[left] + arr[right] > target) {
            right--;
        }

        // If sum is equal
        else {
            let cnt1 = 0, cnt2 = 0;
            const ele1 = arr[left], ele2 = arr[right];
          
            // Count frequency of first element of the pair
            while (left <= right && arr[left] === ele1) {
                left++;
                cnt1++;
            }
          
            // Count frequency of second element of the pair
            while (left <= right && arr[right] === ele2) {
                right--;
                cnt2++;
            }
           
            // If both the elements are same, then count of
            // pairs = the number of ways to choose 2 
            // elements among cnt1 elements
            if (ele1 === ele2) 
                res += (cnt1 * (cnt1 - 1)) / 2;
          
            // If the elements are different, then count of
            // pairs = product of the count of both elements
            else 
                res += (cnt1 * cnt2);
        }
    }
    return res;
}

// Driver Code
const arr = [-1, 1, 5, 5, 7];
const target = 6;

console.log(countPairs(arr, target));

Output
3



Count Pairs with given Sum in Sorted Array
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