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# Find a triplet in an array whose sum is closest to a given number

Given an array arr[] of N integers and an integer X, the task is to find three integers in arr[] such that the sum is closest to X.

Examples:

```Input: arr[] = {-1, 2, 1, -4}, X = 1
Output: 2
Explanation:
Sums of triplets:
(-1) + 2 + 1 = 2
(-1) + 2 + (-4) = -3
2 + 1 + (-4) = -1
2 is closest to 1.

Input: arr[] = {1, 2, 3, 4, -5}, X = 10
Output: 9
Explanation:
Sums of triplets:
1 + 2 + 3 = 6
2 + 3 + 4 = 9
1 + 3 + 4 = 7
...
9 is closest to 10.```
Recommended Practice

Naive Approach: Using Recursion

Here we will generate all subsets of the given array then we will choose that subset whose size is 3 means that is forming a triplet. In that, we will choose that triplet’s sum whose sum is closest to X.

Code-

## C++

 `#include ``using` `namespace` `std;` `// Naive recursive function``void` `findcloseTriplet(``int` `arr[], ``int` `n, ``int` `x, ``int` `count,``                      ``int` `sum, ``int` `ind, ``int``& ans, ``int``& minm)``{` `    ``// Return on reaching to the end of array``    ``// Here if we picked three element till now then check``    ``// that sum is closest to our "x" or not``    ``if` `(ind == n) {``        ``if` `(count == 3) {``            ``if` `(``abs``(x - sum) < minm) {``                ``minm = ``abs``(x - sum);``                ``ans = sum;``            ``}``        ``}``        ``return``;``    ``}` `    ``// Pick this number``    ``findcloseTriplet(arr, n, x, count + 1, sum + arr[ind],``                     ``ind + 1, ans, minm);` `    ``// Don't pick this number``    ``findcloseTriplet(arr, n, x, count, sum, ind + 1, ans,``                     ``minm);``}` `int` `main()``{``    ``int` `arr[] = { -1, 2, 1, -4 };``    ``int` `x = 1;``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `minm = INT_MAX;``    ``int` `ans;``    ``findcloseTriplet(arr, n, x, 0, 0, 0, ans, minm);``    ``cout << ans << endl;` `    ``return` `0;``}`

## Java

 `import` `java.util.*;` `public` `class` `Main {``    ``// Naive recursive function``    ``public` `static` `void` `findCloseTriplet(``int``[] arr, ``int` `n, ``int` `x, ``int` `count, ``int` `sum, ``int` `ind, ``int``[] ans, ``int``[] minm) {``        ``// Return on reaching to the end of array``        ``// Here if we picked three element till now then check``        ``// that sum is closest to our "x" or not``        ``if` `(ind == n) {``            ``if` `(count == ``3``) {``                ``if` `(Math.abs(x - sum) < minm[``0``]) {``                    ``minm[``0``] = Math.abs(x - sum);``                    ``ans[``0``] = sum;``                ``}``            ``}``            ``return``;``        ``}` `        ``// Pick this number``        ``findCloseTriplet(arr, n, x, count + ``1``, sum + arr[ind], ind + ``1``, ans, minm);` `        ``// Don't pick this number``        ``findCloseTriplet(arr, n, x, count, sum, ind + ``1``, ans, minm);``    ``}` `    ``public` `static` `void` `main(String[] args) {``        ``int``[] arr = { -``1``, ``2``, ``1``, -``4` `};``        ``int` `x = ``1``;``        ``int` `n = arr.length;``        ``int``[] minm = { Integer.MAX_VALUE };``        ``int``[] ans = ``new` `int``[``1``];``        ``findCloseTriplet(arr, n, x, ``0``, ``0``, ``0``, ans, minm);``        ``System.out.println(ans[``0``]);``    ``}``}`

