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# Find a triplet that sum to a given value

Given an array and a value, find if there is a triplet in array whose sum is equal to the given value. If there is such a triplet present in array, then print the triplet and return true. Else return false.

Examples:

Input: array = {12, 3, 4, 1, 6, 9}, sum = 24;
Output: 12, 3, 9
Explanation: There is a triplet (12, 3 and 9) present
in the array whose sum is 24.
Input: array = {1, 2, 3, 4, 5}, sum = 9
Output: 5, 3, 1
Explanation: There is a triplet (5, 3 and 1) present
in the array whose sum is 9.

Recommended Practice

Method 1: This is the naive approach towards solving the above problem.

• Approach: A simple method is to generate all possible triplets and compare the sum of every triplet with the given value. The following code implements this simple method using three nested loops.
• Algorithm:
1. Given an array of length n and a sum s
2. Create three nested loop first loop runs from start to end (loop counter i), second loop runs from i+1 to end (loop counter j) and third loop runs from j+1 to end (loop counter k)
3. The counter of these loops represents the index of 3 elements of the triplets.
4. Find the sum of ith, jth and kth element. If the sum is equal to given sum. Print the triplet and break.
5. If there is no triplet, then print that no triplet exist.

Implementation:

## C++

 `#include ``using` `namespace` `std;` `// returns true if there is triplet with sum equal``// to 'sum' present in A[]. Also, prints the triplet``bool` `find3Numbers(``int` `A[], ``int` `arr_size, ``int` `sum)``{``    ``// Fix the first element as A[i]``    ``for` `(``int` `i = 0; i < arr_size - 2; i++)``    ``{` `        ``// Fix the second element as A[j]``        ``for` `(``int` `j = i + 1; j < arr_size - 1; j++)``        ``{` `            ``// Now look for the third number``            ``for` `(``int` `k = j + 1; k < arr_size; k++)``            ``{``                ``if` `(A[i] + A[j] + A[k] == sum)``                ``{``                    ``cout << ``"Triplet is "` `<< A[i] <<``                        ``", "` `<< A[j] << ``", "` `<< A[k];``                    ``return` `true``;``                ``}``            ``}``        ``}``    ``}` `    ``// If we reach here, then no triplet was found``    ``return` `false``;``}` `/* Driver code */``int` `main()``{``    ``int` `A[] = { 1, 4, 45, 6, 10, 8 };``    ``int` `sum = 22;``    ``int` `arr_size = ``sizeof``(A) / ``sizeof``(A);``    ``find3Numbers(A, arr_size, sum);``    ``return` `0;``}` `// This is code is contributed by rathbhupendra`

## C

 `#include ` `// returns true if there is triplet with sum equal``// to 'sum' present in A[]. Also, prints the triplet``bool` `find3Numbers(``int` `A[], ``int` `arr_size, ``int` `sum)``{``    ``int` `l, r;` `    ``// Fix the first element as A[i]``    ``for` `(``int` `i = 0; i < arr_size - 2; i++) {` `        ``// Fix the second element as A[j]``        ``for` `(``int` `j = i + 1; j < arr_size - 1; j++) {` `            ``// Now look for the third number``            ``for` `(``int` `k = j + 1; k < arr_size; k++) {``                ``if` `(A[i] + A[j] + A[k] == sum) {``                    ``printf``(``"Triplet is %d, %d, %d"``,``                           ``A[i], A[j], A[k]);``                    ``return` `true``;``                ``}``            ``}``        ``}``    ``}` `    ``// If we reach here, then no triplet was found``    ``return` `false``;``}` `/* Driver program to test above function */``int` `main()``{``    ``int` `A[] = { 1, 4, 45, 6, 10, 8 };``    ``int` `sum = 22;``    ``int` `arr_size = ``sizeof``(A) / ``sizeof``(A);``    ``find3Numbers(A, arr_size, sum);``    ``return` `0;``}`

