Burst balloons
Given positive integer balloon values, you burst them one by one. Bursting balloon i earns left * value[i] * right coins, where left and right are the values of the adjacent surviving balloons (treat out-of-range neighbors as 1). After bursting, neighbors become adjacent. Return the maximum coins from bursting all balloons. The list has at least one balloon.
Implement
burst_balloons(nums: list[int]) → intExamples
in
[[3,1,5,8]]out167What 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
[[3,1,5,8]]167not run yetsample