## Python3

 `# Python3 code for the approach` `# Naive recursive function``def` `find_close_triplet(arr, n, x, count, ``sum``, ind, ans, minm):``    ``# Return on reaching to the end of array``    ``# Here if we picked three element till now then check``    ``# that sum is closest to our "x" or not``    ``if` `ind ``=``=` `n:``        ``if` `count ``=``=` `3``:``            ``if` `abs``(x ``-` `sum``) < minm[``0``]:``                ``minm[``0``] ``=` `abs``(x ``-` `sum``)``                ``ans[``0``] ``=` `sum``        ``return` `    ``# Pick this number``    ``find_close_triplet(arr, n, x, count ``+` `1``, ``sum` `+` `arr[ind], ind ``+` `1``, ans, minm)` `    ``# Don't pick this number``    ``find_close_triplet(arr, n, x, count, ``sum``, ind ``+` `1``, ans, minm)` `# Driver's code``if` `__name__ ``=``=` `"__main__"``:``  ``#Input array``  ``arr ``=` `[``-``1``, ``2``, ``1``, ``-``4``]``  ``x ``=` `1``  ``n ``=` `len``(arr)``  ``minm ``=` `[``float``(``'inf'``)]``  ``ans ``=` `[``0``]``  ` `  ``# Function Call``  ``find_close_triplet(arr, n, x, ``0``, ``0``, ``0``, ans, minm)``  ``print``(ans[``0``])`

## C#

 `using` `System;` `public` `class` `Program``{``// Naive recursive function``public` `static` `void` `FindCloseTriplet(``int``[] arr, ``int` `n, ``int` `x, ``int` `count, ``int` `sum, ``int` `ind, ``ref` `int` `ans, ``ref` `int` `minm)``{``// Return on reaching to the end of array``// Here if we picked three element till now then check``// that sum is closest to our "x" or not``if` `(ind == n)``{``if` `(count == 3)``{``if` `(Math.Abs(x - sum) < minm)``{``minm = Math.Abs(x - sum);``ans = sum;``}``}``return``;``}``      ``// Pick this number``    ``FindCloseTriplet(arr, n, x, count + 1, sum + arr[ind], ind + 1, ``ref` `ans, ``ref` `minm);` `    ``// Don't pick this number``    ``FindCloseTriplet(arr, n, x, count, sum, ind + 1, ``ref` `ans, ``ref` `minm);``}` `public` `static` `void` `Main()``{``    ``int``[] arr = ``new` `int``[] { -1, 2, 1, -4 };``    ``int` `x = 1;``    ``int` `n = arr.Length;``    ``int` `minm = ``int``.MaxValue;``    ``int` `ans = 0;``    ``FindCloseTriplet(arr, n, x, 0, 0, 0, ``ref` `ans, ``ref` `minm);``    ``Console.WriteLine(ans);``}``}`

## Javascript

 `// Naive recursive function``function` `findcloseTriplet(arr, n, x, count, sum, ind, ans, minm) {``    ``// Return on reaching to the end of array``    ``// Here if we picked three element till now then check``    ``// that sum is closest to our "x" or not``    ``if` `(ind == n) {``        ``if` `(count == 3) {``            ``if` `(Math.abs(x - sum) < minm[0]) {``                ``minm[0] = Math.abs(x - sum);``                ``ans[0] = sum;``            ``}``        ``}``        ``return``;``    ``}` `    ``// Pick this number``    ``findcloseTriplet(arr, n, x, count + 1, sum + arr[ind], ind + 1, ans, minm);` `    ``// Don't pick this number``    ``findcloseTriplet(arr, n, x, count, sum, ind + 1, ans, minm);``}` `let arr = [-1, 2, 1, -4];``let x = 1;``let n = arr.length;``let minm = [Number.MAX_VALUE];``let ans = [0];``findcloseTriplet(arr, n, x, 0, 0, 0, ans, minm);``console.log(ans[0]);`

Output-

`2`

Complexity Analysis:

• Time complexity: O(2n),because of Recursion
• Auxiliary Space: O(n),Recursion stack space

Simple Approach: The naive approach is to explore all the subsets of size 3 and keep a track of the difference between X and the sum of this subset. Then return the subset whose difference between its sum and X is minimum.

Algorithm:

1. Create three nested loops with counter i, j and k respectively.
2. The first loop will start from start to end, the second loop will run from i+1 to end, the third loop will run from j+1 to end.