## Java

 `// Java program to find a triplet``class` `FindTriplet {` `    ``// returns true if there is triplet with sum equal``    ``// to 'sum' present in A[]. Also, prints the triplet``    ``boolean` `find3Numbers(``int` `A[], ``int` `arr_size, ``int` `sum)``    ``{``        ``int` `l, r;` `        ``// Fix the first element as A[i]``        ``for` `(``int` `i = ``0``; i < arr_size - ``2``; i++) {` `            ``// Fix the second element as A[j]``            ``for` `(``int` `j = i + ``1``; j < arr_size - ``1``; j++) {` `                ``// Now look for the third number``                ``for` `(``int` `k = j + ``1``; k < arr_size; k++) {``                    ``if` `(A[i] + A[j] + A[k] == sum) {``                        ``System.out.print(``"Triplet is "` `+ A[i] + ``", "` `+ A[j] + ``", "` `+ A[k]);``                        ``return` `true``;``                    ``}``                ``}``            ``}``        ``}` `        ``// If we reach here, then no triplet was found``        ``return` `false``;``    ``}` `    ``// Driver program to test above functions``    ``public` `static` `void` `main(String[] args)``    ``{``        ``FindTriplet triplet = ``new` `FindTriplet();``        ``int` `A[] = { ``1``, ``4``, ``45``, ``6``, ``10``, ``8` `};``        ``int` `sum = ``22``;``        ``int` `arr_size = A.length;` `        ``triplet.find3Numbers(A, arr_size, sum);``    ``}``}`

## Python3

 `# Python3 program to find a triplet``# that sum to a given value` `# returns true if there is triplet with``# sum equal to 'sum' present in A[].``# Also, prints the triplet``def` `find3Numbers(A, arr_size, ``sum``):` `    ``# Fix the first element as A[i]``    ``for` `i ``in` `range``( ``0``, arr_size``-``2``):` `        ``# Fix the second element as A[j]``        ``for` `j ``in` `range``(i ``+` `1``, arr_size``-``1``):``            ` `            ``# Now look for the third number``            ``for` `k ``in` `range``(j ``+` `1``, arr_size):``                ``if` `A[i] ``+` `A[j] ``+` `A[k] ``=``=` `sum``:``                    ``print``(``"Triplet is"``, A[i],``                          ``", "``, A[j], ``", "``, A[k])``                    ``return` `True``    ` `    ``# If we reach here, then no``    ``# triplet was found``    ``return` `False` `# Driver program to test above function``A ``=` `[``1``, ``4``, ``45``, ``6``, ``10``, ``8``]``sum` `=` `22``arr_size ``=` `len``(A)``find3Numbers(A, arr_size, ``sum``)` `# This code is contributed by Smitha Dinesh Semwal`

## C#

 `// C# program to find a triplet``// that sum to a given value``using` `System;` `class` `GFG {``    ``// returns true if there is``    ``// triplet with sum equal``    ``// to 'sum' present in A[].``    ``// Also, prints the triplet``    ``static` `bool` `find3Numbers(``int``[] A,``                             ``int` `arr_size,``                             ``int` `sum)``    ``{``        ``// Fix the first``        ``// element as A[i]``        ``for` `(``int` `i = 0;``             ``i < arr_size - 2; i++) {` `            ``// Fix the second``            ``// element as A[j]``            ``for` `(``int` `j = i + 1;``                 ``j < arr_size - 1; j++) {` `                ``// Now look for``                ``// the third number``                ``for` `(``int` `k = j + 1;``                     ``k < arr_size; k++) {``                    ``if` `(A[i] + A[j] + A[k] == sum) {``                        ``Console.WriteLine(``"Triplet is "` `+ A[i] + ``", "` `+ A[j] + ``", "` `+ A[k]);``                        ``return` `true``;``                    ``}``                ``}``            ``}``        ``}` `        ``// If we reach here,``        ``// then no triplet was found``        ``return` `false``;``    ``}` `    ``// Driver Code``    ``static` `public` `void` `Main()``    ``{``        ``int``[] A = { 1, 4, 45, 6, 10, 8 };``        ``int` `sum = 22;``        ``int` `arr_size = A.Length;` `        ``find3Numbers(A, arr_size, sum);``    ``}``}` `// This code is contributed by m_kit`

## PHP

 ``

## Javascript

 ``

Output

`Triplet is 4, 10, 8`
• Complexity Analysis:
• Time Complexity: O(n3).
There are three nested loops traversing the array, so the time complexity is O(n^3)
• Space Complexity: O(1).
As no extra space is required.

Method 2: using recursion

Implementation steps:

• Define a function that will take the input array, the target sum, and any other necessary parameters as input and return the triplet that adds up to the target sum or a message indicating that no such triplet exists.
• Implement the function to check all possible subsets of the array to find a triplet with a sum of the target value. This can be done using recursion.
• Call the function with the input array and target sum as arguments and print the result.

