// 1049. Last Stone Weight II

//

// 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

//

// Runtime 2 ms Beats 94.10%

// Memory 40.2 MB Beats 83.25%

class Solution {

public int lastStoneWeightII(int[] stones) {

boolean[] canGetSum = new boolean[1501];

canGetSum[0] = true;

int sum = 0;

for (int i = 0; i < stones.length; i++) {

sum = sum + stones[i];

for (int j = Math.min(sum, 1500); j >= stones[i]; j--) {

canGetSum[j] = canGetSum[j] || canGetSum[j - stones[i]];

}

}

for (int i = sum / 2; i >= 0; i--) {

if (canGetSum[i] == true) {

return sum - i * 2;

}

}

return -1;

}

}