3 Sum - Find All Triplets with Zero Sum
Last Updated :
23 Jul, 2025
Given an array arr[], the task is to find all possible indices {i, j, k} of triplet {arr[i], arr[j], arr[k]} such that their sum is equal to zero and all indices in a triplet should be distinct (i != j, j != k, k != i). We need to return indices of a triplet in sorted order, i.e., i < j < k.
Examples :
Input: arr[] = {0, -1, 2, -3, 1}
Output: {{0, 1, 4}, {2, 3, 4}}
Explanation: Two triplets with sum 0 are:
arr[0] + arr[1] + arr[4] = 0 + (-1) + 1 = 0
arr[2] + arr[3] + arr[4] = 2 + (-3) + 1 = 0
Input: arr[] = {1, -2, 1, 0, 5}
Output: {{0, 1, 2}}
Explanation: Only triplet which satisfies the condition is arr[0] + arr[1] + arr[2] = 1 + (-2) + 1 = 0
Input: arr[] = {2, 3, 1, 0, 5}
Output: {{}}
Explanation: There is no triplet with sum 0
[Naive Approach] Using Three Nested Loops - O(n^3) Time and O(1) Space
The simplest approach is to generate all possible triplets using three nested loops and if the sum of any triplet is equal to zero then add it to the result.
C++
// C++ program to find triplet having sum zero using
// three nested loops
#include <iostream>
#include <vector>
using namespace std;
vector<vector<int>> findTriplets(vector<int> &arr) {
vector<vector<int>> res;
int n = arr.size();
// Generating all triplets
for (int i = 0; i < n - 2; i++) {
for (int j = i + 1; j < n - 1; j++) {
for (int k = j + 1; k < n; k++) {
// If the sum of a triplet equals to zero
// then add it's indices to the result
if (arr[i] + arr[j] + arr[k] == 0)
res.push_back({i, j, k});
}
}
}
return res;
}
int main() {
vector<int> arr = {0, -1, 2, -3, 1};
vector<vector<int>> res = findTriplets(arr);
for(int i = 0; i < res.size(); i++)
cout << res[i][0] << " " << res[i][1] << " " << res[i][2] << endl;
return 0;
}
// C program to find triplet having sum zero using
// three nested loops
#include <stdio.h>
#include <stdlib.h>
#define MAX_LIMIT 100
void findTriplets(int arr[], int n, int res[][3], int* count) {
*count = 0;
// Generating all triplets
for (int i = 0; i < n - 2; i++) {
for (int j = i + 1; j < n - 1; j++) {
for (int k = j + 1; k < n; k++) {
// If the sum of triplet equals zero
// then add it's indexes to reuslt
if (arr[i] + arr[j] + arr[k] == 0) {
res[*count][0] = i;
res[*count][1] = j;
res[*count][2] = k;
(*count)++;
}
}
}
}
}
int main() {
int arr[] = {0, -1, 2, -3, 1};
int n = sizeof(arr) / sizeof(arr[0]);
// res array to store all triplets
int res[MAX_LIMIT][3];
// Variable to store number of triplets found
int count = 0;
findTriplets(arr, n, res, &count);
for (int i = 0; i < count; i++)
printf("%d %d %d\n", res[i][0], res[i][1], res[i][2]);
return 0;
}
// Java program to find triplet having sum zero using
// three nested loops
import java.util.ArrayList;
import java.util.List;
class GfG {
static ArrayList<ArrayList<Integer>> findTriplets(int[] arr) {
ArrayList<ArrayList<Integer>> res = new ArrayList<>();
int n = arr.length;
// Generating all triplets
for (int i = 0; i < n - 2; i++) {
for (int j = i + 1; j < n - 1; j++) {
for (int k = j + 1; k < n; k++) {
// If the sum of triplet equals to zero
// then add it's indexes to the result
if (arr[i] + arr[j] + arr[k] == 0) {
ArrayList<Integer> triplet = new ArrayList<>();
triplet.add(i);
triplet.add(j);
triplet.add(k);
res.add(triplet);
}
}
}
}
return res;
}
public static void main(String[] args) {
int[] arr = {0, -1, 2, -3, 1};