Implementations:

## C++

 `#include ``using` `namespace` `std;` `// The check function recursively checks all possible subsets of the array``//  to see if any of them sum to the target value``bool` `check(``int` `n,``int` `arr[],``int` `target, vector<``int``> triplet)``{``      ``// If the target value has been reached and the length of the triplet is 3,``    ``// then a triplet with a sum of target has been found and is printed``    ``if``(target==0 and triplet.size()==3){``        ``cout< triplet;``  ` `      ``//calling function``    ``if``(!check(n,arr,target,triplet)){``      ``cout<<``"does not exist"``<

## Java

 `import` `java.util.ArrayList;` `public` `class` `Main``{``  ` `    ``// The check function recursively checks all possible subsets of the array``    ``// to see if any of them sum to the target value``    ``public` `static` `boolean` `check(``int` `n, ``int``[] arr, ``int` `target, ArrayList triplet) {``        ` `      ``// If the target value has been reached and the length of the triplet is 3,``        ``// then a triplet with a sum of target has been found and is printed``        ``if` `(target == ``0` `&& triplet.size() == ``3``) {``            ``System.out.println(triplet.get(``0``) + ``" "` `+ triplet.get(``1``) + ``" "` `+ triplet.get(``2``));``            ``return` `true``;``        ``}` `        ``// If there are no more elements in the array or the target value has gone negative,``        ``// then a triplet with a sum of target cannot be found``        ``if` `(n == ``0` `|| target < ``0` `|| triplet.size() == ``3``)``            ``return` `false``;` `        ``triplet.add(arr[n - ``1``]);` `        ``// Recursively check if target can be reached by including the current element in the triplet``        ``// or by excluding the current element and checking the remaining elements``        ``boolean` `a = check(n - ``1``, arr, target - arr[n - ``1``], triplet);` `        ``triplet.remove(triplet.size() - ``1``);` `        ``boolean` `b = check(n - ``1``, arr, target, triplet);` `        ``return` `a | b;``    ``}` `    ``public` `static` `void` `main(String[] args) {``        ``// array length``        ``int` `n = ``6``;` `        ``// array``        ``int``[] arr = {``1``, ``4``, ``45``, ``6``, ``10``, ``8``};` `        ``// target value``        ``int` `target = ``22``;` `        ``// taking empty array for triplet``        ``ArrayList triplet = ``new` `ArrayList<>();` `        ``// calling function``        ``if` `(!check(n, arr, target, triplet)) {``            ``System.out.println(``"does not exist"``);``        ``}``    ``}``}`

## Python

 `# python program for above approach` `array ``=` `[``1``, ``4``, ``45``, ``6``, ``10``, ``8``]``target ``=` `22``n ``=` `len``(array)` `# The check function recursively checks all possible subsets of the array``# to see if any of them sum to the target value``def` `check(array, n, target, triplet):``    ``# If the target value has been reached and the length of the triplet is 3,``    ``# then a triplet with a sum of target has been found and is printed``    ``if` `target ``=``=` `0` `and` `len``(triplet) ``=``=` `3``:``        ``print``(triplet)``        ``return` `True``    ` `    ``# If there are no more elements in the array or the target value has gone negative,``    ``# then a triplet with a sum of target cannot be found``    ``if` `n ``=``=` `0` `or` `target < ``0` `or` `len``(triplet) ``=``=` `3``:``        ``return` `False``    ` `    ``# Recursively check if target can be reached by including the current element in the triplet``    ``# or by excluding the current element and checking the remaining elements``    ``return` `(check(array, n``-``1``, target``-``array[n``-``1``], triplet``+``[array[n``-``1``]])``            ``or` `check(array, n``-``1``, target, triplet))` `# Call the check function with an empty triplet to start checking all possible subsets of the array``if` `not` `check(array, n, target, []):``    ``print``(``'does not exist'``)``    ` `# this code is contributed by bhardwajji`

