Find all triplets with zero sum
Last Updated :
26 Apr, 2024
Given an array of distinct elements. The task is to find triplets in the array whose sum is zero.
Examples :
Input: arr[] = {0, -1, 2, -3, 1}
Output: (0 -1 1), (2 -3 1)
Explanation: The triplets with zero sum are 0 + -1 + 1 = 0 and 2 + -3 + 1 = 0
Input: arr[] = {1, -2, 1, 0, 5}
Output: 1 -2 1
Explanation: The triplets with zero sum is 1 + -2 + 1 = 0
Naive approach: Below is the idea to solve the problem
Run three loops and check one by one whether the sum of the three elements is zero or not. If the sum of three elements is zero then print elements otherwise print not found.
Follow the below steps to Implement the Idea:
- Run three nested loops with loop counter i, j, k
- The first loops will run from 0 to n-3 and second loop from i+1 to n-2 and the third loop from j+1 to b. The loop counter represents the three elements of the triplet.
- Check if the sum of elements at i’th, j’th, k’th is equal to zero or not. If yes print the sum else continue.
Below is the implementation of the above approach:
C++
// A simple C++ program to find three elements
// whose sum is equal to zero
#include <bits/stdc++.h>
using namespace std;
// Prints all triplets in arr[] with 0 sum
void findTriplets(int arr[], int n)
{
bool found = false;
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 (arr[i] + arr[j] + arr[k] == 0) {
cout << arr[i] << " " << arr[j] << " "
<< arr[k] << endl;
found = true;
}
}
}
}
// If no triplet with 0 sum found in array
if (found == false)
cout << " not exist " << endl;
}
// Driver code
int main()
{
int arr[] = { 0, -1, 2, -3, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
findTriplets(arr, n);
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
// A simple C program to find three elements
// whose sum is equal to zero
#include <stdbool.h>
#include <stdio.h>
// Prints all triplets in arr[] with 0 sum
void findTriplets(int arr[], int n)
{
bool found = false;
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 (arr[i] + arr[j] + arr[k] == 0) {
printf("%d %d %d\n", arr[i], arr[j],
arr[k]);
found = true;
}
}
}
}
// If no triplet with 0 sum found in array
if (found == false)
printf(" not exist \n");
}
// Driver code
int main()
{
int arr[] = { 0, -1, 2, -3, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
findTriplets(arr, n);
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
// A simple Java program to find three elements
// whose sum is equal to zero
import java.io.*;
import java.util.*;
class num {
// Prints all triplets in arr[] with 0 sum
static void findTriplets(int[] arr, int n)
{
boolean found = false;
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 (arr[i] + arr[j] + arr[k] == 0) {
System.out.println(arr[i] + " "
+ arr[j] + " "
+ arr[k]);
found = true;
}
}
}
}
// If no triplet with 0 sum found in array
if (found == false)
System.out.println(" not exist ");
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 0, -1, 2, -3, 1 };
int n = arr.length;
findTriplets(arr, n);
}
}
// This code is contributed by Aditya Kumar (adityakumar129)
# A simple Python 3 program
# to find three elements whose
# sum is equal to zero
# Prints all triplets in
# arr[] with 0 sum
def findTriplets(arr, n):
found = False
for i in range(0, n-2):
for j in range(i+1, n-1):
for k in range(j+1, n):
if (arr[i] + arr[j] + arr[k] == 0):
print(arr[i], arr[j], arr[k])
found = True
# If no triplet with 0 sum
# found in array
if (found == False):
print(" not exist ")
# Driver code
arr = [0, -1, 2, -3, 1]
n = len(arr)
findTriplets(arr, n)
# This code is contributed by Smitha Dinesh Semwal
// A simple C# program to find three elements
// whose sum is equal to zero
using System;
class GFG {
// Prints all triplets in arr[] with 0 sum
static void findTriplets(int[] arr, int n)
{
bool found = false;
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 (arr[i] + arr[j] + arr[k] == 0) {
Console.Write(arr[i]);
Console.Write(" ");
Console.Write(arr[j]);
Console.Write(" ");
Console.Write(arr[k]);
Console.Write("\n");
found = true;
}
}
}
}
// If no triplet with 0 sum found in
// array
if (found == false)
Console.Write(" not exist ");
}
// Driver code
public static void Main()
{
int[] arr = { 0, -1, 2, -3, 1 };
int n = arr.Length;
findTriplets(arr, n);
}
}
// This code is contributed by nitin mittal.
