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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

Binary Indexed Tree or Segment Tree.

  • 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;
    }
}

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

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

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

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

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

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

  • 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


  • type BinaryIndexedTree struct {
	n int
	c []int
}

func newBinaryIndexedTree(n int) *BinaryIndexedTree {
	c := make([]int, n+1)
	return &BinaryIndexedTree{n, c}
}

func (this *BinaryIndexedTree) lowbit(x int) int {
	return x & -x
}

func (this *BinaryIndexedTree) update(x, delta int) {
	for x <= this.n {
		this.c[x] += delta
		x += this.lowbit(x)
	}
}

func (this *BinaryIndexedTree) query(x int) int {
	s := 0
	for x > 0 {
		s += this.c[x]
		x -= this.lowbit(x)
	}
	return s
}

func goodTriplets(nums1 []int, nums2 []int) int64 {
	n := len(nums1)
	pos := make([]int, n)
	for i, v := range nums2 {
		pos[v] = i + 1
	}
	tree := newBinaryIndexedTree(n)
	var ans int64
	for _, num := range nums1 {
		p := pos[num]
		left := tree.query(p)
		right := n - p - (tree.query(n) - tree.query(p))
		ans += int64(left) * int64(right)
		tree.update(p, 1)
	}
	return ans
}

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