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1049. Last Stone Weight II

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You are given an array of integers stones where stones[i] is the weight of the ith stone.

We are playing a game with the stones. On each turn, we choose any two stones and smash them together. Suppose the stones have weights x and y with x <= y. The result of this smash is:

  • If x == y, both stones are destroyed, and
  • If x != y, the stone of weight x is destroyed, and the stone of weight y has new weight y - x.

At the end of the game, there is at most one stone left.

Return the smallest possible weight of the left stone. If there are no stones left, return 0.

Example 1:

Input: stones = [2,7,4,1,8,1]
Output: 1
Explanation:
We can combine 2 and 4 to get 2, so the array converts to [2,7,1,8,1] then,
we can combine 7 and 8 to get 1, so the array converts to [2,1,1,1] then,
we can combine 2 and 1 to get 1, so the array converts to [1,1,1] then,
we can combine 1 and 1 to get 0, so the array converts to [1], then that's the optimal value.

Example 2:

Input: stones = [31,26,33,21,40]
Output: 5

Constraints:

  • 1 <= stones.length <= 30
  • 1 <= stones[i] <= 100

Hints open one at a time

Hint 1
Think of the final answer as a sum of weights with + or - sign symbols infront of each weight. Actually, all sums with 1 of each sign symbol are possible.
Hint 2
Use dynamic programming: for every possible sum with N stones, those sums +x or -x is possible with N+1 stones, where x is the value of the newest stone. (This overcounts sums that are all positive or all negative, but those don't matter.)
19:04
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1class Solution:
2 def lastStoneWeightII(self, stones: List[int]) -> int:
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Starter codePython 3Spaces: 4
Examples
Input
[2,7,4,1,8,1]
Input
[31,26,33,21,40]
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