Find subarray with given sum with negatives allowed in constant space
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 had already discussed an approach to find subarray with a given sum in an array containing both positive and negative elements using Hashing
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 increment 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 in the input array, the task reduces to find 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 similarily 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 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 trailing elements while curr_sum is greater than the sum and keep decrementing target sum by abs(minimum element).
Below is the implementation of the above approach:
C++
// 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 - 1) { 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; } |
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Java
// Java program to check if subarray with sum // exists when negative elements are also present class GFG { // Function to check if subarray with sum // exists when negative elements are also present static void subArraySum(int arr[], int n, int sum) { int minEle = Integer.MAX_VALUE; // Find minimum element in the array for (int i = 0; i < n; i++) minEle = Math.min(arr[i], minEle); // Initialize curr_sum as value of // first element and starting point as 0 int curr_sum = arr[0] + Math.abs(minEle); int 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) * Math.abs(minEle) > targetSum && start < i - 1) { curr_sum = curr_sum - arr[start] - Math.abs(minEle); start++; } // If curr_sum becomes equal to sum, // then return true if (curr_sum - (i - start) * Math.abs(minEle) == targetSum) { System.out.println("Sum found between indexes " + start + " and " + (i - 1)); return; } // Add this element to curr_sum if (i < n) { curr_sum = curr_sum + arr[i] + Math.abs(minEle); } } // If we reach here, then no subarray System.out.println("No subarray found"); } // Driver Code public static void main (String[] args) { int arr[] = { 10, -12, 2, -2, -20, 10 }; int n = arr.length; int sum = -10; subArraySum(arr, n, sum); } } // This code is contributed by AnkitRai01 |
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Python3
# Python3 program to check if subarray with summ # exists when negative elements are also present # Function to check if subarray with summ # exists when negative elements are also present def subArraySum(arr, n, summ): minEle = 10**9 # Find minimum element in the array for i in range(n): minEle = min(arr[i], minEle) # Initialize curr_summ as value of # first element and starting poas 0 curr_summ = arr[0] + abs(minEle) start = 0 # Starting window length will be 1, # For generating new target summ, # add abs(minEle) to summ only 1 time targetSum = summ # Add elements one by one to curr_summ # and if the curr_summ exceeds the # updated summ, then remove starting element for i in range(1, n + 1): # If curr_summ exceeds the summ, # then remove the starting elements while (curr_summ - (i - start) * abs(minEle) > targetSum and start < i - 1): curr_summ = curr_summ - arr[start] - abs(minEle) start += 1 # If curr_summ becomes equal to summ, # then return true if (curr_summ - (i - start) * abs(minEle) == targetSum): print("Sum found between indexes", start, "and", i - 1) return # Add this element to curr_summ if (i < n): curr_summ = curr_summ + arr[i] + abs(minEle) # If we reach here, then no subarray print("No subarray found") # Driver Code arr = [10, -12, 2, -2, -20, 10] n = len(arr) summ = -10 subArraySum(arr, n, summ) # This code is contributed by Mohit Kumar |
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C#
// C# program to check if subarray // with sum exists when negative // elements are also present using System; class GFG { // Function to check if subarray // with sum exists when negative // elements are also present static void subArraySum(int []arr, int n, int sum) { int minEle = int.MaxValue; int i; // Find minimum element in the array for (i = 0; i < n; i++) minEle = Math.Min(arr[i], minEle); // Initialize curr_sum as value of // first element and starting point as 0 int curr_sum = arr[0] + Math.Abs(minEle); int start = 0; // 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) * Math.Abs(minEle) > targetSum && start < i - 1) { curr_sum = curr_sum - arr[start] - Math.Abs(minEle); start++; } // If curr_sum becomes equal to sum, // then return true if (curr_sum - (i - start) * Math.Abs(minEle) == targetSum) { Console.Write("Sum found between indexes " + start + " and " + (i - 1)); return; } // Add this element to curr_sum if (i < n) { curr_sum = curr_sum + arr[i] + Math.Abs(minEle); } } // If we reach here, then no subarray Console.Write("No subarray found"); } // Driver Code static public void Main () { int []arr = { 10, -12, 2, -2, -20, 10 }; int n = arr.Length; int sum = -10; subArraySum(arr, n, sum); } } // This code is contributed by ajit |
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Output:
Sum found between indexes 1 and 2
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