README_EN.md

2657. Find the Prefix Common Array of Two Arrays

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Description

You are given two 0-indexed integer permutations A and B of length n.

A prefix common array of A and B is an array C such that C[i] is equal to the count of numbers that are present at or before the index i in both A and B.

Return the prefix common array of A and B.

A sequence of n integers is called a permutation if it contains all integers from 1 to n exactly once.

 

Example 1:

Input: A = [1,3,2,4], B = [3,1,2,4]
Output: [0,2,3,4]
Explanation: At i = 0: no number is common, so C[0] = 0.
At i = 1: 1 and 3 are common in A and B, so C[1] = 2.
At i = 2: 1, 2, and 3 are common in A and B, so C[2] = 3.
At i = 3: 1, 2, 3, and 4 are common in A and B, so C[3] = 4.

Example 2:

Input: A = [2,3,1], B = [3,1,2]
Output: [0,1,3]
Explanation: At i = 0: no number is common, so C[0] = 0.
At i = 1: only 3 is common in A and B, so C[1] = 1.
At i = 2: 1, 2, and 3 are common in A and B, so C[2] = 3.

 

Constraints:

• 1 <= A.length == B.length == n <= 50
• 1 <= A[i], B[i] <= n
• It is guaranteed that A and B are both a permutation of n integers.

Solutions

Solution 1: Count + Enumeration

We can use two arrays $cnt1$ and $cnt2$ to record the number of occurrences of each element in arrays $A$ and $B$, and use array $ans$ to record the answer.

Traverse arrays $A$ and $B$, add the number of occurrences of $A[i]$ in $cnt1$ by one, and add the number of occurrences of $B[i]$ in $cnt2$ by one. Then enumerate $j \in [1,n]$, calculate the minimum value of the number of occurrences of each element $j$ in $cnt1$ and $cnt2$, and add it to $ans[i]$.

After the traversal is over, return the answer array $ans$.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Where $n$ is the length of arrays $A$ and $B$.

Python3

class Solution:
    def findThePrefixCommonArray(self, A: List[int], B: List[int]) -> List[int]:
        ans = []
        cnt1 = Counter()
        cnt2 = Counter()
        for a, b in zip(A, B):
            cnt1[a] += 1
            cnt2[b] += 1
            t = sum(min(v, cnt2[x]) for x, v in cnt1.items())
            ans.append(t)
        return ans

Java

class Solution {
    public int[] findThePrefixCommonArray(int[] A, int[] B) {
        int n = A.length;
        int[] ans = new int[n];
        int[] cnt1 = new int[n + 1];
        int[] cnt2 = new int[n + 1];
        for (int i = 0; i < n; ++i) {
            ++cnt1[A[i]];
            ++cnt2[B[i]];
            for (int j = 1; j <= n; ++j) {
                ans[i] += Math.min(cnt1[j], cnt2[j]);
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> findThePrefixCommonArray(vector<int>& A, vector<int>& B) {
        int n = A.size();
        vector<int> ans(n);
        vector<int> cnt1(n + 1), cnt2(n + 1);
        for (int i = 0; i < n; ++i) {
            ++cnt1[A[i]];
            ++cnt2[B[i]];
            for (int j = 1; j <= n; ++j) {
                ans[i] += min(cnt1[j], cnt2[j]);
            }
        }
        return ans;
    }
};

Go

func findThePrefixCommonArray(A []int, B []int) []int {
	n := len(A)
	cnt1 := make([]int, n+1)
	cnt2 := make([]int, n+1)
	ans := make([]int, n)
	for i, a := range A {
		b := B[i]
		cnt1[a]++
		cnt2[b]++
		for j := 1; j <= n; j++ {
			ans[i] += min(cnt1[j], cnt2[j])
		}
	}
	return ans
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

TypeScript

function findThePrefixCommonArray(A: number[], B: number[]): number[] {
    const n = A.length;
    const cnt1: number[] = new Array(n + 1).fill(0);
    const cnt2: number[] = new Array(n + 1).fill(0);
    const ans: number[] = new Array(n).fill(0);
    for (let i = 0; i < n; ++i) {
        ++cnt1[A[i]];
        ++cnt2[B[i]];
        for (let j = 1; j <= n; ++j) {
            ans[i] += Math.min(cnt1[j], cnt2[j]);
        }
    }
    return ans;
}

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