3 Sum - Triplet Sum in Array
Last Updated :
13 Aug, 2025
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Given an array arr[] and an integer sum, check if there is a triplet in the array which sums up to the given target sum.
Examples:
Input: arr[] = [1, 4, 45, 6, 10, 8], target = 13
Output: true
Explanation: The triplet [1, 4, 8] sums up to 13Input: arr[] = [1, 2, 4, 3, 6, 7], target = 10
Output: true
Explanation: The triplets [1, 3, 6] and [1, 2, 7] both sum to 10.Input: arr[] = [40, 20, 10, 3, 6, 7], sum = 24
Output: false
Explanation: No triplet in the array sums to 24.
Table of Content
[Naive Approach] Generating All Triplets - O(n^3) Time and O(1) Space
A simple method is to generate all possible triplets and compare the sum of every triplet with the given target. If the sum is equal to target, return true. Otherwise, return false.
#include <iostream>
#include <vector>
using namespace std;
bool hasTripletSum(vector<int>& arr, int target) {
int n = arr.size();
// Fix the first element as arr[i]
for (int i = 0; i < n - 2; i++) {
// Fix the second element as arr[j]
for (int j = i + 1; j < n - 1; j++) {
// Now look for the third number
for (int k = j + 1; k < n; k++) {
if (arr[i] + arr[j] + arr[k] == target)
return true;
}
}
}
return false;
}
int main() {
vector<int> arr = { 1, 4, 45, 6, 10, 8 };
int target = 13;
if(hasTripletSum(arr, target))
cout << "true";
else
cout << "false";
return 0;
}
#include <stdio.h>
#include <stdbool.h>
bool hasTripletSum(int arr[], int n, int target) {
// Fix the first element as arr[i]
for (int i = 0; i < n - 2; i++) {
// Fix the second element as arr[j]
for (int j = i + 1; j < n - 1; j++) {
// Now look for the third number
for (int k = j + 1; k < n; k++) {
if (arr[i] + arr[j] + arr[k] == target)
return true;
}
}
}
return false;
}
int main() {
int arr[] = { 1, 4, 45, 6, 10, 8 };
int target = 13;
int n = sizeof(arr) / sizeof(arr[0]);
if (hasTripletSum(arr, n, target))
printf("true");
else
printf("false");
return 0;
}
class GfG {
static boolean hasTripletSum(int[] arr, int target) {
int n = arr.length;
// Fix the first element as arr[i]
for (int i = 0; i < n - 2; i++) {
// Fix the second element as arr[j]
for (int j = i + 1; j < n - 1; j++) {
// Now look for the third number
for (int k = j + 1; k < n; k++) {
if (arr[i] + arr[j] + arr[k] == target)
return true; // If a triplet is found
}
}
}
return false;
}
public static void main(String[] args) {
int[] arr = { 1, 4, 45, 6, 10, 8 };
int target = 13;
if (hasTripletSum(arr, target))
System.out.println("true");
else
System.out.println("false");
}
}
def hasTripletSum(arr, target):
n = len(arr)
# Fix the first element as arr[i]
for i in range(n - 2):
# Fix the second element as arr[j]
for j in range(i + 1, n - 1):
# Now look for the third number
for k in range(j + 1, n):
if arr[i] + arr[j] + arr[k] == target:
return True
return False
if __name__ == "__main__":
arr = [1, 4, 45, 6, 10, 8]
target = 13
if hasTripletSum(arr, target):
print("true")
else:
print("false")
using System;
class GfG {
static bool hasTripletSum(int[] arr, int target) {
int n = arr.Length;
// Fix the first element as arr[i]
for (int i = 0; i < n - 2; i++) {
// Fix the second element as arr[j]
for (int j = i + 1; j < n - 1; j++) {
// Now look for the third number
for (int k = j + 1; k < n; k++) {
if (arr[i] + arr[j] + arr[k] == target)
return true;
}
}
}
return false;
}
static void Main() {
int[] arr = { 1, 4, 45, 6, 10, 8 };
int target = 13;
if (hasTripletSum(arr, target))
Console.WriteLine("true");
else
Console.WriteLine("false");
}
}
function hasTripletSum(arr, target) {
let n = arr.length;
// Fix the first element as arr[i]
for (let i = 0; i < n - 2; i++) {
// Fix the second element as arr[j]
for (let j = i + 1; j < n - 1; j++) {
// Now look for the third number
for (let k = j + 1; k < n; k++) {
if (arr[i] + arr[j] + arr[k] === target)
return true;
}
}
}
return false;
}
// Driver code
let arr = [1, 4, 45, 6, 10, 8];
let target = 13;
if (hasTripletSum(arr, target))
console.log("true");
else
console.log("false");
Output
true
[Better Approach] - Hash Set - O(n^2) Time and O(n) Space
The idea is to traverse every element arr[i] in a loop. For every arr[i], use the hashing based solution of 2 Sum Problem to check if there is a pair with sum equal to given sum - arr[i].
