My first instinct was brute force, enumerate every subarray, track min and max, done.
Recognize that for any subarray, the sum of its min and max is determined by the pair of elements that are the min and max. Use a divide-and-conquer strategy: recursively split the array, and for subarrays crossing the midpoint, compute the best (min+max) efficiently by expanding from the midpoint while tracking min and max. This yields O(n log n) time, which is better than O(n^2).
Pro tip: During the interview, explicitly discuss the trade-offs between the O(n log n) divide-and-conquer approach and a potential O(n) monotonic stack solution, showing you understand both time and implementation complexity. Also, clarify that the subarray must be non-empty and handle edge cases like single-element arrays.
Confirm that the array can contain negative numbers, zeros, and that subarrays must be contiguous and non-empty. Ask about input size to gauge the expected time complexity.
Start with the brute-force O(n^2) method, then propose a divide-and-conquer O(n log n) solution. Mention that an O(n) solution might exist using monotonic stacks but is more complex.
Split the array into two halves, recursively find the maximum in each half, and then find the maximum for subarrays that cross the midpoint. For crossing subarrays, expand from the midpoint while maintaining the current min and max, updating the best sum.
Explain that the recurrence T(n) = 2T(n/2) + O(n) leads to O(n log n) time, and the recursion stack uses O(log n) space. Compare with the O(n^2) brute-force approach.
Walk through a small example (e.g., [1, -2, 3, -4]) to verify the algorithm. Discuss edge cases like all negative numbers, single element, and large input.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.