Print subarray with maximum sum

Given an array arr[], the task is to print the subarray having maximum sum.

Examples:

Input: arr[] = {2, 3, -8, 7, -1, 2, 3}
Output:{7, -1, 2, 3}
Explanation: The subarray {7, -1, 2, 3} has the largest sum 11.

Input: arr[] = {-2, -5, 6, -2, -3, 1, 5, -6}
Output: {6, -2, -3, 1, 5}
Explanation: The subarray {6, -2, -3, 1, 5} has the largest sum of 7.

[Naive Approach] By iterating over all subarrays - O(n^2) Time and O(1) Space

The idea is to run two nested loops to iterate over all possible subarrays and find the maximum sum. The outer loop will mark the starting point of a subarray and inner loop will mark the ending point of the subarray. At any time, if we find a subarray whose sum is greater than the maximum sum so far, then we will update the starting and ending point of the maximum sum subarray.

7 -1 2 3 

Time complexity: O(n²), as we are iterating over all possible subarrays.
Auxiliary Space: O(1)

[Expected Approach] Using Kadane's Algorithm - O(n) Time and O(1) Space

The idea is similar to Kadane's Algorithm with the only difference that here, we need to keep track of the start and end of the subarray with maximum sum, that is the result array. Iterate over the array keeping track of the start and end of current subarray and at any point, if the sum of current subarray becomes greater than the result array, update the result array with the current subarray.

For each element, we have two choices:

• Choice 1: Extend the maximum sum subarray ending at the previous element by adding the current element to it. In this case, the ending index of the current subarray increases by 1.
• Choice 2: Start a new subarray starting from the current element. In this case, the starting index of the current subarray updates to the current index.

If the maximum sum ending at an element becomes greater than the result array, we update the start and end of result subarray with the start and end of current subarray respectively.

7 -1 2 3 

Time Complexity: O(n), as we are traversing the array only once.
Auxiliary Space: O(1)