Step by Step Approach:
- Iterate through the array, fixing the first element (arr[i]) for the triplet.
- For each arr[i], use a Hash Set to store potential second elements and run another loop inside it for j from i+1 to n-1.
- Inside a nested loop, check if given sum - arr[i] - arr[j] is present in the hash set. If yes, then print the triplet.
- If no triplet is found in the entire array, the function returns false.
#include <iostream>
#include <vector>
#include <unordered_set>
using namespace std;
bool hasTripletSum(vector<int>& arr, int target) {
int n = arr.size();
// Fix the first element as arr[i]
for (int i = 0; i < n - 2; i++) {
// Hash set to store potential second elements
unordered_set<int> st;
// Fix the third element as arr[j]
for(int j = i + 1; j < n; j++) {
int second = target - arr[i] - arr[j];
// Search for second element in hash set
if(st.find(second) != st.end()) {
return true;
}
// Add arr[j] as a potential second element
st.insert(arr[j]);
}
}
return false;
}
int main() {
vector<int> arr = { 1, 4, 45, 6, 10, 8 };
int target = 13;
if(hasTripletSum(arr, target))
cout << "true";
else
cout << "false";
return 0;
}
import java.util.HashSet;
import java.util.Set;
class GfG {
static boolean hasTripletSum(int[] arr, int target) {
int n = arr.length;
// Fix the first element as arr[i]
for (int i = 0; i < n - 2; i++) {
// Hash set to store potential second elements
Set<Integer> st = new HashSet<>();
// Fix the third element as arr[j]
for (int j = i + 1; j < n; j++) {
int second = target - arr[i] - arr[j];
// Search for second element in hash set
if (st.contains(second)) {
return true;
}
// Add arr[j] as a potential second element
st.add(arr[j]);
}
}
return false;
}
public static void main(String[] args) {
int[] arr = { 1, 4, 45, 6, 10, 8 };
int target = 13;
if (hasTripletSum(arr, target))
System.out.println("true");
else
System.out.println("false");
}
}
def hasTripletSum(arr, target):
n = len(arr)
# Fix the first element as arr[i]
for i in range(n - 2):
# Hash set to store potential second elements
st = set()
# Fix the third element as arr[j]
for j in range(i + 1, n):
second = target - arr[i] - arr[j]
# Search for second element in hash set
if second in st:
return True
# Add arr[j] as a potential second element
st.add(arr[j])
return False
if __name__ == "__main__":
arr = [1, 4, 45, 6, 10, 8]
target = 13
if hasTripletSum(arr, target):
print("true")
else:
print("false")
using System;
using System.Collections.Generic;
class GfG {
static bool hasTripletSum(int[] arr, int target) {
int n = arr.Length;
// Fix the first element as arr[i]
for (int i = 0; i < n - 2; i++) {
// Hash set to store potential second elements
HashSet<int> st = new HashSet<int>();
// Fix the third element as arr[j]
for (int j = i + 1; j < n; j++) {
int second = target - arr[i] - arr[j];
// Search for second element in hash set
if (st.Contains(second))
return true;
// Add arr[j] as a potential second element
st.Add(arr[j]);
}
}
return false;
}
static void Main(string[] args) {
int[] arr = { 1, 4, 45, 6, 10, 8 };
int target = 13;
if (hasTripletSum(arr, target))
Console.WriteLine("true");
else
Console.WriteLine("false");
}
}
function hasTripletSum(arr, target) {
let n = arr.length;
// Fix the first element as arr[i]
for (let i = 0; i < n - 2; i++) {
// Hash set to store potential second elements
let st = new Set();
// Fix the third element as arr[j]
for (let j = i + 1; j < n; j++) {
let second = target - arr[i] - arr[j];
// Search for second element in hash set
if (st.has(second)) {
return true;
}
// Add arr[j] as a potential second element
st.add(arr[j]);
}
}
return false;
}
// Driver code
let arr = [1, 4, 45, 6, 10, 8];
let target = 13;
if (hasTripletSum(arr, target))
console.log("true");
else
console.log("false");
Output
true
[Expected Approach] - Sorting and Two Pointer - O(n^2) Time and O(1) Space
We first sort the array. After sorting, we traverse every element arr[i] in a loop. For every arr[i], use the Two Pointer Technique based solution of 2 Sum Problem to check if there is a pair with sum equal to given sum - arr[i].
