Merge stone piles
You have `stones` piles in a row. Repeatedly merge two ADJACENT piles into one; the cost of a merge is the sum of the two piles' sizes. Keep merging until a single pile remains. Return the minimum possible total merge cost. A single pile (or empty) costs 0.
Implement
min_merge_cost(stones: list[int]) → intExamples
in
[[4,3,3,4]]out28What a strong answer looks like
State your approach and its time/space complexity out loud before you optimize. Handle the edge cases (empty input, duplicates, overflow), and say why you chose this over the brute force. Green tests are the floor, not the grade.
0:00 of about 35 min
solution.py
InputExpectedGot
[[4,3,3,4]]28not run yetsample