# Find subarray with given sum | Set 2 (Handles Negative Numbers)

Given an unsorted array of integers, find a subarray which adds to a given number. If there are more than one subarrays with the sum as the given number, print any of them.

Examples:

```Input: arr[] = {1, 4, 20, 3, 10, 5}, sum = 33
Ouptut: Sum found between indexes 2 and 4

Input: arr[] = {10, 2, -2, -20, 10}, sum = -10
Ouptut: Sum found between indexes 0 to 3

Input: arr[] = {-10, 0, 2, -2, -20, 10}, sum = 20
Ouptut: No subarray with given sum exists
```

We have discussed a solution that do not handles negative integers here. In this post, negative integers are also handled.

A simple solution is to consider all subarrays one by one and check if sum of every subarray is equal to given sum or not. The complexity of this solution would be O(n^2).

An efficient way is to use a map. The idea is to maintain sum of elements encountered so far in a variable (say curr_sum). Let the given number is sum. Now for each element, we check if curr_sum – sum exists in the map or not. If we found it in the map that means, we have a subarray present with given sum, else we insert curr_sum into the map and proceed to next element. If all elements of the array are processed and we didn’t find any subarray with given sum, then subarray doesn’t exists.

Below is the implementation of above idea –

## C++

 `// C++ program to print subarray with sum as given sum ` `#include ` `using` `namespace` `std; ` ` `  `// Function to print subarray with sum as given sum ` `void` `subArraySum(``int` `arr[], ``int` `n, ``int` `sum) ` `{ ` `    ``// create an empty map ` `    ``unordered_map<``int``, ``int``> map; ` ` `  `    ``// Maintains sum of elements so far ` `    ``int` `curr_sum = 0; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``// add current element to curr_sum ` `        ``curr_sum = curr_sum + arr[i]; ` ` `  `        ``// if curr_sum is equal to target sum ` `        ``// we found a subarray starting from index 0 ` `        ``// and ending at index i ` `        ``if` `(curr_sum == sum) ` `        ``{ ` `            ``cout << ``"Sum found between indexes "` `                 ``<< 0 << ``" to "` `<< i << endl; ` `            ``return``; ` `        ``} ` ` `  `        ``// If curr_sum - sum already exists in map ` `        ``// we have found a subarray with target sum ` `        ``if` `(map.find(curr_sum - sum) != map.end()) ` `        ``{ ` `            ``cout << ``"Sum found between indexes "` `                 ``<< map[curr_sum - sum] + 1 ` `                 ``<< ``" to "` `<< i << endl; ` `            ``return``; ` `        ``} ` ` `  `        ``map[curr_sum] = i; ` `    ``} ` ` `  `    ``// If we reach here, then no subarray exists ` `    ``cout << ``"No subarray with given sum exists"``; ` `} ` ` `  `// Driver program to test above function ` `int` `main() ` `{ ` `    ``int` `arr[] = {10, 2, -2, -20, 10}; ` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr); ` `    ``int` `sum = -10; ` ` `  `    ``subArraySum(arr, n, sum); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to print subarray with sum as given sum ` `import` `java.util.*; ` ` `  `class` `GFG { ` ` `  `    ``public` `static` `void` `subArraySum(``int``[] arr, ``int` `n, ``int` `sum) { ` `        ``//cur_sum to keep track of cummulative sum till that point ` `        ``int` `cur_sum = ``0``; ` `        ``int` `start = ``0``; ` `        ``int` `end = -``1``; ` `        ``HashMap hashMap = ``new` `HashMap<>(); ` ` `  `        ``for` `(``int` `i = ``0``; i < n; i++) { ` `            ``cur_sum = cur_sum + arr[i]; ` `            ``//check whether cur_sum - sum = 0, if 0 it means ` `            ``//the sub array is starting from index 0- so stop ` `            ``if` `(cur_sum - sum == ``0``) { ` `                ``start = ``0``; ` `                ``end = i; ` `                ``break``; ` `            ``} ` `            ``//if hashMap already has the value, means we already  ` `            ``// have subarray with the sum - so stop ` `            ``if` `(hashMap.containsKey(cur_sum - sum)) { ` `                ``start = hashMap.get(cur_sum - sum) + ``1``; ` `                ``end = i; ` `                ``break``; ` `            ``} ` `            ``//if value is not present then add to hashmap ` `            ``hashMap.put(cur_sum, i); ` ` `  `        ``} ` `        ``// if end is -1 : means we have reached end without the sum ` `        ``if` `(end == -``1``) { ` `            ``System.out.println(``"No subarray with given sum exists"``); ` `        ``} ``else` `{ ` `            ``System.out.println(``"Sum found between indexes "`  `                            ``+ start + ``" to "` `+ end); ` `        ``} ` ` `  `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) { ` `        ``int``[] arr = {``10``, ``2``, -``2``, -``20``, ``10``}; ` `        ``int` `n = arr.length; ` `        ``int` `sum = -``10``; ` `        ``subArraySum(arr, n, sum); ` ` `  `    ``} ` `} `

