Stone game optimal score difference
Two players alternate taking the stone from either end of a row; piles[i] is the value of stone i. Both play optimally to maximize their own total. Player one goes first. Return the score of player one minus the score of player two under optimal play. 1 <= n <= 300, 0 <= piles[i] <= 10^4.
Implement
stone_game_diff(piles: list[int]) → intExamples
in
[[5,3,4,5]]out1What 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 30 min
solution.py
InputExpectedGot
[[5,3,4,5]]1not run yetsample