3 Sum - Count Triplets With Given Sum In Sorted Array

Last Updated : 10 Feb, 2025

Given a sorted array arr[] and a target value, the task is to find the count of triplets present in the given array having sum equal to the given target.

More specifically, the task is to count triplets (i, j, k) of valid indices, such that arr[i] + arr[j] + arr[k] = target and i < j < k.

Examples:

Input: arr[] = [-3, -1, -1, 0, 1, 2], target = -2
Output: 4
Explanation: Two triplets that add up to -2 are:
arr[0] + arr[3] + arr[4] = (-3) + 0 + (1) = -2
arr[0] + arr[1] + arr[5] = (-3) + (-1) + (2) = -2
arr[0] + arr[2] + arr[5] = (-3) + (-1) + (2) = -2
arr[1] + arr[2] + arr[3] = (-1) + (-1) + (0) = -2

Input: arr[] = [-2, 0, 1, 1, 5], target = 1
Output: 0
Explanation: There is no triplet whose sum is equal to 1.

We have discussed two approaches, that works for both sorted and unsorted arrays, in the post 3 Sum - Count all triplets with given sum. Now in case the given array is already sorted, we can further optimize the space using two pointers technique.

Using Two Pointers Technique

The idea is to fix the first element of the triplet at index i and then using two pointers technique to count the other two elements. For each index i, initialize left pointer to i + 1 and right pointer to n - 1.
For each triplet, calculate the sum as arr[i] + arr[left] + arr[right] and evaluate:
If sum = target: A valid triplet is found. Additionally, handle duplicates by counting the frequencies of arr[left] and arr[right] and updating the result accordingly. Then move both pointers to the next distinct values.
If sum < target: Increment the left pointer to explore larger values and increase the sum.
If sum > target: Decrement the right pointer to explore smaller values and decrease the sum.

C++

// C++ program to count all triplets having sum equal to
// target by exploring all possible triplets

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

int countTriplets(vector<int> &arr, int target) {
    int n = arr.size();
    int res = 0;

    // Iterate through each element as the first element of the triplet
    for (int i = 0; i < n - 2; i++) {
        int left = i + 1, right = n - 1;

        // Use two-pointer approach to find triplets
        while (left < right) {

            // Calculate the sum of the triplet
            int sum = arr[i] + arr[left] + arr[right];

            // If sum is smaller, move to bigger values
            if (sum < target) {
                left += 1;
            }
          
            // If sum is greater, move to smaller values
            else if (sum > target) {
                right -= 1;
            }
            else if (sum == target) {
                int ele1 = arr[left], ele2 = arr[right];
                int cnt1 = 0, cnt2 = 0;

                // Count frequency of the current value at 'left'
                while (left <= right && arr[left] == ele1) {
                    left++;
                    cnt1++;
                }

                // Count frequency of the current value at 'right'
                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() {
    vector<int> arr = {-3, -1, -1, 0, 1, 2};
    int target = -2;

    cout << countTriplets(arr, target) << endl;
    return 0;
}

Java

// Java program to count all triplets having sum equal to
// target by exploring all possible triplets

import java.util.*;
class GfG {

    // Function to count triplets whose sum equals the target
    static int countTriplets(int[] arr, int target) {
        int n = arr.length;
        int res = 0;

        // Iterate through each element as the first element
        // of the triplet
        for (int i = 0; i < n - 2; i++) {
            int left = i + 1, right = n - 1;

            // Use two-pointer approach to find triplets
            while (left < right) {

                // Calculate the sum of the triplet
                int sum = arr[i] + arr[left] + arr[right];

                // If sum is smaller, move to bigger values
                if (sum < target) {
                    left += 1;
                }
                
                // If sum is greater, move to smaller values
                else if (sum > target) {
                    right -= 1;
                }
                
                // If sum is equal, check for frequency of elements
                else if (sum == target) {
                    int ele1 = arr[left], ele2 = arr[right];
                    int cnt1 = 0, cnt2 = 0;

                    // Count frequency of the current value at 'left'
                    while (left <= right && arr[left] == ele1) {
                        left++;
                        cnt1++;
                    }

