# Find the Sub-array with sum closest to 0

Given an array of both positive and negative numbers, the task is to find out the subarray whose sum is closest to 0.
There can be multiple such subarrays, we need to output just 1 of them.
Examples:

```Input : arr[] = {-1, 3, 2, -5, 4}
Output : 1, 3
Subarray from index 1 to 3 has sum closest to 0 i.e.
3 + 2 + -5 = 0

Input : {2, -5, 4, -6, 3}
Output : 0, 2
2 + -5 + 4 = 1 closest to 0
```

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

A Naive approach is to consider all subarrays one by one and update indexes of subarray with sum closest to 0.

## C++

 `// C++ program to find subarray with ` `// sum closest to 0 ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the subarray ` `pair<``int``, ``int``> findSubArray(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``int` `start, end, min_sum = INT_MAX; ` ` `  `    ``// Pick a starting point ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// Consider current starting point ` `        ``// as a subarray and update minimum ` `        ``// sum and subarray indexes ` `        ``int` `curr_sum = arr[i]; ` `        ``if` `(min_sum > ``abs``(curr_sum)) { ` `            ``min_sum = ``abs``(curr_sum); ` `            ``start = i; ` `            ``end = i; ` `        ``} ` ` `  `        ``// Try all subarrays starting with i ` `        ``for` `(``int` `j = i + 1; j < n; j++) { ` `            ``curr_sum = curr_sum + arr[j]; ` ` `  `            ``// update minimum sum ` `            ``// and subarray indexes ` `            ``if` `(min_sum > ``abs``(curr_sum)) { ` `                ``min_sum = ``abs``(curr_sum); ` `                ``start = i; ` `                ``end = j; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Return starting and ending indexes ` `    ``pair<``int``, ``int``> p = make_pair(start, end); ` `    ``return` `p; ` `} ` ` `  `// Drivers code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, -5, 4, -6, -3 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``pair<``int``, ``int``> point = findSubArray(arr, n); ` `    ``cout << ``"Subarray starting from "``; ` `    ``cout << point.first << ``" to "` `<< point.second; ` `    ``return` `0; ` `} `

## Java

 `// Java program to find subarray with ` `// sum closest to 0 ` ` `  `class` `GFG ` `{ ` ` `  `    ``static` `class` `Pair  ` `    ``{ ` ` `  `        ``int` `first, second; ` `        ``public` `Pair(``int` `first, ``int` `second) ` `        ``{ ` `            ``this``.first = first; ` `            ``this``.second = second; ` `        ``} ` ` `  `    ``} ` `     `  `    ``// Function to find the subarray ` `    ``static` `Pair findSubArray(``int` `arr[], ``int` `n) ` `    ``{ ` ` `  `        ``int` `start = ``0``, end = ``0``, min_sum = Integer.MAX_VALUE; ` ` `  `        ``// Pick a starting point ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` ` `  `            ``// Consider current starting point ` `            ``// as a subarray and update minimum ` `            ``// sum and subarray indexes ` `            ``int` `curr_sum = arr[i]; ` `            ``if` `(min_sum > Math.abs(curr_sum))  ` `            ``{ ` `                ``min_sum = Math.abs(curr_sum); ` `                ``start = i; ` `                ``end = i; ` `            ``} ` ` `  `            ``// Try all subarrays starting with i ` `            ``for` `(``int` `j = i + ``1``; j < n; j++)  ` `            ``{ ` `                ``curr_sum = curr_sum + arr[j]; ` ` `  `                ``// update minimum sum ` `                ``// and subarray indexes ` `                ``if` `(min_sum > Math.abs(curr_sum))  ` `                ``{ ` `                    ``min_sum = Math.abs(curr_sum); ` `                    ``start = i; ` `                    ``end = j; ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``// Return starting and ending indexes ` `        ``Pair p = ``new` `Pair(start, end); ` `        ``return` `p; ` `    ``} ` ` `  `    ``// Drivers code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int` `arr[] = {``2``, -``5``, ``4``, -``6``, -``3``}; ` `        ``int` `n = arr.length; ` ` `  `        ``Pair point = findSubArray(arr, n); ` `        ``System.out.println(``"Subarray starting from "` `                ``+ point.first + ``" to "` `+ point.second); ` `    ``} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