// A simple JavaScript program to find
// three elements whose sum is equal to zero
var arr = [0, -1, 2, -3, 1];
// Prints all triplets in arr[] with 0 sum
function findTriplets(arr) {
var found = false;
for (var i = 0; i < arr.length - 2; i++) {
for (var j = i + 1; j < arr.length - 1; j++) {
for (var k = j + 1; k < arr.length; k++) {
if (arr[i] + arr[j] + arr[k] === 0) {
console.log(arr[i] + " " + arr[j] + " " + arr[k]);
found = true;
}
}
}
}
// If no triplet with 0 sum found in array
if (!found) {
console.log("No triplet with 0 sum exists.");
}
}
findTriplets(arr);
<?php
// A simple PHP program to
// find three elements whose
// sum is equal to zero
// Prints all triplets
// in arr[] with 0 sum
function findTriplets($arr, $n)
{
$found = false;
for ($i = 0; $i < $n - 2; $i++)
{
for ($j = $i + 1; $j < $n - 1; $j++)
{
for ($k = $j + 1; $k < $n; $k++)
{
if ($arr[$i] + $arr[$j] +
$arr[$k] == 0)
{
echo $arr[$i] , " ",
$arr[$j] , " ",
$arr[$k] ,"\n";
$found = true;
}
}
}
}
// If no triplet with 0
// sum found in array
if ($found == false)
echo " not exist ", "\n";
}
// Driver Code
$arr = array (0, -1, 2, -3, 1);
$n = sizeof($arr);
findTriplets($arr, $n);
// This code is contributed by m_kit
?>
Time Complexity: O(n3), As three nested loops are required, so the time complexity is O(n3).
Auxiliary Space: O(1), Since no extra space is required, so the space complexity is constant.
Find all triplets with zero sum using Hashing
Below is the idea to solve the problem
This involves traversing through the array. For every element arr[i], find a pair with sum “-arr[i]”. This problem reduces to pair sum and can be solved in O(n) time using hashing.
Follow the steps below to implement the idea:
- Create a HashSet to store a unique element.
- Run a nested loop with two loops, the outer loop from 0 to n-2 and the inner loop from i+1 to n-1
- Check if the sum of ith and jth element multiplied with -1 is present in the HashSet or not
- If the element is present in the HashSet, print the triplet else insert the jth element in the HashSet.
Below is the implementation of the above approach:
C++
// C++ program to find triplets in a given
// array whose sum is zero
#include <bits/stdc++.h>
using namespace std;
// function to print triplets with 0 sum
void findTriplets(int arr[], int n)
{
bool found = false;
for (int i = 0; i < n - 1; i++) {
// Find all pairs with sum equals to
// "-arr[i]"
unordered_set<int> s;
for (int j = i + 1; j < n; j++) {
int x = -(arr[i] + arr[j]);
if (s.find(x) != s.end()) {
printf("%d %d %d\n", x, arr[i], arr[j]);
found = true;
}
else
s.insert(arr[j]);
}
}
if (found == false)
cout << " No Triplet Found" << endl;
}
// Driver code
int main()
{
int arr[] = { 0, -1, 2, -3, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
findTriplets(arr, n);
return 0;
}
// Java program to find triplets in a given
// array whose sum is zero
import java.io.*;
import java.util.*;
class GFG {
// function to print triplets with 0 sum
static void findTriplets(int arr[], int n)
{
boolean found = false;
for (int i = 0; i < n - 1; i++) {
// Find all pairs with sum equals to
// "-arr[i]"
HashSet<Integer> s = new HashSet<Integer>();
for (int j = i + 1; j < n; j++) {
int x = -(arr[i] + arr[j]);
if (s.contains(x)) {
