package com.fishercoder.solutions;

/**
 * 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
 * */
public class _1049 {
    public static class Solution1 {
        public int lastStoneWeightII(int[] stones) {
            //TODO: implement it
            return 1;
        }
    }
}

@@ -33,1033 +33,8 @@
 s;%0A%0A
-/**%0A * 1049. Last Stone Weight II%0A *%0A * We have a collection of rocks, each rock has a positive integer weight.%0A * Each turn, we choose any two rocks and smash them together.%0A * Suppose the stones have weights x and y with x %3C= y.  The result of this smash is:%0A * If x == y, both stones are totally destroyed;%0A * If x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y-x.%0A * 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.)%0A *%0A * Example 1:%0A * Input: %5B2,7,4,1,8,1%5D%0A * Output: 1%0A *%0A * Explanation:%0A * We can combine 2 and 4 to get 2 so the array converts to %5B2,7,1,8,1%5D then,%0A * we can combine 7 and 8 to get 1 so the array converts to %5B2,1,1,1%5D then,%0A * we can combine 2 and 1 to get 1 so the array converts to %5B1,1,1%5D then,%0A * we can combine 1 and 1 to get 0 so the array converts to %5B1%5D then that's the optimal value.%0A *%0A * Note:%0A * 1 %3C= stones.length %3C= 30%0A * 1 %3C= stones%5Bi%5D %3C= 100%0A * */%0A
 publ

refactor 1049