统计为蚁群构筑房间的不同顺序

// Problem Statement: Similar to the Java solution above, we want to find the number of different orders to construct rooms with constraints using Java Streams.

import java.util.*;

public class AntRoomConstructionStream {
    private static final int MOD = 1000000007;

    public int waysToBuildRooms(int[] prevRoom) {
        int n = prevRoom.length;
        List<List<Integer>> graph = new ArrayList<>();
        Arrays.setAll(graph, ArrayList::new);
        IntStream.range(1, n).forEach(i -> graph.get(prevRoom[i]).add(i));

        long[] fact = new long[n + 1];
        long[] invFact = new long[n + 1];
        fact[0] = invFact[0] = 1;
        IntStream.rangeClosed(1, n).forEach(i -> {
            fact[i] = fact[i - 1] * i % MOD;
            invFact[i] = pow(fact[i], MOD - 2);
        });

        return (int) dfs(graph, 0, fact, invFact)[1];
    }

    private long[] dfs(List<List<Integer>> graph, int node, long[] fact, long[] invFact) {
        return graph.get(node).stream().mapToLong(next -> dfs(graph, next, fact, invFact)[0]).reduce(new long[]{1, 1}, (acc, x) -> {
            acc[0] += x;
            acc[1] = acc[1] * dfs(graph, node, fact, invFact)[1] % MOD * invFact[(int) x] % MOD;
            return acc;
        }, (x, y) -> x);
    }

    private long pow(long a, long b) {
        long res = 1;
        while (b > 0) {
            if ((b & 1) == 1) res = res * a % MOD;
            a = a * a % MOD;
            b >>= 1;
        }
        return res;
    }
}