2657. Find the Prefix Common Array of Two Arrays
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.
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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