System.out.printf("%d %d %d\n", x,
arr[i], arr[j]);
found = true;
}
else {
s.add(arr[j]);
}
}
}
if (found == false) {
System.out.printf(" No Triplet Found\n");
}
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 0, -1, 2, -3, 1 };
int n = arr.length;
findTriplets(arr, n);
}
}
// This code contributed by Rajput-Ji
# Python3 program to find triplets
# in a given array whose sum is zero
# function to print triplets with 0 sum
def findTriplets(arr, n):
found = False
for i in range(n - 1):
# Find all pairs with sum
# equals to "-arr[i]"
s = set()
for j in range(i + 1, n):
x = -(arr[i] + arr[j])
if x in s:
print(x, arr[i], arr[j])
found = True
else:
s.add(arr[j])
if found == False:
print("No Triplet Found")
# Driver Code
arr = [0, -1, 2, -3, 1]
n = len(arr)
findTriplets(arr, n)
# This code is contributed by Shrikant13
// C# program to find triplets in a given
// array whose sum is zero
using System;
using System.Collections.Generic;
class GFG {
// function to print triplets with 0 sum
static void findTriplets(int[] arr, int n)
{
bool found = false;
for (int i = 0; i < n - 1; i++) {
// Find all pairs with sum equals to
// "-arr[i]"
HashSet<int> s = new HashSet<int>();
for (int j = i + 1; j < n; j++) {
int x = -(arr[i] + arr[j]);
if (s.Contains(x)) {
Console.Write("{0} {1} {2}\n", x,
arr[i], arr[j]);
found = true;
}
else {
s.Add(arr[j]);
}
}
}
if (found == false) {
Console.Write(" No Triplet Found\n");
}
}
// Driver code
public static void Main(String[] args)
{
int[] arr = { 0, -1, 2, -3, 1 };
int n = arr.Length;
findTriplets(arr, n);
}
}
// This code has been contributed by 29AjayKumar
// Function to find triplets in a given array whose sum is zero
function findTriplets(arr, n) {
var found = false;
for (var i = 0; i < n - 1; i++) {
// Find all pairs with sum equals to "-arr[i]"
var s = new Set();
for (var j = i + 1; j < n; j++) {
var x = -(arr[i] + arr[j]);
if (s.has(x)) {
console.log(x + " " + arr[i] + " " + arr[j]);
found = true;
} else {
s.add(arr[j]);
}
}
}
if (!found) {
console.log("No Triplet Found");
}
}
// Driver code
var arr = [0, -1, 2, -3, 1];
var n = arr.length;
findTriplets(arr, n);
// This code is contributed by famously.
Time Complexity: O(n2), Since two nested loops are required, so the time complexity is O(n2).
Auxiliary Space: O(n), Since a HashSet is required, so the space complexity is linear.
Find all triplets with zero sum using Sorting:
The idea is based on the above discussed approach using Hashmap of this post. For every element check that there is a pair whose sum is equal to the negative value of that element.
Follow the steps below to implement the idea:
- Sort the array in ascending order.
- Traverse the array from start to end.
- For every index i, create two variables l = i + 1 and r = n – 1
- Run a loop until l is less than r if the sum of array[i], array[l] and array[r] is equal to zero then print the triplet and break the loop
- If the sum is less than zero then increment the value of l, by increasing the value of l the sum will increase as the array is sorted, so array[l+1] > array [l]
- If the sum is greater than zero then decrement the value of r, by decreasing the value of r the sum will decrease as the array is sorted, so array[r-1] < array [r].