3. Check if the difference of the sum of the ith, jth and kth element with the given sum is less than the current minimum or not. Update the current minimum
4. Print the closest sum.

Implementation:

## C++14

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the sum of a``// triplet which is closest to x``int` `solution(vector<``int``>& arr, ``int` `x)``{``    ``// To store the closest sum``    ``int` `closestSum = INT_MAX;` `    ``// Run three nested loops each loop``    ``// for each element of triplet``    ``for` `(``int` `i = 0; i < arr.size() ; i++)``    ``{``        ``for``(``int` `j =i + 1; j < arr.size(); j++)``        ``{``            ``for``(``int` `k =j + 1; k < arr.size(); k++)``            ``{``                ``//update the closestSum``                ``if``(``abs``(x - closestSum) > ``abs``(x - (arr[i] + arr[j] + arr[k])))``                    ``closestSum = (arr[i] + arr[j] + arr[k]);``            ``}``        ``}``    ``}``    ``// Return the closest sum found``    ``return` `closestSum;``}` `// Driver code``int` `main()``{``    ``vector<``int``> arr = { -1, 2, 1, -4 };``    ``int` `x = 1;``    ``cout << solution(arr, x);` `    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``class` `GFG{``    ` `// Function to return the sum of a``// triplet which is closest to x``public` `static` `int` `solution(``int` `arr[], ``int` `x)``{``    ` `    ``// To store the closest sum``    ``int` `closestSum = Integer.MAX_VALUE;``  ` `    ``// Run three nested loops each loop ``    ``// for each element of triplet``    ``for``(``int` `i = ``0``; i < arr.length ; i++) ``    ``{``        ``for``(``int` `j = i + ``1``; j < arr.length; j++)``        ``{``            ``for``(``int` `k = j + ``1``; k < arr.length; k++)``            ``{``                ` `                ``// Update the closestSum``                ``if` `(Math.abs(x - closestSum) >``                    ``Math.abs(x - (arr[i] + arr[j] + arr[k])))``                    ``closestSum = (arr[i] + arr[j] + arr[k]);``            ``} ``        ``}``    ``}``    ` `    ``// Return the closest sum found``    ``return` `closestSum;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { -``1``, ``2``, ``1``, -``4` `};``    ``int` `x = ``1``;``    ` `    ``System.out.print(solution(arr, x));``}``}` `// This code is contributed by divyeshrabadiya07`

## Python3

 `# Python3 implementation of the above approach``import` `sys` `# Function to return the sum of a``# triplet which is closest to x``def` `solution(arr, x):` `    ``# To store the closest sum``    ``closestSum ``=` `sys.maxsize` `    ``# Run three nested loops each loop``    ``# for each element of triplet``    ``for` `i ``in` `range` `(``len``(arr)) :``        ``for` `j ``in` `range``(i ``+` `1``, ``len``(arr)):``            ``for` `k ``in` `range``(j ``+` `1``, ``len``( arr)):``            ` `                ``# Update the closestSum``                ``if``(``abs``(x ``-` `closestSum) >``                ``abs``(x ``-` `(arr[i] ``+``                ``arr[j] ``+` `arr[k]))):``                    ``closestSum ``=` `(arr[i] ``+``                                    ``arr[j] ``+` `arr[k])``            ` `    ``# Return the closest sum found``    ``return` `closestSum` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``arr ``=` `[ ``-``1``, ``2``, ``1``, ``-``4` `]``    ``x ``=` `1``    ` `    ``print``(solution(arr, x))` `# This code is contributed by chitranayal`

## C#

 `// C# implementation of the above approach``using` `System;``using` `System.Collections;``using` `System.Collections.Generic;` `class` `GFG{` `// Function to return the sum of a``// triplet which is closest to x``static` `int` `solution(ArrayList arr, ``int` `x)``{``    ` `    ``// To store the closest sum``    ``int` `closestSum = ``int``.MaxValue;` `    ``// Run three nested loops each loop``    ``// for each element of triplet``    ``for``(``int` `i = 0; i < arr.Count; i++)``    ``{``        ``for``(``int` `j = i + 1; j < arr.Count; j++)``        ``{``            ``for``(``int` `k = j + 1; k < arr.Count; k++)``            ``{``                ``if` `(Math.Abs(x - closestSum) >``                    ``Math.Abs(x - ((``int``)arr[i] +``                   ``(``int``)arr[j] + (``int``)arr[k])))``                ``{``                    ``closestSum = ((``int``)arr[i] +``                                  ``(``int``)arr[j] +``                                  ``(``int``)arr[k]);``                ``}``            ``}``        ``}``    ``}``    ` `    ``// Return the closest sum found``    ``return` `closestSum;``}` `// Driver code``public` `static` `void` `Main(``string``[] args)``{``    ``ArrayList arr = ``new` `ArrayList(){ -1, 2, 1, -4 };``    ``int` `x = 1;``    ``Console.Write(solution(arr, x));``}``}` `// This code is contributed by rutvik_56`

## Javascript

 ``

Output:

`2`

Complexity Analysis:

• Time complexity: O(N3).
