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
Output: Sum found between indexes 2 and 4

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

Input: arr[] = {-10, 0, 2, -2, -20, 10}, sum = 20
Output: 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 the sum of every subarray is equal to the 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 the sum of elements encountered so far in a variable (say curr_sum). Let the given number is the 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 the given sum, else we insert curr_sum into the map and proceed to the next element. If all elements of the array are processed and we didn’t find any subarray with the given sum, then subarray doesn’t exist.

Below image is a dry run of the above approach:

Below is the implementation of the above approach :

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to print subarray with sum as given sum
#include<bits/stdc++.h>
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[0]);
    int sum = -10;
  
    subArraySum(arr, n, sum);
  
    return 0;
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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<Integer, Integer> 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);
  
    }
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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] =
    
    # 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
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

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
chevron_right


Output:
Sum found between indexes 0 to 3

Time complexity of above solution is O(N) if we perform hashing with the help of an array. In case the elements cannot be hashed in an array we use a hash map as shown in the above code.

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

This article is contributed by Aditya Goel. 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 write comments if you find anything incorrect, or you want to share more information about the topic discussed above.






Article Tags :
Practice Tags :