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
bool hasTripletSum(vector<int>& arr, int target) {
int n = arr.size();
sort(arr.begin(), arr.end());
// Fix the first element as arr[i]
for (int i = 0; i < n - 2; i++) {
// Initialize left and right pointers with
// start and end of remaining subarray
int l = i + 1, r = n - 1;
int requiredSum = target - arr[i];
while(l < r) {
if(arr[l] + arr[r] == requiredSum)
return true;
if(arr[l] + arr[r] < requiredSum)
l++;
else if(arr[l] + arr[r] > requiredSum)
r--;
}
}
return false;
}
int main() {
vector<int> arr = { 1, 4, 45, 6, 10, 8 };
int target = 13;
if(hasTripletSum(arr, target))
cout << "true";
else
cout << "false";
return 0;
}
#include <stdio.h>
int compare(const void *a, const void *b) {
return (*(int*)a - *(int*)b);
}
int hasTripletSum(int arr[], int n, int target) {
qsort(arr, n, sizeof(int), compare);
// Fix the first element as arr[i]
for (int i = 0; i < n - 2; i++) {
// Initialize left and right pointers with
// start and end of remaining subarray
int l = i + 1, r = n - 1;
int requiredSum = target - arr[i];
while (l < r) {
if (arr[l] + arr[r] == requiredSum)
return 1;
if (arr[l] + arr[r] < requiredSum)
l++;
else if (arr[l] + arr[r] > requiredSum)
r--;
}
}
return 0;
}
int main() {
int arr[] = { 1, 4, 45, 6, 10, 8 };
int target = 13;
int n = sizeof(arr) / sizeof(arr[0]);
if (hasTripletSum(arr, n, target))
printf("true");
else
printf("false");
return 0;
}
import java.util.Arrays;
class GfG {
static boolean hasTripletSum(int[] arr, int target) {
int n = arr.length;
Arrays.sort(arr);
// Fix the first element as arr[i]
for (int i = 0; i < n - 2; i++) {
// Initialize left and right pointers with
// start and end of remaining subarray
int l = i + 1, r = n - 1;
int requiredSum = target - arr[i];
while (l < r) {
if (arr[l] + arr[r] == requiredSum)
return true;
if (arr[l] + arr[r] < requiredSum)
l++;
else if (arr[l] + arr[r] > requiredSum)
r--;
}
}
return false;
}
public static void main(String[] args) {
int[] arr = { 1, 4, 45, 6, 10, 8 };
int target = 13;
if (hasTripletSum(arr, target))
System.out.println("true");
else
System.out.println("false");
}
}
def hasTripletSum(arr, target):
n = len(arr)
arr.sort()
# Fix the first element as arr[i]
for i in range(n - 2):
# Initialize left and right pointers with
# start and end of remaining subarray
l = i + 1
r = n - 1
requiredSum = target - arr[i]
while l < r:
if arr[l] + arr[r] == requiredSum:
return True
if arr[l] + arr[r] < requiredSum:
l += 1
else:
r -= 1
return False
if __name__ == "__main__":
arr = [1, 4, 45, 6, 10, 8]
target = 13
if hasTripletSum(arr, target):
print("true")
else:
print("false")
using System;
class GfG {
static bool hasTripletSum(int[] arr, int target) {
int n = arr.Length;
Array.Sort(arr);
// Fix the first element as arr[i]
for (int i = 0; i < n - 2; i++) {
// Initialize left and right pointers with
// start and end of remaining subarray
int l = i + 1, r = n - 1;
int requiredSum = target - arr[i];
while (l < r) {
if (arr[l] + arr[r] == requiredSum)
return true;
if (arr[l] + arr[r] < requiredSum)
l++;
else if (arr[l] + arr[r] > requiredSum)
r--;
}
}
return false;
}
static void Main(string[] args) {
int[] arr = { 1, 4, 45, 6, 10, 8 };
int target = 13;
if (hasTripletSum(arr, target))
Console.WriteLine("true");
else
Console.WriteLine("false");
}
}
function hasTripletSum(arr, target) {
let n = arr.length;
arr.sort((a, b) => a - b);
// Fix the first element as arr[i]
for (let i = 0; i < n - 2; i++) {
// Initialize left and right pointers with
// start and end of remaining subarray
let l = i + 1, r = n - 1;
let requiredSum = target - arr[i];
while (l < r) {
if (arr[l] + arr[r] == requiredSum)
return true;
if (arr[l] + arr[r] < requiredSum)
l++;
else if (arr[l] + arr[r] > requiredSum)
r--;
}
}
return false;
}
// Driver code
let arr = [1, 4, 45, 6, 10, 8];
let target = 13;
if (hasTripletSum(arr, target))
console.log("true");
else
console.log("false");
Output
true
Related Problems
- 3 Sum – Find All Triplets with Zero Sum
- 3 Sum – Find all Triplets with Given Sum
- 3 Sum – Triplet Sum Closest to Target
- 3 Sum – Pythagorean Triplet in an array
- 3 Sum – All Distinct Triplets with given Sum
- Pythagorean Triplet with given sum
- Count triplets with sum smaller than a given value
Please refer 3Sum - Complete Tutorial for all list of problems on triplets in an array.
Triplet Sum in Array Explained
Triplet Sum in Array Explained