## C#

 `using` `System;``using` `System.Collections.Generic;` `class` `Program {``  ``static` `bool` `Check(``int` `n, ``int``[] arr, ``int` `target,``                    ``List<``int``> triplet)``  ``{` `    ``// If the target value has been reached and the``    ``// length of the triplet is 3, then a triplet with a``    ``// sum of target has been found and is printed``    ``if` `(target == 0 && triplet.Count == 3) {``      ``Console.WriteLine(triplet + ``" "` `+ triplet``                        ``+ ``" "` `+ triplet);``      ``return` `true``;``    ``}` `    ``// If there are no more elements in the array or the``    ``// target value has gone negative, then a triplet``    ``// with a sum of target cannot be found``    ``if` `(n == 0 || target < 0 || triplet.Count == 3)``      ``return` `false``;` `    ``triplet.Add(arr[n - 1]);` `    ``// Recursively check if target can be reached by``    ``// including the current element in the triplet or``    ``// by excluding the current element and checking the``    ``// remaining elements``    ``bool` `a = Check(n - 1, arr, target - arr[n - 1],``                   ``triplet);` `    ``triplet.RemoveAt(triplet.Count - 1);` `    ``bool` `b = Check(n - 1, arr, target, triplet);` `    ``return` `a || b;``  ``}` `  ``static` `void` `Main(``string``[] args)``  ``{``    ``// array length``    ``int` `n = 6;` `    ``// array``    ``int``[] arr = { 1, 4, 45, 6, 10, 8 };` `    ``// target value``    ``int` `target = 22;` `    ``// taking empty list for triplet``    ``List<``int``> triplet = ``new` `List<``int``>();` `    ``// calling function``    ``if` `(!Check(n, arr, target, triplet)) {``      ``Console.WriteLine(``"does not exist"``);``    ``}` `    ``Console.ReadLine();``  ``}``}` `// This code is contributed by user_dtewbxkn77n`

## Javascript

 `let array = [1, 4, 45, 6, 10, 8];``let target = 22;``let n = array.length;` `// The check function recursively checks all possible subsets of the array``// to see if any of them sum to the target value``function` `check(array, n, target, triplet) {``    ``// If the target value has been reached and the length of the triplet is 3,``    ``// then a triplet with a sum of target has been found and is printed``    ``if` `(target == 0 && triplet.length == 3) {``        ``console.log(triplet);``        ``return` `true``;``    ``}``    ` `    ``// If there are no more elements in the array or the target value has gone negative,``    ``// then a triplet with a sum of target cannot be found``    ``if` `(n == 0 || target < 0 || triplet.length == 3) {``        ``return` `false``;``    ``}``    ` `    ``// Recursively check if target can be reached by including the current element in the triplet``    ``// or by excluding the current element and checking the remaining elements``    ``return` `(check(array, n-1, target-array[n-1], [...triplet,array[n-1]])``            ``|| check(array,n-1,target,[...triplet]));``}` `// Call the check function with an empty triplet to start checking all possible subsets of the array``if` `(!check(array,n,target,[])) {``    ``console.log(``'does not exist'``);``}`

Output

`[8, 10, 4]`

Time complexity: O(2^N), because this solution made recursive tree.
Auxiliary Space: O(1)

Method 3: This method uses sorting to increase the efficiency of the code.

• Approach: By Sorting the array the efficiency of the algorithm can be improved. This efficient approach uses the two-pointer technique. Traverse the array and fix the first element of the triplet. Now use the Two Pointers algorithm to find if there is a pair whose sum is equal to x – array[i]. Two pointers algorithm take linear time so it is better than a nested loop.
• Algorithm :
1. Sort the given array.
2. Loop over the array and fix the first element of the possible triplet, arr[i].
3. Then fix two pointers, one at i + 1 and the other at n – 1. And look at the sum,
1. If the sum is smaller than the required sum, increment the first pointer.
2. Else, If the sum is bigger, Decrease the end pointer to reduce the sum.
3. Else, if the sum of elements at two-pointer is equal to given sum then print the triplet and break.
• Implementation:

## C++

 `// C++ program to find a triplet``#include ``using` `namespace` `std;` `// returns true if there is triplet with sum equal``// to 'sum' present in A[]. Also, prints the triplet``bool` `find3Numbers(``int` `A[], ``int` `arr_size, ``int` `sum)``{``    ``int` `l, r;``    ``/* Sort the elements */``    ``sort(A, A + arr_size);``    ``/* Now fix the first element one by one and find the``       ``other two elements */``    ``for` `(``int` `i = 0; i < arr_size - 2; i++) {` `        ``// To find the other two elements, start two index``        ``// variables from two corners of the array and move``        ``// them toward each other``        ``l = i + 1; ``// index of the first element in the``        ``// remaining elements``        ``r = arr_size - 1; ``// index of the last element``        ``while` `(l < r) {``            ``if` `(A[i] + A[l] + A[r] == sum) {``                ``printf``(``"Triplet is %d, %d, %d"``, A[i], A[l],A[r]);``                ``return` `true``;``            ``}``            ``else` `if` `(A[i] + A[l] + A[r] < sum)``                ``l++;``            ``else` `// A[i] + A[l] + A[r] > sum``                ``r--;``        ``}``    ``}``    ``// If we reach here, then no triplet was found``    ``return` `false``;``}` `/* Driver program to test above function */``int` `main()``{``    ``int` `A[] = { 1, 4, 45, 6, 10, 8 };``    ``int` `sum = 22;``    ``int` `arr_size = ``sizeof``(A) / ``sizeof``(A);``    ``find3Numbers(A, arr_size, sum);``    ``return` `0;``}` `// This code is contributed by Aditya Kumar (adityakumar129)`