ArrayList<ArrayList<Integer>> res = findTriplets(arr);
for (List<Integer> triplet : res)
System.out.println(triplet.get(0) + " " + triplet.get(1)
+ " " + triplet.get(2));
}
}
# Python program to find triplet with sum zero
# using three nested loops
def findTriplets(arr):
res = []
n = len(arr)
# Generating all triplets
for i in range(n - 2):
for j in range(i + 1, n - 1):
for k in range(j + 1, n):
# If the sum of triplet equals to zero
# then add it's indexes to the result
if arr[i] + arr[j] + arr[k] == 0:
res.append([i, j, k])
return res
arr = [0, -1, 2, -3, 1]
res = findTriplets(arr)
for triplet in res:
print(triplet[0], triplet[1], triplet[2])
// C# program to find triplet with sum zero using
// three nested loops
using System;
using System.Collections.Generic;
class GfG {
static List<List<int>> FindTriplets(int[] arr) {
List<List<int>> res = new List<List<int>>();
int n = arr.Length;
// Generating all triplets
for (int i = 0; i < n - 2; i++) {
for (int j = i + 1; j < n - 1; j++) {
for (int k = j + 1; k < n; k++) {
// If the sum of triplet equals to zero
// then add it's indexes to the result
if (arr[i] + arr[j] + arr[k] == 0) {
res.Add(new List<int> { i, j, k });
}
}
}
}
return res;
}
public static void Main() {
int[] arr = { 0, -1, 2, -3, 1 };
List<List<int>> res = FindTriplets(arr);
foreach (var triplet in res) {
Console.WriteLine($"{triplet[0]} {triplet[1]} {triplet[2]}");
}
}
}
// JavaScript program to find triplet with sum zero
// using three nested loops
function findTriplets(arr) {
const res = [];
const n = arr.length;
// Generating all triplets
for (let i = 0; i < n - 2; i++) {
for (let j = i + 1; j < n - 1; j++) {
for (let k = j + 1; k < n; k++) {
// If the sum of triplet equals to zero
// then add it's indexes to the result
if (arr[i] + arr[j] + arr[k] === 0) {
res.push([i, j, k]);
}
}
}
}
return res;
}
const arr = [0, -1, 2, -3, 1];
const res = findTriplets(arr);
res.forEach(triplet => {
console.log(triplet[0] + " " + triplet[1] + " " + triplet[2]);
});
Time Complexity: O(n3), As three nested loops are used.
Auxiliary Space: O(1)
[Expected Approach] Using Hash Map - O(n^3) Time and O(n) Space
- The idea is to use a hash map to store indices of each element and efficiently find triplets that sum to zero.
- We iterate through all pairs
(j, k), compute the required third element as -(arr[j] + arr[k]), and check if it exists in the map with a valid index i < j. - If found, we store
{i, j, k} in the result. To ensure unique triplets, the map maintains only indices less than the current j. - In the worst case, this approach also takes O(n^3) time but in the average case, it is much faster than Naive approach as we are iterating over only those triplets whose sum is equal to target.
C++
// C++ program to find all triplets with zero sum
#include <bits/stdc++.h>
using namespace std;
vector<vector<int>> findTriplets(vector<int> &arr) {
// Map to store indices for each value
unordered_map<int, vector<int>> map;
// Resultant array
vector<vector<int>> ans;
// Check for all pairs i, j
for (int j=0; j<arr.size(); j++) {
for (int k=j+1; k<arr.size(); k++) {
// Value of third index should be
int val = -1*(arr[j]+arr[k]);
// If such indices exists
if (map.find(val)!=map.end()) {
// Append the i, j, k
for (auto i: map[val]) {
ans.push_back({i, j, k});
}
}
}
// After j'th index is traversed
// We can use it as i.
