# 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: Please try your approach on {IDE} first, before moving on to the solution.

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 loop with counter i, j and k respectively.
2. First loop will start from start to end, second loop will run from i+1 to end, third loop will run from j+1 to end.
3. Check if the difference of sum of ith, jth and kth element with given sum is less than the current minimum or not. Update the current minimum
4. Print the closest sum.
• Implementation:
 `// 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 closets 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; ` `} `

Output:

```2
```
• Complexity Analysis:
• Time complexity: O(N3).
There are three nested loops traversing the array, so time complexity is O(n^3).
• Space Compelxity: 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. 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,
• 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++ 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 closets sum ` `    ``int` `closestSum = INT_MAX; ` ` `  `    ``// 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 the sum is more closer than ` `            ``// the current closest sum ` `            ``if` `(``abs``(1LL*x - sum) < ``abs``(1LL*x - closestSum)) { ` `                ``closestSum = sum; ` `            ``} ` ` `  `            ``// If sum is greater then 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 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 closets sum ` `    ``int` `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 then 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)  ` `{ ` `    ``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 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 closets 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 then 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# 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 closets 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 then 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 `

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 Compelxity: O(1).
As no extra space is required.

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

Check out this Author's contributed articles.

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :