3 Sum - Triplet Sum in Array
Last Updated :
02 Jan, 2025
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Given an array arr[] of size n and an integer sum, the task is to 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.
// C++ Program to check for triplet sum by generating
// all triplets
#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;
}
}
}
// If we reach here, then no triplet was found
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;
}
// C Program to check for triplet sum by generating
// all triplets
#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;
}
}
}
// If we reach here, then no triplet was found
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;
}
// Java Program to check for triplet sum by generating
// all triplets
import java.util.*;
class GfG {
// Function to check if there exists a triplet with the given sum
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
}
}
}
// If we reach here, then no triplet was 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");
}
}
# Python Program to check for triplet sum by generating
# all triplets
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
# If we reach here, then no triplet was found
return False
if __name__ == "__main__":
arr = [1, 4, 45, 6, 10, 8]
target = 13
if hasTripletSum(arr, target):
print("true")
else:
print("false")
// C# Program to check for triplet sum by generating
// all triplets
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;
}
}
}
// If we reach here, then no triplet was found
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");
}
}
// JavaScript Program to check for triplet sum by generating
// all triplets
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;
}
}
}
// If we reach here, then no triplet was found
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] . Below is the 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.
// C++ Program to check for triplet sum using Hash Set
#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]);
}
}
// If we reach here, then no triplet was found
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;
}
// Java Program to check for triplet sum using Hash Set
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]);
}
}
// If we reach here, then no triplet was 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");
}
}
# Python Program to check for triplet sum using Hash Set
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])
# If we reach here, then no triplet was found
return False
if __name__ == "__main__":
arr = [1, 4, 45, 6, 10, 8]
target = 13
if hasTripletSum(arr, target):
print("true")
else:
print("false")
// C# Program to check for triplet sum using Hash Set
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]);
}
}
// If we reach here, then no triplet was found
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");
}
}
// JavaScript Program to check for triplet sum using Hash Set
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]);
}
}
// If we reach here, then no triplet was found
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].
// C++ Program to check for triplet sum using Sorting
// and Two Pointer Technique
#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--;
}
}
// If we reach here, then no triplet was found
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;
}
// C Program to check for triplet sum using Sorting
// and Two Pointer Technique
#include <stdio.h>
// Function to compare integers for sorting (used by qsort)
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--;
}
}
// If we reach here, then no triplet was found
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;
}
// Java Program to check for triplet sum using Sorting
// and Two Pointer Technique
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--;
}
}
// If we reach here, then no triplet was 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");
}
}
# Python Program to check for triplet sum using Sorting
# and Two Pointer Technique
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
# If we reach here, then no triplet was found
return False
if __name__ == "__main__":
arr = [1, 4, 45, 6, 10, 8]
target = 13
if hasTripletSum(arr, target):
print("true")
else:
print("false")
// C# Program to check for triplet sum using Sorting
// and Two Pointer Technique
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--;
}
}
// If we reach here, then no triplet was found
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");
}
}
// JavaScript Program to check for triplet sum using Sorting
// and Two Pointer Technique
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--;
}
}
// If we reach here, then no triplet was found
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.
