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
Explantion: Sum of elements between indices
2 and 4 is 20 + 3 + 10 = 33

Input: arr[] = {10, 2, -2, -20, 10}, sum = -10
Output: Sum found between indexes 0 to 3
Explantion: Sum of elements between indices
0 and 3 is 10 + 2 - 2 - 20 = -10

Input: arr[] = {-10, 0, 2, -2, -20, 10}, sum = 20
Output: No subarray with given sum exists
Explantion: There is no subarray with the given sum

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

Approach: The idea is to store the sum of elements of every prefix of the array in a hashmap, i.e for every index store the sum of elements upto that index hashmap. So to check if there is a subarray with sum equal to s, check for every index i, and sum upto that index as x. If there is a prefix with sum equal to x – s, then the subarray with given sum is found.

Algorithm:



  1. create a Hashmap (hm) to store key value pair, i.e, key = prefix sum and value = its index and a variable to store the current sum (sum = 0) and the sum of subarray as s
  2. Traverse through the array from start to end.
  3. For every element update the sum, i.e sum = sum + array[i]
  4. If sum is equal to s then print that the subarray with given sum is from 0 to i
  5. If there is any key in the HashMap which is equal to sum – s then print that the subarray with given sum is from hm[sum – s] to i
  6. Put the sum and index in the hashmap as key-value pair.

Dry-run of the above approach:

Implementation:

C++

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// 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;
}

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Java

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// 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);
  
    }
}

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Python3

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# 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

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C#

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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

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Output:

Sum found between indexes 0 to 3

Complexity Analysis:

  • Time complexity: O(N).
    If hashing is performed with the help of an array then this the time complexity. In case the elements cannot be hashed in an array a hash map can also be used as shown in the above code.
  • Auxiliary space: O(n).
    As a HashMap is needed, this takes a linear space.


Find subarray with given sum with negatives allowed in constant space

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