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2657. Find the Prefix Common Array of Two Arrays

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 <= 50
• 1 <= A[i], B[i] <= n
• It is guaranteed that A and B are both a permutation of n integers.

Solutions

• Java
• C++
• Python
• Go
• TypeScript

• 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;
      }
  }
  

• 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;
      }
  };
  

• class Solution:
      def findThePrefixCommonArray(self, A: List[int], B: List[int]) -> List[int]:
          ans = []
          cnt1 = Counter()
          cnt2 = Counter()
          for a, b in zip(A, B):
              cnt1[a] += 1
              cnt2[b] += 1
              t = sum(min(v, cnt2[x]) for x, v in cnt1.items())
              ans.append(t)
          return ans
  
  

• func findThePrefixCommonArray(A []int, B []int) []int {
  	n := len(A)
  	cnt1 := make([]int, n+1)
  	cnt2 := make([]int, n+1)
  	ans := make([]int, n)
  	for i, a := range A {
  		b := B[i]
  		cnt1[a]++
  		cnt2[b]++
  		for j := 1; j <= n; j++ {
  			ans[i] += min(cnt1[j], cnt2[j])
  		}
  	}
  	return ans
  }
  
  func min(a, b int) int {
  	if a < b {
  		return a
  	}
  	return b
  }
  

• function findThePrefixCommonArray(A: number[], B: number[]): number[] {
      const n = A.length;
      const cnt1: number[] = new Array(n + 1).fill(0);
      const cnt2: number[] = new Array(n + 1).fill(0);
      const ans: number[] = new Array(n).fill(0);
      for (let i = 0; i < n; ++i) {
          ++cnt1[A[i]];
          ++cnt2[B[i]];
          for (let j = 1; j <= n; ++j) {
              ans[i] += Math.min(cnt1[j], cnt2[j]);
          }
      }
      return ans;
  }
  
  

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