统计数组中好三元组数目

// Java Stream solution for counting good triplets
// Problem: Given two arrays nums1 and nums2, find the number of good triplets (x, y, z)
// where the order of elements in both arrays is the same.

// Approach:
// 1. Use a stream to map the elements of nums1 and nums2 to their indices.
// 2. Filter the triplets in nums1 and check if the indices in nums2 follow the same order.
// 3. Count such triplets.

import java.util.HashMap;
import java.util.Map;
import java.util.stream.IntStream;

public class GoodTripletsStream {
    public long countGoodTriplets(int[] nums1, int[] nums2) {
        int n = nums1.length;
        Map<Integer, Integer> posInNums1 = new HashMap<>();
        Map<Integer, Integer> posInNums2 = new HashMap<>();

        // Store the positions of elements in nums1 and nums2
        IntStream.range(0, n).forEach(i -> {
            posInNums1.put(nums1[i], i);
            posInNums2.put(nums2[i], i);
        });

        // Count good triplets
        return IntStream.range(0, n)
                .boxed()
                .flatMap(i -> IntStream.range(i + 1, n)
                        .boxed()
                        .flatMap(j -> IntStream.range(j + 1, n)
                                .mapToObj(k -> new int[]{i, j, k})))
                .filter(triplet -> {
                    int pos1 = posInNums2.get(nums1[triplet[0]]);
                    int pos2 = posInNums2.get(nums1[triplet[1]]);
                    int pos3 = posInNums2.get(nums1[triplet[2]]);
                    return pos1 < pos2 && pos2 < pos3;
                })
                .count();
    }
}