## C

 `// C program to find a triplet``#include ``#include ``#include ` `int` `cmpfunc(``const` `void``* a, ``const` `void``* b)``{``    ``return` `(*(``int``*)a - *(``int``*)b);``}` `// returns true if there is triplet with sum equal``// to 'sum' present in A[]. Also, prints the triplet``bool` `find3Numbers(``int` `A[], ``int` `arr_size, ``int` `sum)``{``    ``int` `l, r;``  ` `    ``/* Sort the elements */``    ``qsort``(A, arr_size, ``sizeof``(``int``), cmpfunc);``  ` `    ``/* Now fix the first element one by one and find the``       ``other two elements */``    ``for` `(``int` `i = 0; i < arr_size - 2; i++)``    ``{``      ` `        ``// To find the other two elements, start two index``        ``// variables from two corners of the array and move``        ``// them toward each other``        ``l = i + 1; ``// index of the first element in the``        ``// remaining elements``        ``r = arr_size - 1; ``// index of the last element``        ``while` `(l < r) {``            ``if` `(A[i] + A[l] + A[r] == sum) {``                ``printf``(``"Triplet is %d, %d, %d"``, A[i], A[l],``                       ``A[r]);``                ``return` `true``;``            ``}``            ``else` `if` `(A[i] + A[l] + A[r] < sum)``                ``l++;``            ``else` `// A[i] + A[l] + A[r] > sum``                ``r--;``        ``}``    ``}``  ` `    ``// If we reach here, then no triplet was found``    ``return` `false``;``}` `/* Driver program to test above function */``int` `main()``{``    ``int` `A[] = { 1, 4, 45, 6, 10, 8 };``    ``int` `sum = 22;``    ``int` `arr_size = ``sizeof``(A) / ``sizeof``(A);``    ``find3Numbers(A, arr_size, sum);``    ``return` `0;``}` `// This code is contributed by Aditya Kumar (adityakumar129)`

## Java

 `// Java program to find a triplet``class` `FindTriplet {` `    ``// returns true if there is triplet with sum equal``    ``// to 'sum' present in A[]. Also, prints the triplet``    ``boolean` `find3Numbers(``int` `A[], ``int` `arr_size, ``int` `sum)``    ``{``        ``int` `l, r;` `        ``/* Sort the elements */``        ``quickSort(A, ``0``, arr_size - ``1``);` `        ``/* Now fix the first element one by one and find the``           ``other two elements */``        ``for` `(``int` `i = ``0``; i < arr_size - ``2``; i++) {` `            ``// To find the other two elements, start two``            ``// index variables from two corners of the array``            ``// and move them toward each other``            ``l = i + ``1``; ``// index of the first element in the``                       ``// remaining elements``            ``r = arr_size - ``1``; ``// index of the last element``            ``while` `(l < r) {``                ``if` `(A[i] + A[l] + A[r] == sum) {``                    ``System.out.print(``"Triplet is "` `+ A[i] + ``", "` `+ A[l] + ``", "` `+ A[r]);``                    ``return` `true``;``                ``}``                ``else` `if` `(A[i] + A[l] + A[r] < sum)``                    ``l++;` `                ``else` `// A[i] + A[l] + A[r] > sum``                    ``r--;``            ``}``        ``}` `        ``// If we reach here, then no triplet was found``        ``return` `false``;``    ``}` `    ``int` `partition(``int` `A[], ``int` `si, ``int` `ei)``    ``{``        ``int` `x = A[ei];``        ``int` `i = (si - ``1``);``        ``int` `j;` `        ``for` `(j = si; j <= ei - ``1``; j++) {``            ``if` `(A[j] <= x) {``                ``i++;``                ``int` `temp = A[i];``                ``A[i] = A[j];``                ``A[j] = temp;``            ``}``        ``}``        ``int` `temp = A[i + ``1``];``        ``A[i + ``1``] = A[ei];``        ``A[ei] = temp;``        ``return` `(i + ``1``);``    ``}` `    ``/* Implementation of Quick Sort``    ``A[] --> Array to be sorted``    ``si  --> Starting index``    ``ei  --> Ending index``     ``*/``    ``void` `quickSort(``int` `A[], ``int` `si, ``int` `ei)``    ``{``        ``int` `pi;` `        ``/* Partitioning index */``        ``if` `(si < ei) {``            ``pi = partition(A, si, ei);``            ``quickSort(A, si, pi - ``1``);``            ``quickSort(A, pi + ``1``, ei);``        ``}``    ``}` `    ``// Driver program to test above functions``    ``public` `static` `void` `main(String[] args)``    ``{``        ``FindTriplet triplet = ``new` `FindTriplet();``        ``int` `A[] = { ``1``, ``4``, ``45``, ``6``, ``10``, ``8` `};``        ``int` `sum = ``22``;``        ``int` `arr_size = A.length;` `        ``triplet.find3Numbers(A, arr_size, sum);``    ``}``}`

