Step 1 — Subproblem: For each position i, we ask: what is the maximum sum of a subarray ending at i? Either we extend the best subarray ending at i-1 by adding arr[i], or we start fresh with just arr[i]. So currentSum(i) = max(currentSum(i-1) + arr[i], arr[i]).

Step 2 — Initialization: Set currentSum = arr[0] and maxSum = arr[0]. These represent the best sum ending at index 0 and the best sum seen so far.

Step 3 — Loop: For i from 1 to n-1, update currentSum = max(currentSum + arr[i], arr[i]). If we take the first option, we extend the previous subarray; if we take arr[i] alone, we restart. Then update maxSum = max(maxSum, currentSum).

Step 4 — Result: After the loop, maxSum holds the maximum sum of any contiguous subarray. We can track start/end indices by remembering when we restarted and when we beat the previous max. Time O(n), space O(1).