Last stone weight II

Time: O(2^N); Space: O(2^N); medium

We have a collection of rocks, each rock has a positive integer weight.

Each turn, we choose any two rocks 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 totally destroyed;

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

At the end, there is at most 1 stone left. Return the smallest possible weight of this stone (the weight is 0 if there are no stones left.)

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.

Constraints:

  • 1 <= len(stones) <= 30

  • 1 <= stones[i] <= 100

Hints:

  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.

  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.)

1. Dynamic programming [O(2^N), O(2^N)]

[1]:

class Solution1(object):
    """
    Time: O(2^N)
    Space: O(2^N)
    """
    def lastStoneWeightII(self, stones):
        """
        :type stones: List[int]
        :rtype: int
        """
        dp = {0}

        for stone in stones:
            dp |= {stone+i for i in dp}
        S = sum(stones)

        return min(abs(i-(S-i)) for i in dp)

[2]:

s = Solution1()

stones = [2,7,4,1,8,1]
assert s.lastStoneWeightII(stones) == 1