## Python3

 `# Python3 program to find a triplet` `# returns true if there is triplet``# with sum equal to 'sum' present``# in A[]. Also, prints the triplet``def` `find3Numbers(A, arr_size, ``sum``):` `    ``# Sort the elements``    ``A.sort()` `    ``# Now fix the first element``    ``# one by one and find the``    ``# other two elements``    ``for` `i ``in` `range``(``0``, arr_size``-``2``):``    `  `        ``# To find the other two elements,``        ``# start two index variables from``        ``# two corners of the array and``        ``# move them toward each other``        ` `        ``# index of the first element``        ``# in the remaining elements``        ``l ``=` `i ``+` `1``        ` `        ``# index of the last element``        ``r ``=` `arr_size``-``1``        ``while` `(l < r):``        ` `            ``if``( A[i] ``+` `A[l] ``+` `A[r] ``=``=` `sum``):``                ``print``(``"Triplet is"``, A[i],``                     ``', '``, A[l], ``', '``, A[r]);``                ``return` `True``            ` `            ``elif` `(A[i] ``+` `A[l] ``+` `A[r] < ``sum``):``                ``l ``+``=` `1``            ``else``: ``# A[i] + A[l] + A[r] > sum``                ``r ``-``=` `1` `    ``# If we reach here, then``    ``# no triplet was found``    ``return` `False` `# Driver program to test above function``A ``=` `[``1``, ``4``, ``45``, ``6``, ``10``, ``8``]``sum` `=` `22``arr_size ``=` `len``(A)` `find3Numbers(A, arr_size, ``sum``)` `# This is contributed by Smitha Dinesh Semwal`

## C#

 `// C# program to find a triplet``using` `System;` `class` `GFG {` `    ``// returns true if there is triplet``    ``// with sum equal to 'sum' present``    ``// in A[]. Also, prints the triplet``    ``bool` `find3Numbers(``int``[] A, ``int` `arr_size,``                      ``int` `sum)``    ``{``        ``int` `l, r;` `        ``/* Sort the elements */``        ``quickSort(A, 0, arr_size - 1);` `        ``/* Now fix the first element``    ``one by one and find the``    ``other two elements */``        ``for` `(``int` `i = 0; i < arr_size - 2; i++) {` `            ``// To find the other two elements,``            ``// start two index variables from``            ``// two corners of the array and``            ``// move them toward each other``            ``l = i + 1; ``// index of the first element``            ``// in the remaining elements``            ``r = arr_size - 1; ``// index of the last element``            ``while` `(l < r) {``                ``if` `(A[i] + A[l] + A[r] == sum) {``                    ``Console.Write(``"Triplet is "` `+ A[i] + ``", "` `+ A[l] + ``", "` `+ A[r]);``                    ``return` `true``;``                ``}``                ``else` `if` `(A[i] + A[l] + A[r] < sum)``                    ``l++;` `                ``else` `// A[i] + A[l] + A[r] > sum``                    ``r--;``            ``}``        ``}` `        ``// If we reach here, then``        ``// no triplet was found``        ``return` `false``;``    ``}` `    ``int` `partition(``int``[] A, ``int` `si, ``int` `ei)``    ``{``        ``int` `x = A[ei];``        ``int` `i = (si - 1);``        ``int` `j;` `        ``for` `(j = si; j <= ei - 1; j++) {``            ``if` `(A[j] <= x) {``                ``i++;``                ``int` `temp = A[i];``                ``A[i] = A[j];``                ``A[j] = temp;``            ``}``        ``}``        ``int` `temp1 = A[i + 1];``        ``A[i + 1] = A[ei];``        ``A[ei] = temp1;``        ``return` `(i + 1);``    ``}` `    ``/* Implementation of Quick Sort``A[] --> Array to be sorted``si --> Starting index``ei --> Ending index``*/``    ``void` `quickSort(``int``[] A, ``int` `si, ``int` `ei)``    ``{``        ``int` `pi;` `        ``/* Partitioning index */``        ``if` `(si < ei) {``            ``pi = partition(A, si, ei);``            ``quickSort(A, si, pi - 1);``            ``quickSort(A, pi + 1, ei);``        ``}``    ``}` `    ``// Driver Code``    ``static` `void` `Main()``    ``{``        ``GFG triplet = ``new` `GFG();``        ``int``[] A = ``new` `int``[] { 1, 4, 45, 6, 10, 8 };``        ``int` `sum = 22;``        ``int` `arr_size = A.Length;` `        ``triplet.find3Numbers(A, arr_size, sum);``    ``}``}` `// This code is contributed by mits`

