2766. Find The Prefix Common Array Of Two Arrays¶

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2766. Find the Prefix Common Array of Two Arrays

Medium

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¶

class Solution {
    public int[] findThePrefixCommonArray(int[] A, int[] B) {
        int n = A.length;
        HashSet<Integer> set1 = new HashSet<>();
        HashSet<Integer> set2 = new HashSet<>();
        int res[] = new int[n];
        for (int i = 0; i < n; i++) {
            set1.add(A[i]); set2.add(B[i]);
            int count = 0;
            for (int j = 1; j <= 50; j++) if (set1.contains(j) && set2.contains(j)) count++;
            res[i] = count;
        }   
        return res;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here