## Python3

 `# Python3 program to print subarray with sum as given sum  ` ` `  `# Function to print subarray with sum as given sum  ` `def` `subArraySum(arr, n, ``Sum``):  ` `  `  `    ``# create an empty map  ` `    ``Map` `=` `{}  ` `   `  `    ``# Maintains sum of elements so far  ` `    ``curr_sum ``=` `0`  `   `  `    ``for` `i ``in` `range``(``0``,n):  ` `      `  `        ``# add current element to curr_sum  ` `        ``curr_sum ``=` `curr_sum ``+` `arr[i]  ` `   `  `        ``# if curr_sum is equal to target sum  ` `        ``# we found a subarray starting from index 0  ` `        ``# and ending at index i  ` `        ``if` `curr_sum ``=``=` `Sum``:  ` `          `  `            ``print``(``"Sum found between indexes 0 to"``, i) ` `            ``return`  `          `  `   `  `        ``# If curr_sum - sum already exists in map  ` `        ``# we have found a subarray with target sum  ` `        ``if` `(curr_sum ``-` `Sum``) ``in` `Map``:  ` `          `  `            ``print``(``"Sum found between indexes"``, ` `                   ``Map``[curr_sum ``-` `Sum``] ``+` `1``, ``"to"``, i)  ` `             `  `            ``return`  `   `  `        ``Map``[curr_sum] ``=` `i  ` `   `  `    ``# If we reach here, then no subarray exists  ` `    ``print``(``"No subarray with given sum exists"``)  ` `  `  `   `  `# Driver program to test above function  ` `if` `__name__ ``=``=` `"__main__"``:  ` `  `  `    ``arr ``=` `[``10``, ``2``, ``-``2``, ``-``20``, ``10``]  ` `    ``n ``=` `len``(arr)  ` `    ``Sum` `=` `-``10`  `   `  `    ``subArraySum(arr, n, ``Sum``)  ` `   `  `# This code is contributed by Rituraj Jain `

## C#

 `using` `System; ` `using` `System.Collections.Generic; ` ` `  `// C# program to print subarray with sum as given sum  ` ` `  `public` `class` `GFG ` `{ ` ` `  `    ``public` `static` `void` `subArraySum(``int``[] arr, ``int` `n, ``int` `sum) ` `    ``{ ` `        ``//cur_sum to keep track of cummulative sum till that point  ` `        ``int` `cur_sum = 0; ` `        ``int` `start = 0; ` `        ``int` `end = -1; ` `        ``Dictionary<``int``, ``int``> hashMap = ``new` `Dictionary<``int``, ``int``>(); ` ` `  `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `            ``cur_sum = cur_sum + arr[i]; ` `            ``//check whether cur_sum - sum = 0, if 0 it means  ` `            ``//the sub array is starting from index 0- so stop  ` `            ``if` `(cur_sum - sum == 0) ` `            ``{ ` `                ``start = 0; ` `                ``end = i; ` `                ``break``; ` `            ``} ` `            ``//if hashMap already has the value, means we already   ` `            ``// have subarray with the sum - so stop  ` `            ``if` `(hashMap.ContainsKey(cur_sum - sum)) ` `            ``{ ` `                ``start = hashMap[cur_sum - sum] + 1; ` `                ``end = i; ` `                ``break``; ` `            ``} ` `            ``//if value is not present then add to hashmap  ` `            ``hashMap[cur_sum] = i; ` ` `  `        ``} ` `        ``// if end is -1 : means we have reached end without the sum  ` `        ``if` `(end == -1) ` `        ``{ ` `            ``Console.WriteLine(``"No subarray with given sum exists"``); ` `        ``} ` `        ``else` `        ``{ ` `            ``Console.WriteLine(``"Sum found between indexes "` `+ start + ``" to "` `+ end); ` `        ``} ` ` `  `    ``} ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `Main(``string``[] args) ` `    ``{ ` `        ``int``[] arr = ``new` `int``[] {10, 2, -2, -20, 10}; ` `        ``int` `n = arr.Length; ` `        ``int` `sum = -10; ` `        ``subArraySum(arr, n, sum); ` ` `  `    ``} ` `} ` ` `  `// This code is contributed by Shrikant13 `

Output:

```Sum found between indexes 0 to 3
```

Time complexity of above solution is O(n) as we are doing only one traversal of the array.

Auxiliary space
used by the program is O(n).