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 <= 501 <= A[i], B[i] <= nIt is guaranteed that A and B are both a permutation of n integers.class Solution {
public int[] findThePrefixCommonArray(int[] A, int[] B) {
int n = A.length, cumulative = 0;
int[] visited = new int[n + 1], prefixCommon = new int[n];
for (int i = 0; i < n; i++) {
if (++visited[A[i]] == 2)
cumulative++;
if (++visited[B[i]] == 2)
cumulative++;
prefixCommon[i] = cumulative;
}
return prefixCommon;
}
}