连通两组点的最小成本

/*
 * Solution Overview:
 * Similar to the Java solution, but utilizing Java Streams for concise and readable code. We will still use dynamic programming to keep track of the minimum costs for each combination of connected points.
 */

import java.util.List;
import java.util.stream.IntStream;
import java.util.Arrays;

public class SolutionStream {
    public int connectTwoGroups(List<List<Integer>> cost) {
        int size1 = cost.size();
        int size2 = cost.get(0).size();
        int maxMask = 1 << size1;
        int[] dp = new int[maxMask];
        Arrays.fill(dp, Integer.MAX_VALUE);
        dp[0] = 0;

        // Using Java Streams to calculate minimum costs
        IntStream.range(0, maxMask).forEach(mask ->
            IntStream.range(0, size2).forEach(j -> {
                int currentCost = IntStream.range(0, size1)
                    .filter(i -> (mask & (1 << i)) != 0)
                    .map(i -> cost.get(i).get(j))
                    .sum();
                IntStream.iterate(mask, subMask -> (subMask - 1) & mask)
                    .limit(1 << size1)
                    .forEach(subMask -> dp[mask] = Math.min(dp[mask], dp[subMask] + currentCost));
            })
        );

        return Arrays.stream(dp).min().orElse(Integer.MAX_VALUE);
    }
}