## PHP

 ` sum``                ``\$r``--;``        ``}``    ``}` `    ``// If we reach here, then``    ``// no triplet was found``    ``return` `false;``}` `// Driver Code``\$A` `= ``array` `(1, 4, 45, 6, 10, 8);``\$sum` `= 22;``\$arr_size` `= sizeof(``\$A``);` `find3Numbers(``\$A``, ``\$arr_size``, ``\$sum``);` `// This code is contributed by ajit``?>`

## Javascript

 ``

Output

`Triplet is 4, 8, 10`
• Complexity Analysis:
• Time complexity: O(N^2).
There are only two nested loops traversing the array, so time complexity is O(n^2). Two pointers algorithm takes O(n) time and the first element can be fixed using another nested traversal.
• Space Complexity: O(1).
As no extra space is required.

Method 4: This is a Hashing-based solution.

• Approach: This approach uses extra space but is simpler than the two-pointers approach. Run two loops outer loop from start to end and inner loop from i+1 to end. Create a hashmap or set to store the elements in between i+1 to n-1. So if the given sum is x, check if there is a number in the set which is equal to x – arr[i] – arr[j]. If yes print the triplet.

• Algorithm:
1. Traverse the array from start to end. (loop counter i)
2. Create a HashMap or set to store unique pairs.
3. Run another loop from i+1 to end of the array. (loop counter j)
4. If there is an element in the set which is equal to x- arr[i] – arr[j], then print the triplet (arr[i], arr[j], x-arr[i]-arr[j]) and break
5. Insert the jth element in the set.
• Implementation:

## C++

 `// C++ program to find a triplet using Hashing``#include ``using` `namespace` `std;` `// returns true if there is triplet with sum equal``// to 'sum' present in A[]. Also, prints the triplet``bool` `find3Numbers(``int` `A[], ``int` `arr_size, ``int` `sum)``{``    ``// Fix the first element as A[i]``    ``for` `(``int` `i = 0; i < arr_size - 2; i++)``    ``{` `        ``// Find pair in subarray A[i+1..n-1]``        ``// with sum equal to sum - A[i]``        ``unordered_set<``int``> s;``        ``int` `curr_sum = sum - A[i];``        ``for` `(``int` `j = i + 1; j < arr_size; j++)``        ``{``            ``if` `(s.find(curr_sum - A[j]) != s.end())``            ``{``                ``printf``(``"Triplet is %d, %d, %d"``, A[i],``                       ``A[j], curr_sum - A[j]);``                ``return` `true``;``            ``}``            ``s.insert(A[j]);``        ``}``    ``}` `    ``// If we reach here, then no triplet was found``    ``return` `false``;``}` `/* Driver program to test above function */``int` `main()``{``    ``int` `A[] = { 1, 4, 45, 6, 10, 8 };``    ``int` `sum = 22;``    ``int` `arr_size = ``sizeof``(A) / ``sizeof``(A);` `    ``find3Numbers(A, arr_size, sum);` `    ``return` `0;``}`