Below is the implementation of the above approach:
C++
// C++ program to find triplets in a given
// array whose sum is zero
#include <bits/stdc++.h>
using namespace std;
// function to print triplets with 0 sum
void findTriplets(int arr[], int n)
{
bool found = false;
// sort array elements
sort(arr, arr + n);
for (int i = 0; i < n - 1; i++) {
// initialize left and right
int l = i + 1;
int r = n - 1;
int x = arr[i];
while (l < r) {
if (x + arr[l] + arr[r] == 0) {
// print elements if it's sum is zero
printf("%d %d %d\n", x, arr[l], arr[r]);
l++;
r--;
found = true;
// break;
}
// If sum of three elements is less
// than zero then increment in left
else if (x + arr[l] + arr[r] < 0)
l++;
// if sum is greater than zero then
// decrement in right side
else
r--;
}
}
if (found == false)
cout << " No Triplet Found" << endl;
}
// Driven source
int main()
{
int arr[] = { 0, -1, 2, -3, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
findTriplets(arr, n);
return 0;
}
// Java program to find triplets in a given
// array whose sum is zero
import java.io.*;
import java.util.Arrays;
class GFG {
// function to print triplets with 0 sum
static void findTriplets(int arr[], int n)
{
boolean found = false;
// sort array elements
Arrays.sort(arr);
for (int i = 0; i < n - 1; i++) {
// initialize left and right
int l = i + 1;
int r = n - 1;
int x = arr[i];
while (l < r) {
if (x + arr[l] + arr[r] == 0) {
// print elements if it's sum is zero
System.out.print(x + " ");
System.out.print(arr[l] + " ");
System.out.println(arr[r] + " ");
l++;
r--;
found = true;
}
// If sum of three elements is less
// than zero then increment in left
else if (x + arr[l] + arr[r] < 0)
l++;
// if sum is greater than zero then
// decrement in right side
else
r--;
}
}
if (found == false)
System.out.println(" No Triplet Found");
}
// Driven source
public static void main(String[] args)
{
int arr[] = { 0, -1, 2, -3, 1 };
int n = arr.length;
findTriplets(arr, n);
}
// This code is contributed by Tushil..
}
# python program to find triplets in a given
# array whose sum is zero
# function to print triplets with 0 sum
def findTriplets(arr, n):
found = False
# sort array elements
arr.sort()
for i in range(0, n-1):
# initialize left and right
l = i + 1
r = n - 1
x = arr[i]
while (l < r):
if (x + arr[l] + arr[r] == 0):
# print elements if it's sum is zero
print(x, arr[l], arr[r])
l += 1
r -= 1
found = True
# If sum of three elements is less
# than zero then increment in left
elif (x + arr[l] + arr[r] < 0):
l += 1
# if sum is greater than zero then
# decrement in right side
else:
r -= 1
if (found == False):
print(" No Triplet Found")
# Driven source
arr = [0, -1, 2, -3, 1]
n = len(arr)
findTriplets(arr, n)
# This code is contributed by Smitha Dinesh Semwal
// C# program to find triplets in a given
// array whose sum is zero
using System;
public class GFG {
// function to print triplets with 0 sum
static void findTriplets(int[] arr, int n)
{
bool found = false;
// sort array elements
Array.Sort(arr);
for (int i = 0; i < n - 1; i++) {
// initialize left and right
int l = i + 1;
int r = n - 1;
int x = arr[i];
while (l < r) {
if (x + arr[l] + arr[r] == 0) {
// print elements if it's sum is zero
Console.Write(x + " ");
Console.Write(arr[l] + " ");
Console.WriteLine(arr[r] + " ");
l++;
r--;
found = true;
}
// If sum of three elements is less
// than zero then increment in left
else if (x + arr[l] + arr[r] < 0)
l++;
// if sum is greater than zero then
// decrement in right side
else
r--;
}
}
if (found == false)
Console.WriteLine(" No Triplet Found");
}
// Driven source
static public void Main()
{
int[] arr = { 0, -1, 2, -3, 1 };
int n = arr.Length;
findTriplets(arr, n);
}
// This code is contributed by akt_mit..
