1049. Last Stone Weight II

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: [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.



Note:

    1 <= stones.length <= 30
    1 <= stones[i] <= 100
  • 01 Knapsack problem
    • cost array == value array
    • capacity == sum / 2

public int lastStoneWeightII(int[] stones) {
    int sum = 0;
    for(int i : stones) sum += i;
    int V = sum / 2;
    int[] dp = new int[V+1];
    for(int stone : stones)
        for(int j = V; j >= stone; j--)
            dp[j] = Math.max(dp[j], dp[j-stone]+stone);
    return sum - 2 * dp[V];
}