## Python3

 `# Python 3 program to find subarray with ` `# sum closest to 0 ` `import` `sys ` ` `  `# Function to find the subarray ` `def` `findSubArray(arr, n): ` `    ``min_sum ``=` `sys.maxsize ` ` `  `    ``# Pick a starting point ` `    ``for` `i ``in` `range``(n): ` `         `  `        ``# Consider current starting point ` `        ``# as a subarray and update minimum ` `        ``# sum and subarray indexes ` `        ``curr_sum ``=` `arr[i] ` `        ``if` `(min_sum > ``abs``(curr_sum)): ` `            ``min_sum ``=` `abs``(curr_sum) ` `            ``start ``=` `i ` `            ``end ``=` `i ` ` `  `        ``# Try all subarrays starting with i ` `        ``for` `j ``in` `range``(i ``+` `1``, n, ``1``): ` `            ``curr_sum ``=` `curr_sum ``+` `arr[j] ` ` `  `            ``# update minimum sum ` `            ``# and subarray indexes ` `            ``if` `(min_sum > ``abs``(curr_sum)): ` `                ``min_sum ``=` `abs``(curr_sum) ` `                ``start ``=` `i ` `                ``end ``=` `j ` ` `  `    ``# Return starting and ending indexes ` `    ``p ``=` `[start, end] ` `    ``return` `p ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr ``=` `[``2``, ``-``5``, ``4``, ``-``6``, ``-``3``] ` `    ``n ``=` `len``(arr) ` ` `  `    ``point ``=` `findSubArray(arr, n) ` `    ``print``(``"Subarray starting from "``, end ``=` `"") ` `    ``print``(point[``0``], ``"to"``, point[``1``]) ` ` `  `# This code is contributed by ` `# Surendra_Gangwar `

## C#

 `// C# program to find subarray with ` `// sum closest to 0 ` `using` `System; ` `     `  `class` `GFG ` `{ ` ` `  `    ``public` `class` `Pair  ` `    ``{ ` ` `  `        ``public` `int` `first, second; ` `        ``public` `Pair(``int` `first, ``int` `second) ` `        ``{ ` `            ``this``.first = first; ` `            ``this``.second = second; ` `        ``} ` ` `  `    ``} ` `     `  `    ``// Function to find the subarray ` `    ``static` `Pair findSubArray(``int` `[]arr, ``int` `n) ` `    ``{ ` ` `  `        ``int` `start = 0, end = 0, min_sum = ``int``.MaxValue; ` ` `  `        ``// Pick a starting point ` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` ` `  `            ``// Consider current starting point ` `            ``// as a subarray and update minimum ` `            ``// sum and subarray indexes ` `            ``int` `curr_sum = arr[i]; ` `            ``if` `(min_sum > Math.Abs(curr_sum))  ` `            ``{ ` `                ``min_sum = Math.Abs(curr_sum); ` `                ``start = i; ` `                ``end = i; ` `            ``} ` ` `  `            ``// Try all subarrays starting with i ` `            ``for` `(``int` `j = i + 1; j < n; j++)  ` `            ``{ ` `                ``curr_sum = curr_sum + arr[j]; ` ` `  `                ``// update minimum sum ` `                ``// and subarray indexes ` `                ``if` `(min_sum > Math.Abs(curr_sum))  ` `                ``{ ` `                    ``min_sum = Math.Abs(curr_sum); ` `                    ``start = i; ` `                    ``end = j; ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``// Return starting and ending indexes ` `        ``Pair p = ``new` `Pair(start, end); ` `        ``return` `p; ` `    ``} ` ` `  `    ``// Drivers code ` `    ``public` `static` `void` `Main(String[] args)  ` `    ``{ ` `        ``int` `[]arr = {2, -5, 4, -6, -3}; ` `        ``int` `n = arr.Length; ` ` `  `        ``Pair point = findSubArray(arr, n); ` `        ``Console.WriteLine(``"Subarray starting from "` `                ``+ point.first + ``" to "` `+ point.second); ` `    ``} ` `} ` ` `  `// This code is contributed by Princi Singh `

Output:

```Subarray starting from 0 to 2
```

Time Complexity: O(n2)

An Efficient method is to perform following steps:-

1. Maintain a Prefix sum array . Also maintain indexes in the prefix sum array.
2. Sort the prefix sum array on the basis of sum.
3. Find the two elements in a prefix sum array with minimum difference.
`i.e.  Find min(pre_sum[i] - pre_sum[i-1]) `
4. Return indexes of pre_sum with minimum difference.
5. Subarray with (lower_index+1, upper_index) will have the sum closest to 0.
6. Taking lower_index+1 because on subtracting value at lower_index we get the sum closest to 0. That’s why lower_index need not to be included.

