range(start, len(nums)): subarray_sum += nums[end] max_abs_sum = max(max_abs_sum, abs(subarray_sum)) return max_abs_sum

Complexity

⏰ Time complexity: O(n^2) due to nested loops
🧺 Space complexity: O(1)

Method 3 - Using prefix sum but naive

The prefix sum approach is a way to optimise subarray sums by precomputing cumulative sums for the array. Once we have the prefix sums, the sum of any subarray between indices i and j can be computed in O(1), which avoids redundant calculations. $$ \text{subarraySum} = \text{prefixSum}[j+1] - \text{prefixSum}[i] $$

Here is the approach:

Compute the prefix_sum array in ( O(n) ).
Use two nested loops:
- Outer loop: Fix a starting index for the subarray (i).
- Inner loop: Compute all subarray sums starting from i by using the

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