✨ อัปเดตล่าสุด: web//https://www.geeksforgeeks.org/find triplets array whose sum equal zero

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

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Method 1: This is a simple method that takes O(n³) time to arrive at the result.

Approach: The naive approach run three loops and check one by one that sum of three elements is zero or not. If the sum of three elements is zero then print elements otherwise print not found.
Algorithm:
1. Run three nested loops with loop counter i, j, k
2. The three 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 n-1. The loop counter represents the three elements of the triplet.
3. 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.

Implementation:

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 = true; 
    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; 
} 

Java

// A simple Java program to find three elements 
// whose sum is equal to zero 
class num{ 
// Prints all triplets in arr[] with 0 sum 
static void findTriplets(int[] arr, int n) 
{ 
    boolean found = true; 
    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.print(arr[i]); 
                    System.out.print(" "); 
                    System.out.print(arr[j]); 
                    System.out.print(" "); 
                    System.out.print(arr[k]); 
                    System.out.print("\n"); 
                    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 
//Smitha Dinesh Semwal 

Python3

# 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 = True
    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     

C#

// 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 = true; 
        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. 

PHP

<?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) 
{ 
    $found = true; 
    for ($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 
?> 

Output:

0 -1 1
2 -3 1

Complexity Analysis:
- 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 time complexity is constant.

Method 2: The second method uses the process of Hashing to arrive at the result and is solved at a lesser time of O(n²).

Approach: This involves traversing through the array. For every element arr[i], find a pair with sum “-arr[i]”. This problem reduces to pairs sum and can be solved in O(n) time using hashing.
Algorithm:
1. Create a hashap to store a key value pair.
2. Run a nested loop with two loops, outer loop from 0 to n-2 and the inner loop from i+1 to n-1
3. Check if the sum of ith and jth element multiplied with -1 is present in the hashmap or not
4. If the element is present in the hashmap, print the triplet else insert the j’th element in the hashmap.

Implementation:

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

// Java program to find triplets in a given 
// array whose sum is zero 
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

# 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#

// 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 

Output:

-1 0 1
-3 2 1

Complexity Analysis:
- Time Complexity: O(n²).
  Since two nested loops is required, so the time complexity is O(n²).
- Auxiliary Space: O(n).
  Since a hashmap is required, so the time complexity is linear.

Method 3: This method uses Sorting to arrive at the correct result and is solved in O(n²) time.

Approach: The above method requires extra space. The idea is based on method 2 of this post. For every element check that there is a pair whose sum is equal to the negative value of that element.
Algorithm:
1. Sort the array in ascending order.
2. Traverse the array from start to end.
3. For every index i, create two variables l = i + 1 and r = n – 1
4. Run a loop until l is less than r, if the sum of array[l], array[r] is equal to zero then print the triplet and break the loop
5. If the sum is less than zero then increment value of l, by increasing value of l the sum will increase as the array is sorted, so array[l+1] > array [l]
6. If the sum is greater than zero then decrement value of r, by increasing value of l the sum will decrease as the array is sorted, so array[r-1] < array [r].

Implementation:

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; 
            } 
  
            // 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 than 
            // 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

// Java  program to find triplets in a given 
// array whose sum is zero 
import java.util.Arrays;  
import java.io.*; 
  
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 than 
            // 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..     
} 

Python3

# 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 than 
            # 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#

// 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 than 
            // 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..  
} 

PHP

<?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 than 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 
?> 

Output :

-3 1 2
-1 0 1

Complexity Analysis:
- Time Complexity : O(n²).
  only two nested loops is required, so the time complexity is O(n²).
- Auxiliary Space : O(1), no extra space is required, so the time complexity is constant.

This article is contributed by DANISH_RAZA. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

Oct	NOV	Dec
	25
2019	2020	2021

Find all triplets with zero sum

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

C++

Java

Python3

C#

PHP

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