## C++

 `// C++ program to find subarray with sum  ` `// closest to 0 ` `#include ` `using` `namespace` `std; ` ` `  `struct` `prefix { ` `    ``int` `sum; ` `    ``int` `index; ` `}; ` ` `  `// Sort on the basis of sum ` `bool` `comparison(prefix a, prefix b) ` `{ ` `    ``return` `a.sum < b.sum; ` `} ` ` `  `// Returns subarray with sum closest to 0.  ` `pair<``int``, ``int``> findSubArray(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `start, end, min_diff = INT_MAX; ` ` `  `    ``prefix pre_sum[n + 1]; ` ` `  `    ``// To consider the case of subarray starting ` `    ``// from beginning of the array ` `    ``pre_sum.sum = 0; ` `    ``pre_sum.index = -1; ` ` `  `    ``// Store prefix sum with index ` `    ``for` `(``int` `i = 1; i <= n; i++) { ` `        ``pre_sum[i].sum = pre_sum[i-1].sum + arr[i-1]; ` `        ``pre_sum[i].index = i - 1; ` `    ``} ` ` `  `    ``// Sort on the basis of sum ` `    ``sort(pre_sum, pre_sum + (n + 1), comparison); ` ` `  `    ``// Find two consecutive elements with minimum difference ` `    ``for` `(``int` `i = 1; i <= n; i++) { ` `        ``int` `diff = pre_sum[i].sum - pre_sum[i-1].sum; ` ` `  `        ``// Update minimum difference ` `        ``// and starting and ending indexes ` `        ``if` `(min_diff > diff) { ` `            ``min_diff = diff; ` `            ``start = pre_sum[i-1].index; ` `            ``end = pre_sum[i].index; ` `        ``} ` `    ``} ` ` `  `    ``// Return starting and ending indexes ` `    ``pair<``int``, ``int``> p = make_pair(start + 1, end); ` `    ``return` `p; ` `} ` ` `  `// Drivers code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, 3, -4, -1, 6 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``pair<``int``, ``int``> point = findSubArray(arr, n); ` `    ``cout << ``"Subarray starting from "``; ` `    ``cout << point.first << ``" to "` `<< point.second; ` ` `  `    ``return` `0; ` `} `

## Python3

 `# Python3 program to find subarray  ` `# with sum closest to 0  ` `class` `prefix: ` `     `  `    ``def` `__init__(``self``, ``sum``, index): ` `        ``self``.``sum` `=` `sum` `        ``self``.index ``=` `index ` ` `  `# Returns subarray with sum closest to 0.  ` `def` `findSubArray(arr, n):  ` ` `  `    ``start, end, min_diff ``=` `None``, ``None``, ``float``(``'inf'``)  ` ` `  `    ``pre_sum ``=` `[``None``] ``*` `(n ``+` `1``) ` ` `  `    ``# To consider the case of subarray  ` `    ``# starting from beginning of the array  ` `    ``pre_sum[``0``] ``=` `prefix(``0``, ``-``1``) ` ` `  `    ``# Store prefix sum with index  ` `    ``for` `i ``in` `range``(``1``, n ``+` `1``):  ` `        ``pre_sum[i] ``=` `prefix(pre_sum[i ``-` `1``].``sum` `+`  `                                ``arr[i ``-` `1``], i ``-` `1``)  ` ` `  `    ``# Sort on the basis of sum  ` `    ``pre_sum.sort(key ``=` `lambda` `x: x.``sum``) ` ` `  `    ``# Find two consecutive elements  ` `    ``# with minimum difference  ` `    ``for` `i ``in` `range``(``1``, n ``+` `1``):  ` `        ``diff ``=` `pre_sum[i].``sum` `-` `pre_sum[i ``-` `1``].``sum` ` `  `        ``# Update minimum difference  ` `        ``# and starting and ending indexes  ` `        ``if` `min_diff > diff:  ` `            ``min_diff ``=` `diff  ` `            ``start ``=` `pre_sum[i ``-` `1``].index  ` `            ``end ``=` `pre_sum[i].index  ` `         `  `    ``# Return starting and ending indexes  ` `    ``return` `(start ``+` `1``, end)  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"``:  ` ` `  `    ``arr ``=` `[``2``, ``3``, ``-``4``, ``-``1``, ``6``]  ` `    ``n ``=` `len``(arr)  ` ` `  `    ``point ``=` `findSubArray(arr, n)  ` `    ``print``(``"Subarray starting from"``,  ` `           ``point[``0``], ``"to"``, point[``1``])  ` ` `  `# This code is contributed by Rituraj Jain `

Output:

```Subarray starting from 0 to 3
```

Time Complexity: O(n log n)

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