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Leetcode1049-Last Stone Weight II

Posted on Edited on In Views: Valine:
Solution Report of LeetCode Acceptted

Description

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

Example 1:

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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

This question eaquals to partition an array into 2 subsets whose difference is minimal

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(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

  • Time Complexity: S = Sum(A), N = Length(A)
  • Space Complexity:

Solution 1: 2-D dp array

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

Solution 2: 1-D dp array

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