last-stone-weight-ii

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1049. Last Stone Weight II (Medium)

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

Example 3:

Input: stones = [1,2]
Output: 1

 

Constraints:

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

Related Topics

[Array] [Dynamic Programming]

Hints

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.)