An array is given, find the length of the subarray having maximum sum.
Examples:
Input : a[] = {1, -2, 1, 1, -2, 1}
Output : Length of the subarray is 2
Explanation: Subarray with consecutive elements
and maximum sum will be {1, 1}. So length is 2
Input : ar[] = { -2, -3, 4, -1, -2, 1, 5, -3 }
Output : Length of the subarray is 5
Explanation: Subarray with consecutive elements
and maximum sum will be {4, -1, -2, 1, 5}
This problem is mainly a variation of the Largest Sum Contiguous Subarray Problem.
The idea is to update starting index whenever the sum ending here becomes less than 0.
Java
// Java program to print length of the largest // contiguous array sumclass GFG { // Main driver method static int maxSubArraySum(int a[], int size) { int max_so_far = Integer.MIN_VALUE, max_ending_here = 0,start = 0, end = 0, s = 0; for (int i = 0; i < size; i++) { max_ending_here += a[i]; if (max_so_far < max_ending_here) { max_so_far = max_ending_here; start = s; end = i; } if (max_ending_here < 0) { max_ending_here = 0; s = i + 1; } } return (end - start + 1); } // Main driver method public static void main(String[] args) { // Declaring and initializing integer array int a[] = { -2, -3, 4, -1, -2, 1, 5, -3 }; int n = a.length; // Calling function so as to print // largest contiguous sum on console System.out.println(maxSubArraySum(a, n)); }} |
5
Time Complexity: O(N) where N is size of the input array. This is because a for loop is executed from 1 to size of the array.
Auxiliary Space: O(1) as no extra space has been taken.
Approach#2: Using Kadane’s algorithm
This approach implements Kadane’s algorithm to find the maximum subarray sum and returns the size of the subarray with the maximum sum.
Algorithm:
- Initialize max_sum, current_sum, start, end, max_start, and max_end to the first element of the array.
- Iterate through the array from the second element.
- If the current element is greater than the sum of the current element and current_sum, update start to the current index.
- Update current_sum as the maximum of the current element and the sum of current element and current_sum.
- If current_sum is greater than max_sum, update max_sum, end to the current index, and max_start and max_end to start and end respectively.
- Return max_end – max_start + 1 as the size of the subarray with maximum sum.
Below is the implementation of the approach:
Java
// Java code for the approachimport java.util.*;public class GFG { // Function to find the maximum subarray sum static int maxSubarraySum(List<Integer> a) { int n = a.size(); int max_sum = a.get(0); int current_sum = a.get(0); int start = 0; int end = 0; int max_start = 0; int max_end = 0; // Traverse the list for (int i = 1; i < n; i++) { // If the current element is greater than the // sum so far plus the current element, then // update the start index to the current index if (a.get(i) > current_sum + a.get(i)) { start = i; } // Update the current sum to be either the // current element or the sum so far plus the // current element current_sum = Math.max(a.get(i), current_sum + a.get(i)); // If the current sum is greater than the // maximum sum so far, then update the maximum // sum and its start and end indices if (current_sum > max_sum) { max_sum = current_sum; end = i; max_start = start; max_end = end; } } // Return the length of the maximum subarray return max_end - max_start + 1; } // Main driver method public static void main(String[] args) { // Storing as list List<Integer> a = Arrays.asList(-2, -3, 4, -1, -2, 1, 5, -3); System.out.println(maxSubarraySum(a)); }} |
5
Time Complexity: O(n), where n is the length of array
Auxiliary Space: O(1)
Note: The above code assumes that there is at least one positive element in the array. If all the elements are negative, the code needs to be modified to return the maximum element in the array.
Please refer complete article on Size of The Subarray With Maximum Sum for more details!
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