Three nested loops are traversing in the array, so time complexity is O(n^3).
• Space Complexity: O(1).
As no extra space is required.

Efficient 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 the closest number to x – array[i]. Update the closest sum. The two-pointers algorithm takes 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,
• If the sum is smaller than the sum we need to get to, we increase the first pointer.
• Else, If the sum is bigger, Decrease the end pointer to reduce the sum.
• Update the closest sum found so far.

Implementation:

## C++14

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the sum of a``// triplet which is closest to x``int` `solution(vector<``int``>& arr, ``int` `x)``{` `    ``// Sort the array``    ``sort(arr.begin(), arr.end());` `    ``// To store the closest sum``  ``//not using INT_MAX to avoid overflowing condition``    ``int` `closestSum = 1000000000;` `    ``// Fix the smallest number among``    ``// the three integers``    ``for` `(``int` `i = 0; i < arr.size() - 2; i++) {` `        ``// Two pointers initially pointing at``        ``// the last and the element``        ``// next to the fixed element``        ``int` `ptr1 = i + 1, ptr2 = arr.size() - 1;` `        ``// While there could be more pairs to check``        ``while` `(ptr1 < ptr2) {` `            ``// Calculate the sum of the current triplet``            ``int` `sum = arr[i] + arr[ptr1] + arr[ptr2];``            ` `              ``// if sum is equal to x, return sum as``              ``if` `(sum == x)``              ``return` `sum;``            ``// If the sum is more closer than``            ``// the current closest sum``            ``if` `(``abs``(x - sum) < ``abs``(x - closestSum)) {``                ``closestSum = sum;``            ``}` `            ``// If sum is greater than x then decrement``            ``// the second pointer to get a smaller sum``            ``if` `(sum > x) {``                ``ptr2--;``            ``}` `            ``// Else increment the first pointer``            ``// to get a larger sum``            ``else` `{``                ``ptr1++;``            ``}``        ``}``    ``}` `    ``// Return the closest sum found``    ``return` `closestSum;``}` `// Driver code``int` `main()``{``    ``vector<``int``> arr = { -1, 2, 1, -4 };``    ``int` `x = 1;``    ``cout << solution(arr, x);` `    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``import` `static` `java.lang.Math.abs;``import` `java.util.*;` `class` `GFG``{` `// Function to return the sum of a``// triplet which is closest to x``static` `int` `solution(Vector arr, ``int` `x)``{` `    ``// Sort the array``    ``Collections.sort(arr);` `    ``// To store the closest sum``      ``// Assigning long to avoid overflow condition``      ``// when array has negative integers``    ``long` `closestSum = Integer.MAX_VALUE;` `    ``// Fix the smallest number among``    ``// the three integers``    ``for` `(``int` `i = ``0``; i < arr.size() - ``2``; i++)``    ``{` `        ``// Two pointers initially pointing at``        ``// the last and the element``        ``// next to the fixed element``        ``int` `ptr1 = i + ``1``, ptr2 = arr.size() - ``1``;` `        ``// While there could be more pairs to check``        ``while` `(ptr1 < ptr2)``        ``{` `            ``// Calculate the sum of the current triplet``            ``int` `sum = arr.get(i) + arr.get(ptr1) + arr.get(ptr2);` `            ``// If the sum is more closer than``            ``// the current closest sum``            ``if` `(abs(x - sum) < abs(x - closestSum))``            ``{``                ``closestSum = sum;``            ``}` `            ``// If sum is greater than x then decrement``            ``// the second pointer to get a smaller sum``            ``if` `(sum > x)``            ``{``                ``ptr2--;``            ``}` `            ``// Else increment the first pointer``            ``// to get a larger sum``            ``else``            ``{``                ``ptr1++;``            ``}``        ``}``    ``}` `    ``// Return the closest sum found``    ``return` `(``int``)closestSum;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``Vector arr = ``new` `Vector(Arrays.asList( -``1``, ``2``, ``1``, -``4` `));``    ``int` `x = ``1``;``    ``System.out.println(solution(arr, x));``}``}` `/* This code is contributed by PrinciRaj1992 */`

