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 pos1_v as the index of the value v in nums1 and pos2_v 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 pos1_x < pos1_y < pos1_z and pos2_x < pos2_y < pos2_z.

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 pos1_x < pos1_y < pos1_z. 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 pos2_x < pos2_y < pos2_z. 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: Binary Indexed Tree (Fenwick Tree)

For this problem, we first use pos to record the position of each number in nums2, and then process each element in nums1 sequentially.

Consider the number of good triplets with the current number as the middle number. The first number must have already been traversed and must appear earlier than the current number in nums2. The third number must not yet have been traversed and must appear later than the current number in nums2.

Take nums1 = [4,0,1,3,2] and nums2 = [4,1,0,2,3] as an example. Consider the traversal process:

First, process 4. At this point, the state of nums2 is [4,X,X,X,X]. The number of values before 4 is 0, and the number of values after 4 is 4. Therefore, 4 as the middle number forms 0 good triplets.
Next, process 0. The state of nums2 becomes [4,X,0,X,X]. The number of values before 0 is 1, and the number of values after 0 is 2. Therefore, 0 as the middle number forms 2 good triplets.
Next, process 1. The state of nums2 becomes [4,1,0,X,X]. The number of values before 1 is 1, and the number of values after 1 is 2. Therefore, 1 as the middle number forms 2 good triplets.
...
Finally, process 2. The state of nums2 becomes [4,1,0,2,3]. The number of values before 2 is 4, and the number of values after 2 is 0. Therefore, 2 as the middle number forms 0 good triplets.

We can use a Binary Indexed Tree (Fenwick Tree) to update the occurrence of numbers at each position in nums2, and quickly calculate the number of 1s to the left of each number and the number of 0s to the right of each number.

A Binary Indexed Tree, also known as a Fenwick Tree, efficiently supports the following operations:

Point Update update(x, delta): Add a value delta to the number at position x in the sequence.
Prefix Sum Query query(x): Query the sum of the sequence in the range [1, ..., x], i.e., the prefix sum at position x.

Both operations have a time complexity of \(O(\log n)\). Therefore, the overall time complexity is \(O(n \log n)\), where \(n\) is the length of the array \(\textit{nums1}\). The space complexity is \(O(n)\).

Python3JavaC++GoTypeScript

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

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;
    }
};

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
}

class BinaryIndexedTree {
    private c: number[];
    private n: number;

    constructor(n: number) {
        this.n = n;
        this.c = Array(n + 1).fill(0);
    }

    private static lowbit(x: number): number {
        return x & -x;
    }

    update(x: number, delta: number): void {
        while (x <= this.n) {
            this.c[x] += delta;
            x += BinaryIndexedTree.lowbit(x);
        }
    }

    query(x: number): number {
        let s = 0;
        while (x > 0) {
            s += this.c[x];
            x -= BinaryIndexedTree.lowbit(x);
        }
        return s;
    }
}

function goodTriplets(nums1: number[], nums2: number[]): number {
    const n = nums1.length;
    const pos = new Map<number, number>();
    nums2.forEach((v, i) => pos.set(v, i + 1));

    const tree = new BinaryIndexedTree(n);
    let ans = 0;

    for (const num of nums1) {
        const p = pos.get(num)!;
        const left = tree.query(p);
        const total = tree.query(n);
        const right = n - p - (total - left);
        ans += left * right;
        tree.update(p, 1);
    }

    return ans;
}

Solution 2: Segment Tree

We can also use a segment tree to solve this problem. A segment tree is a data structure that efficiently supports range queries and updates. The basic idea is to divide an interval into multiple subintervals, with each subinterval represented by a node.

The segment tree divides the entire interval into multiple non-overlapping subintervals, with the number of subintervals not exceeding log(width). To update the value of an element, we only need to update log(width) intervals, all of which are contained within a larger interval that includes the element.

Each node of the segment tree represents an interval.
The segment tree has a unique root node, representing the entire range, such as [1, N].
Each leaf node of the segment tree represents a unit interval [x, x].
For each internal node [l, r], its left child represents [l, mid], and its right child represents [mid + 1, r], where mid = ⌊(l + r) / 2⌋ (floor division).

The time complexity is \(O(n \log n)\), where \(n\) is the length of the array \(\textit{nums1}\). The space complexity is \(O(n)\).