# 2657. Find the Prefix Common Array of Two Arrays

## 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.
At i = 1: 1 and 3 are common in A and B, so C = 2.
At i = 2: 1, 2, and 3 are common in A and B, so C = 3.
At i = 3: 1, 2, 3, and 4 are common in A and B, so C = 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.
At i = 1: only 3 is common in A and B, so C = 1.
At i = 2: 1, 2, and 3 are common in A and B, so C = 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

• 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;
}
}

• 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;
}
};

• 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


• 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
}

• 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;
}