Find subarray with given sum with negatives allowed in constant space

Hello Everyone,

Given an unsorted array of integers, find a subarray that adds to a given number. If there is more than one subarray with the sum of 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.

In this article, an approach without using any extra space is discussed.
The idea is to modify the array to contain only positive elements:

  • Find the smallest negative element in the array and increase every value in the array by the absolute value of the smallest negative element found.

We may notice that after doing the above modification, the sum of every subarray will also increase by:

(number of elements in the subarray) * (absolute value of min element)

After doing the above modification to the input array, the task is reduced to finding if there is any subarray in the given array of only positive numbers with the new target sum. This can be done using the sliding window technique in O(N) time and constant space.
Every time while adding elements to the window, increment the target sum by the absolute value of the minimum element and, similarly, while removing elements from the current window, decrement the target sum by the absolute value of the minimum element so that for every subarray of length, say K, the updated target sum will be:

targetSum = sum + K*abs(minimum element)

Below is the approach for the same:

  • Initialize a variable curr_sum as the first element.
  • The variable curr_sum indicates the sum of the current subarray.
  • Start from the second element and add all elements one by one to the curr_sum, and keep incrementing the target sum by abs(minimum element).
  • If curr_sum becomes equal to the target sum, then print the solution.
  • If curr_sum exceeds the sum, then remove the trailing elements while decreasing curr_sum is greater than the sum and keep decreasing the target sum by abs(minimum element).

Below is the implementation of the above approach:

// C++ program to check if subarray with sum

// exists when negative elements are also present

#include <bits/stdc++.h>

using namespace std;

// Function to check if subarray with sum

// exists when negative elements are also present

void subArraySum( int arr[], int n, int sum)

{

int minEle = INT_MAX;

// Find minimum element in the array

for ( int i = 0; i < n; i++)

minEle = min(arr[i], minEle);

// Initialize curr_sum as value of first element

// and starting point as 0

int curr_sum = arr[0] + abs (minEle), start = 0, i;

// Starting window length will be 1,

// For generating new target sum,

// add abs(minEle) to sum only 1 time

int targetSum = sum;

// Add elements one by one to curr_sum

// and if the curr_sum exceeds the

// updated sum, then remove starting element

for (i = 1; i <= n; i++) {

// If curr_sum exceeds the sum,

// then remove the starting elements

while (curr_sum - (i - start) * abs (minEle) > targetSum && start < i) {

curr_sum = curr_sum - arr[start] - abs (minEle);

start++;

}

// If curr_sum becomes equal to sum, then return true

if (curr_sum - (i - start) * abs (minEle) == targetSum) {

cout << "Sum found between indexes " << start << " and " << i - 1;

return ;

}

// Add this element to curr_sum

if (i < n) {

curr_sum = curr_sum + arr[i] + abs (minEle);

}

}

// If we reach here, then no subarray

cout << "No subarray found" ;

}

// Driver Code

int main()

{

int arr[] = { 10, -12, 2, -2, -20, 10 };

int n = sizeof (arr) / sizeof (arr[0]);

int sum = -10;

subArraySum(arr, n, sum);

return 0;

}

Output:

Sum found between indexes 1 and 2