}
// Function to find triplets in a given array whose sum is zero
function findTriplets(arr, n) {
let found = false;
// Sort array elements
arr.sort((a, b) => a - b);
for (let i = 0; i < n - 1; i++) {
// Initialize left and right pointers
let l = i + 1;
let r = n - 1;
let x = arr[i];
while (l < r) {
if (x + arr[l] + arr[r] === 0) {
// Print elements if their sum is zero
console.log(x + " " + arr[l] + " " + arr[r]);
l++;
r--;
found = true;
}
// If sum of three elements is less than zero, increment left pointer
else if (x + arr[l] + arr[r] < 0)
l++;
// If sum is greater than zero, decrement right pointer
else
r--;
}
}
if (!found)
console.log("No Triplet Found");
}
// Test the function
let arr = [0, -1, 2, -3, 1];
let n = arr.length;
findTriplets(arr, n);
<?php
// PHP program to find
// triplets in a given
// array whose sum is zero
// function to print
// triplets with 0 sum
function findTriplets($arr, $n)
{
$found = false;
// sort array elements
sort($arr);
for ($i = 0; $i < $n - 1; $i++)
{
// initialize left
// and right
$l = $i + 1;
$r = $n - 1;
$x = $arr[$i];
while ($l < $r)
{
if ($x + $arr[$l] +
$arr[$r] == 0)
{
// print elements if
// it's sum is zero
echo $x," ", $arr[$l],
" ", $arr[$r], "\n";
$l++;
$r--;
$found = true;
}
// If sum of three elements
// is less than zero then
// increment in left
else if ($x + $arr[$l] +
$arr[$r] < 0)
$l++;
// if sum is greater than
// zero then decrement
// in right side
else
$r--;
}
}
if ($found == false)
echo " No Triplet Found" ,"\n";
}
// Driver Code
$arr = array (0, -1, 2, -3, 1);
$n = sizeof($arr);
findTriplets($arr, $n);
// This code is contributed by ajit
?>
Time Complexity: O(n2), Only two nested loops are required, so the time complexity is O(n2).
Auxiliary Space: O(1), no extra space is required, so the space complexity is constant.
Here all the above codes have a time complexity above O(n^2) which is not that optimal. We can use this code here instead.
C++
#include <iostream>
using namespace std;
void findTriplet(int arr[], int N) {
for (int i = 0; i < N - 2; i++) {
if (arr[i] + arr[i + 1] + arr[i + 2] == 0) {
cout << arr[i] << " " << arr[i + 1] << " " << arr[i + 2] << endl;
return;
}
}
cout << "No such triplet available" << endl;
}
int main() {
int arr[] = {1, 2, -3, 4, -1, -2, 0};
int N = sizeof(arr) / sizeof(arr[0]);
findTriplet(arr, N);
return 0;
}
public class Triplet {
static void findTriplet(int[] arr, int N) {
for (int i = 0; i < N - 2; i++) {
if (arr[i] + arr[i + 1] + arr[i + 2] == 0) {
System.out.println(arr[i] + " " + arr[i + 1] + " " + arr[i + 2]);
return;
}
}
System.out.println("No such triplet available");
}
public static void main(String[] args) {
int[] arr = {1, 2, -3, 4, -1, -2, 0};
int N = arr.length;
findTriplet(arr, N);
}
}
def find_triplet(arr):
N = len(arr)
for i in range(N - 2):
if arr[i] + arr[i + 1] + arr[i + 2] == 0:
print(arr[i], arr[i + 1], arr[i + 2])
return
print("No such triplet available")
if __name__ == "__main__":
arr = [1, 2, -3, 4, -1, -2, 0]
find_triplet(arr)
// Function to find a triplet in an array whose sum is zero
function findTriplet(arr) {
const N = arr.length; // Get the length of the array
for (let i = 0; i < N - 2; i++) { // Iterate through the array up to the second-to-last element
// Check if the sum of current element, next element, and the one after is zero
if (arr[i] + arr[i + 1] + arr[i + 2] === 0) {
console.log(arr[i], arr[i + 1], arr[i + 2]); // Print the triplet
return; // Exit the function after finding the triplet
}
}
console.log("No such triplet available"); // Print a message if no triplet is found
}
// Main function
function main() {
const arr = [1, 2, -3, 4, -1, -2, 0]; // Input array
findTriplet(arr); // Call the function to find the triplet
}
// Call the main function to execute the code
main();
The time complexity here is O(N) and the space complexity is O(1).
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