## Python3

 `# Python3 implementation of the approach` `import` `sys` `# Function to return the sum of a``# triplet which is closest to x``def` `solution(arr, x) :` `    ``# Sort the array``    ``arr.sort();``    ` `    ``# To store the closest sum``    ``closestSum ``=` `sys.maxsize;` `    ``# Fix the smallest number among``    ``# the three integers``    ``for` `i ``in` `range``(``len``(arr)``-``2``) :` `        ``# Two pointers initially pointing at``        ``# the last and the element``        ``# next to the fixed element``        ``ptr1 ``=` `i ``+` `1``; ptr2 ``=` `len``(arr) ``-` `1``;` `        ``# While there could be more pairs to check``        ``while` `(ptr1 < ptr2) :` `            ``# Calculate the sum of the current triplet``            ``sum` `=` `arr[i] ``+` `arr[ptr1] ``+` `arr[ptr2];` `            ``# If the sum is more closer than``            ``# the current closest sum``            ``if` `(``abs``(x ``-` `sum``) < ``abs``(x ``-` `closestSum)) :``                ``closestSum ``=` `sum``;` `            ``# If sum is greater than x then decrement``            ``# the second pointer to get a smaller sum``            ``if` `(``sum` `> x) :``                ``ptr2 ``-``=` `1``;` `            ``# Else increment the first pointer``            ``# to get a larger sum``            ``else` `:``                ``ptr1 ``+``=` `1``;` `    ``# Return the closest sum found``    ``return` `closestSum;`  `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``arr ``=` `[ ``-``1``, ``2``, ``1``, ``-``4` `];``    ``x ``=` `1``;``    ``print``(solution(arr, x));` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `// Function to return the sum of a``// triplet which is closest to x``static` `int` `solution(List<``int``> arr, ``int` `x)``{` `    ``// Sort the array``    ``arr.Sort();` `    ``// To store the closest sum``    ``int` `closestSum = ``int``.MaxValue;` `    ``// Fix the smallest number among``    ``// the three integers``    ``for` `(``int` `i = 0; i < arr.Count - 2; i++)``    ``{` `        ``// Two pointers initially pointing at``        ``// the last and the element``        ``// next to the fixed element``        ``int` `ptr1 = i + 1, ptr2 = arr.Count - 1;` `        ``// While there could be more pairs to check``        ``while` `(ptr1 < ptr2)``        ``{` `            ``// Calculate the sum of the current triplet``            ``int` `sum = arr[i] + arr[ptr1] + arr[ptr2];` `            ``// If the sum is more closer than``            ``// the current closest sum``            ``if` `(Math.Abs(x - sum) <``                ``Math.Abs(x - closestSum))``            ``{``                ``closestSum = sum;``            ``}` `            ``// If sum is greater than x then decrement``            ``// the second pointer to get a smaller sum``            ``if` `(sum > x)``            ``{``                ``ptr2--;``            ``}` `            ``// Else increment the first pointer``            ``// to get a larger sum``            ``else``            ``{``                ``ptr1++;``            ``}``        ``}``    ``}` `    ``// Return the closest sum found``    ``return` `closestSum;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]ar = { -1, 2, 1, -4 };``    ``List<``int``> arr = ``new` `List<``int``>(ar);``    ``int` `x = 1;``    ``Console.WriteLine(solution(arr, x));``}``}` `// This code is contributed by Princi Singh`

## Javascript

 ``

Output:

`2`

Complexity Analysis:

• Time complexity: O(N2)
There are only two nested loops traversing the array, so time complexity is O(n^2). Two pointer algorithm take O(n) time and the first element can be fixed using another nested traversal.
• Space Complexity: O(1)
As no extra space is required.