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2179. Count Good Triplets in an Array

Description

You are given two 0-indexed arrays nums1 and nums2 of length n, both of which are permutations of [0, 1, ..., n - 1].

A good triplet is a set of 3 distinct values which are present in increasing order by position both in nums1 and nums2. In other words, if we consider pos1v as the index of the value v in nums1 and pos2v as the index of the value v in nums2, then a good triplet will be a set (x, y, z) where 0 <= x, y, z <= n - 1, such that pos1x < pos1y < pos1z and pos2x < pos2y < pos2z.

Return the total number of good triplets.

 

Example 1:

Input: nums1 = [2,0,1,3], nums2 = [0,1,2,3]
Output: 1
Explanation: 
There are 4 triplets (x,y,z) such that pos1x < pos1y < pos1z. They are (2,0,1), (2,0,3), (2,1,3), and (0,1,3). 
Out of those triplets, only the triplet (0,1,3) satisfies pos2x < pos2y < pos2z. Hence, there is only 1 good triplet.

Example 2:

Input: nums1 = [4,0,1,3,2], nums2 = [4,1,0,2,3]
Output: 4
Explanation: The 4 good triplets are (4,0,3), (4,0,2), (4,1,3), and (4,1,2).

 

Constraints:

  • n == nums1.length == nums2.length
  • 3 <= n <= 105
  • 0 <= nums1[i], nums2[i] <= n - 1
  • nums1 and nums2 are permutations of [0, 1, ..., n - 1].

Solutions

Solution 1

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class BinaryIndexedTree:
    def __init__(self, n):
        self.n = n
        self.c = [0] * (n + 1)

    @staticmethod
    def lowbit(x):
        return x & -x

    def update(self, x, delta):
        while x <= self.n:
            self.c[x] += delta
            x += BinaryIndexedTree.lowbit(x)

    def query(self, x):
        s = 0
        while x > 0:
            s += self.c[x]
            x -= BinaryIndexedTree.lowbit(x)
        return s


class Solution:
    def goodTriplets(self, nums1: List[int], nums2: List[int]) -> int:
        pos = {v: i for i, v in enumerate(nums2, 1)}
        ans = 0
        n = len(nums1)
        tree = BinaryIndexedTree(n)
        for num in nums1:
            p = pos[num]
            left = tree.query(p)
            right = n - p - (tree.query(n) - tree.query(p))
            ans += left * right
            tree.update(p, 1)
        return ans

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class Solution {
    public long goodTriplets(int[] nums1, int[] nums2) {
        int n = nums1.length;
        int[] pos = new int[n];
        BinaryIndexedTree tree = new BinaryIndexedTree(n);
        for (int i = 0; i < n; ++i) {
            pos[nums2[i]] = i + 1;
        }
        long ans = 0;
        for (int num : nums1) {
            int p = pos[num];
            long left = tree.query(p);
            long right = n - p - (tree.query(n) - tree.query(p));
            ans += left * right;
            tree.update(p, 1);
        }
        return ans;
    }
}

class BinaryIndexedTree {
    private int n;
    private int[] c;

    public BinaryIndexedTree(int n) {
        this.n = n;
        c = new int[n + 1];
    }

    public void update(int x, int delta) {
        while (x <= n) {
            c[x] += delta;
            x += lowbit(x);
        }
    }

    public int query(int x) {
        int s = 0;
        while (x > 0) {
            s += c[x];
            x -= lowbit(x);
        }
        return s;
    }

    public static int lowbit(int x) {
        return x & -x;
    }
}

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class BinaryIndexedTree {
public:
    int n;
    vector<int> c;

    BinaryIndexedTree(int _n)
        : n(_n)
        , c(_n + 1