                    // Count frequency of the current value at 'right'
                    while (left <= right && arr[right] == ele2) {
                        right--;
                        cnt2++;
                    }

                    // If both the elements are the 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 = {-3, -1, -1, 0, 1, 2};
        int target = -2;

        System.out.println(countTriplets(arr, target));
    }
}

Python

# Python program to count all triplets having sum equal to
# target by exploring all possible triplets

def countTriplets(arr, target):
    n = len(arr)
    res = 0

    # Iterate through each element as the first element of the triplet
    for i in range(n - 2):
        left = i + 1
        right = n - 1

        # Use two-pointer approach to find triplets
        while left < right:

            # Calculate the sum of the triplet
            sum = arr[i] + arr[left] + arr[right]

            # If sum is smaller, move to bigger values
            if sum < target:
                left += 1

            # If sum is greater, move to smaller values
            elif sum > target:
                right -= 1

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

                # Count frequency of the current value at 'left'
                while left <= right and arr[left] == ele1:
                    left += 1
                    cnt1 += 1

                # Count frequency of the current value at 'right'
                while left <= right and arr[right] == ele2:
                    right -= 1
                    cnt2 += 1

                # If both the elements are the 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 = [-3, -1, -1, 0, 1, 2]
    target = -2
    print(countTriplets(arr, target))

// C# program to count all triplets having sum equal to
// target by exploring all possible triplets

using System;
using System.Collections.Generic;

class GfG {
  
    // Function to count all triplets having sum equal to target
    static int countTriplets(int[] arr, int target) {
        int n = arr.Length;
        int res = 0;

        // Iterate through each element as the first element
        // of the triplet
        for (int i = 0; i < n - 2; i++) {
            int left = i + 1, right = n - 1;

            // Use two-pointer approach to find triplets
            while (left < right) {
              
                // Calculate the sum of the triplet
                int sum = arr[i] + arr[left] + arr[right];

                // If sum is smaller, move to bigger values
                if (sum < target) {
                    left += 1;
                }

                // If sum is greater, move to smaller values
                else if (sum > target) {
                    right -= 1;
                }
                else if (sum == target) {
                    int ele1 = arr[left], ele2 = arr[right];
                    int cnt1 = 0, cnt2 = 0;

                    // Count frequency of the current value at 'left'
                    while (left <= right && arr[left] == ele1) {
                        left++;
                        cnt1++;
                    }

                    // Count frequency of the current value at 'right'
                    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() {
        int[] arr = { -3, -1, -1, 0, 1, 2 };
        int target = -2;

        Console.WriteLine(countTriplets(arr, target));
    }
}

JavaScript

// JavaScript program to count all triplets having sum equal to
// target by exploring all possible triplets

// JavaScript program to count all triplets having sum equal to
// target by exploring all possible triplets

function countTriplets(arr, target) {
    let n = arr.length;
    let res = 0;

    // Iterate through each element as the first element of the triplet
    for (let i = 0; i < n - 2; i++) {
        let left = i + 1, right = n - 1;

        // Use two-pointer approach to find triplets
        while (left < right) {

            // Calculate the sum of the triplet
            let sum = arr[i] + arr[left] + arr[right];

            // If sum is smaller, move to bigger values
            if (sum < target) {
                left += 1;
            }
          
            // If sum is greater, move to smaller values
            else if (sum > target) {
                right -= 1;
            }
            else if (sum === target) {
                let ele1 = arr[left], ele2 = arr[right];
                let cnt1 = 0, cnt2 = 0;

                // Count frequency of the current value at 'left'
                while (left <= right && arr[left] === ele1) {
                    left++;
                    cnt1++;
                }

                // Count frequency of the current value at 'right'
                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
let arr = [-3, -1, -1, 0, 1, 2];
let target = -2;

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

Output

Time Complexity: O(n^2). For each index i, we use two pointers that traverse at most n elements combined.
Auxiliary Space: O(1)

3 Sum – All Distinct Triplets with given Sum in an Array

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Using Two Pointers Technique

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