map[arr[j]].push_back(j);
}
return ans;
}
int main() {
vector<int> arr = {0, -1, 2, -3, 1};
vector<vector<int>> res = findTriplets(arr);
for (int i = 0; i < res.size(); i++)
cout << res[i][0] << " " << res[i][1] << " " << res[i][2] << endl;
return 0;
}
// Java program to find all triplets with zero sum
import java.util.*;
class GfG {
// Function to find all triplets with zero sum
static List<List<Integer>> findTriplets(int[] arr) {
// Map to store indices for each value
HashMap<Integer, List<Integer>> map = new HashMap<>();
// Resultant list
List<List<Integer>> ans = new ArrayList<>();
// Check for all pairs i, j
for (int j = 0; j < arr.length; j++) {
for (int k = j + 1; k < arr.length; k++) {
// Value of third index should be
int val = -1 * (arr[j] + arr[k]);
// If such indices exist
if (map.containsKey(val)) {
// Append the i, j, k
for (int i : map.get(val)) {
ans.add(Arrays.asList(i, j, k));
}
}
}
// After j'th index is traversed
// We can use it as i.
map.putIfAbsent(arr[j], new ArrayList<>());
map.get(arr[j]).add(j);
}
return ans;
}
public static void main(String[] args) {
int[] arr = {0, -1, 2, -3, 1};
List<List<Integer>> res = findTriplets(arr);
for (List<Integer> triplet : res) {
System.out.println(triplet.get(0) + " " + triplet.get(1) + " " + triplet.get(2));
}
}
}
# Python program to find all triplets with zero sum
# Function to find all triplets with zero sum
def findTriplets(arr):
# Map to store indices for each value
map = {}
# Resultant array
ans = []
# Check for all pairs i, j
for j in range(len(arr)):
for k in range(j + 1, len(arr)):
# Value of third index should be
val = -1 * (arr[j] + arr[k])
# If such indices exist
if val in map:
# Append the i, j, k
for i in map[val]:
ans.append([i, j, k])
# After j'th index is traversed
# We can use it as i.
if arr[j] not in map:
map[arr[j]] = []
map[arr[j]].append(j)
return ans
if __name__ == "__main__":
arr = [0, -1, 2, -3, 1]
res = findTriplets(arr)
for triplet in res:
print(triplet[0], triplet[1], triplet[2])
// C# program to find all triplets with zero sum
using System;
using System.Collections.Generic;
class GfG {
// Function to find all triplets with zero sum
static List<List<int>> findTriplets(int[] arr) {
// Dictionary to store indices for each value
Dictionary<int, List<int>> map = new Dictionary<int, List<int>>();
// Resultant list
List<List<int>> ans = new List<List<int>>();
// Check for all pairs i, j
for (int j = 0; j < arr.Length; j++) {
for (int k = j + 1; k < arr.Length; k++) {
// Value of third index should be
int val = -1 * (arr[j] + arr[k]);
// If such indices exist
if (map.ContainsKey(val)) {
// Append the i, j, k
foreach (int i in map[val]) {
ans.Add(new List<int> { i, j, k });
}
}
}
// After j'th index is traversed
// We can use it as i.
if (!map.ContainsKey(arr[j])) {
map[arr[j]] = new List<int>();
}
map[arr[j]].Add(j);
}
return ans;
}
static void Main(string[] args) {
int[] arr = { 0, -1, 2, -3, 1 };
List<List<int>> res = findTriplets(arr);
foreach (var triplet in res) {
Console.WriteLine(triplet[0] + " " + triplet[1] + " " + triplet[2]);
}
}
}
// JavaScript program to find all triplets with zero sum
// Function to find all triplets with zero sum
function findTriplets(arr) {
// Map to store indices for each value
let map = new Map();
// Resultant array
let ans = [];
// Check for all pairs i, j
for (let j = 0; j < arr.length; j++) {
for (let k = j + 1; k < arr.length; k++) {
// Value of third index should be
let val = -1 * (arr[j] + arr[k]);
// If such indices exist
if (map.has(val)) {
// Append the i, j, k
for (let i of map.get(val)) {
ans.push([i, j, k]);
}
}
}
// After j'th index is traversed
// We can use it as i.
if (!map.has(arr[j])) {
map.set(arr[j], []);
}
map.get(arr[j]).push(j);
}
return ans;
}
const arr = [0, -1, 2, -3, 1];
const res = findTriplets(arr);
for (let triplet of res) {
console.log(triplet[0], triplet[1], triplet[2]);
}
Time Complexity: O(n3), Since two nested loops are used and traversing through the map may take O(n) in worst case.
Auxiliary Space: O(n), Since a HashMap is used to store all value index pairs.
Please refer 3Sum - Complete Tutorial for all list of problems on triplets in an array.
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