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:
- 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
- Traverse through the array from start to end.
- For every element update the sum, i.e sum = sum + array[i]
- If sum is equal to s then print that the subarray with given sum is from 0 to i
- 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
- Put the sum and index in the hashmap as key-value pair.
Dry-run of the above approach:
Implementation:
C++
// 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
// 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
# 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 |
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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 |
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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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