Created
April 19, 2021 00:02
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class Solution { | |
// A variant of the knapsack problem | |
// dp[i][j]: is there a subset of stones (0 to i-1) can sum up to weight j | |
// dp[i][j] = dp[i-1][j] || dp[i-1][j-nums[i]] | |
// result: last stone weight = the shortest distance between any two nodes among dp[n][0] ... dp[n][W] | |
// | |
// Time: O(n * totalWeight) | |
// Space: O(n * totalWeight) | |
public int lastStoneWeightII(int[] stones) { | |
int totalWeight = Arrays.stream(stones).sum(); | |
boolean[][] dp = new boolean[stones.length + 1][totalWeight/2 + 1]; | |
dp[0][0] = true; | |
for (int i = 1; i <= stones.length; i++) { | |
int currStone = stones[i-1]; | |
for (int j = 0; j < dp[0].length; j++) { | |
if (dp[i-1][j] || (j >= currStone && dp[i-1][j-currStone])) { | |
dp[i][j] = true; | |
} | |
} | |
} | |
for (int i = dp[0].length - 1; i >= 0; i--) { | |
if (dp[stones.length][i]) { | |
return totalWeight - i - i; | |
} | |
} | |
return 0; | |
} | |
} |
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