## Java

 `// Java program to find a triplet using Hashing``import` `java.util.*;` `class` `GFG {` `    ``// returns true if there is triplet``    ``// with sum equal to 'sum' present``    ``// in A[]. Also, prints the triplet``    ``static` `boolean` `find3Numbers(``int` `A[],``                                ``int` `arr_size, ``int` `sum)``    ``{``        ``// Fix the first element as A[i]``        ``for` `(``int` `i = ``0``; i < arr_size - ``2``; i++) {` `            ``// Find pair in subarray A[i+1..n-1]``            ``// with sum equal to sum - A[i]``            ``HashSet s = ``new` `HashSet();``            ``int` `curr_sum = sum - A[i];``            ``for` `(``int` `j = i + ``1``; j < arr_size; j++)``            ``{``                ``if` `(s.contains(curr_sum - A[j]))``                ``{``                    ``System.out.print(``"Triplet is "` `+ A[i] + ``", "` `+ A[j] + ``", "` `+ (curr_sum - A[j]));``                    ``return` `true``;``                ``}``                ``s.add(A[j]);``            ``}``        ``}` `        ``// If we reach here, then no triplet was found``        ``return` `false``;``    ``}` `    ``/* Driver code */``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `A[] = { ``1``, ``4``, ``45``, ``6``, ``10``, ``8` `};``        ``int` `sum = ``22``;``        ``int` `arr_size = A.length;` `        ``find3Numbers(A, arr_size, sum);``    ``}``}` `// This code has been contributed by 29AjayKumar``// This code is improved by Susobhan Akhuli`

## Python3

 `# Python3 program to find a triplet using Hashing``# returns true if there is triplet with sum equal``# to 'sum' present in A[]. Also, prints the triplet``def` `find3Numbers(A, arr_size, ``sum``):``    ``for` `i ``in` `range``(``0``, arr_size``-``1``):``        ``# Find pair in subarray A[i + 1..n-1]``        ``# with sum equal to sum - A[i]``        ``s ``=` `set``()``        ``curr_sum ``=` `sum` `-` `A[i]``        ``for` `j ``in` `range``(i ``+` `1``, arr_size):``            ``if` `(curr_sum ``-` `A[j]) ``in` `s:``                ``print``(``"Triplet is"``, A[i],``                        ``", "``, A[j], ``", "``, curr_sum``-``A[j])``                ``return` `True``            ``s.add(A[j])``    ` `    ``return` `False` `# Driver program to test above function``A ``=` `[``1``, ``4``, ``45``, ``6``, ``10``, ``8``]``sum` `=` `22``arr_size ``=` `len``(A)``find3Numbers(A, arr_size, ``sum``)` `# This is contributed by Yatin gupta`

## C#

 `// C# program to find a triplet using Hashing``using` `System;``using` `System.Collections.Generic;``public` `class` `GFG {` `    ``// returns true if there is triplet``    ``// with sum equal to 'sum' present``    ``// in A[]. Also, prints the triplet``    ``static` `bool` `find3Numbers(``int``[] A,``                             ``int` `arr_size, ``int` `sum)``    ``{``        ``// Fix the first element as A[i]``        ``for` `(``int` `i = 0; i < arr_size - 2; i++) {` `            ``// Find pair in subarray A[i+1..n-1]``            ``// with sum equal to sum - A[i]``            ``HashSet<``int``> s = ``new` `HashSet<``int``>();``            ``int` `curr_sum = sum - A[i];``            ``for` `(``int` `j = i + 1; j < arr_size; j++)``            ``{``                ``if` `(s.Contains(curr_sum - A[j]))``                ``{``                    ``Console.Write(``"Triplet is {0}, {1}, {2}"``, A[i],``                                  ``A[j], curr_sum - A[j]);``                    ``return` `true``;``                ``}``                ``s.Add(A[j]);``            ``}``        ``}` `        ``// If we reach here, then no triplet was found``        ``return` `false``;``    ``}` `    ``/* Driver code */``    ``public` `static` `void` `Main()``    ``{``        ``int``[] A = { 1, 4, 45, 6, 10, 8 };``        ``int` `sum = 22;``        ``int` `arr_size = A.Length;` `        ``find3Numbers(A, arr_size, sum);``    ``}``}` `/* This code contributed by PrinciRaj1992 */`

## Javascript

 ``

Output

`Triplet is 4, 8, 10`

Time complexity: O(N^2)
Auxiliary Space: O(N), since n extra space has been taken

You can watch the explanation of the problem on YouTube discussed By Geeks For Geeks Team.

You can also refer this video present on Youtube.
How to print all triplets with given sum?
Refer Find all triplets with zero sum