2657. Find the Prefix Common Array of Two Arrays

You are given two 0-indexed integer permutations A and B of length n.

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.

JAVA

  • 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;
        }
    }

PYTHON

  • class Solution:
        def findThePrefixCommonArray(self, A: List[int], B: List[int]) -> List[int]:
            n = len(A)
            ans = [0] * n
            cnt1 = [0] * (n + 1)
            cnt2 = [0] * (n + 1)
            
            for i in range(n):
                cnt1[A[i]] += 1
                cnt2[B[i]] += 1
                ans[i] = sum(min(cnt1[j], cnt2[j]) for j in range(1, n + 1))
            
            return ans

C++

  • 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;
        }
    };

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