2657. Find the Prefix Common Array of Two Arrays

Problem

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.

Solution (Java)

class Solution {
    public int[] findThePrefixCommonArray(int[] A, int[] B) {
                int []com = new int[A.length];
        HashSet<Integer> setA = new HashSet<>();
        HashSet<Integer> setB = new HashSet<>();
        for (int i = 0; i < A.length; i++)
        {
            setA.add(A[i]);
            setB.add(B[i]);
            HashSet<Integer> tmp = new HashSet<>(setA);
            tmp.retainAll(setB);
            com[i] = tmp.size();
        }
        return com;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).