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 <= 501 <= A[i], B[i] <= nIt is guaranteed that A and B are both a permutation of n integers.
Solutions
This function finds the prefix common array between two arrays A and B. It uses a frequency array to track which numbers have appeared in both arrays: incrementing the count for elements from A and decrementing for elements from B. The key insight is that when freq[A[i]] transitions from negative to positive (or reaches 0), it means A[i] was previously seen in B, indicating a new common element; similarly, when freq[B[i]] transitions from positive to negative (or reaches 0), it means B[i] was previously seen in A. The count variable accumulates these matches, and for each index i, the result stores how many elements are common to both the prefix A[0...i] and B[0...i].
/**
* @param {number[]} A
* @param {number[]} B
* @return {number[]}
*/
var findThePrefixCommonArray = function(A, B) {
const n = A.length;
const freq = Array(n + 1).fill(0);
const res = Array(n);
let count = 0;
for (let i = 0; i < n; i++) {
freq[A[i]]++;
if (freq[A[i]] <= 0) count++;
freq[B[i]]--;
if (freq[B[i]] >= 0) count++;
res[i] = count;
}
return res;
};