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
Solution:
class Solution {
/**
This question eaquals to partition an array into 2 subsets whose difference is minimal
(1) S1 + S2 = S
(2) S1 - S2 = diff
==> -> diff = S - 2 * S2 ==> minimize diff equals to maximize S2
Now we should find the maximum of S2 , range from 0 to S / 2, using dp can solve this
dp[i][j] = {true if some subset from 1st to j'th has a sum equal to sum i, false otherwise}
i ranges from (sum of all elements) {1..n}
j ranges from {1..n}
same as 494. Target Sum
*/
public int lastStoneWeightII(int[] stones) {
int S = 0, S2 = 0;
for (int s : stones) S += s;
int n = stones.length;
boolean[][] dp = new boolean[S + 1][n + 1];
for (int i = 0; i <= n; i++) {
dp[0][i] = true;
}
for (int i = 1; i <= n; i++) {
for (int s = 1; s <= S / 2; s++) {
if (dp[s][i - 1] || (s >= stones[i - 1] && dp[s - stones[i - 1]][i - 1])) {
dp[s][i] = true;
S2 = Math.max(S2, s);
}
}
}
return S - 2 * S2;
}
}