Find the Prefix Common Array of Two Arrays(Array,Hash Table,Bit Manipulation)

https://leetcode.com/problems/find-the-prefix-common-array-of-two-arrays/description/?envType=daily-question&envId=2025-01-14

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.

글 쓰다가 글 날라가서 대충 쓴다..

 

1. Python

# Super Naive way
class Solution:
    def findThePrefixCommonArray(self, A: List[int], B: List[int]) -> List[int]:
        # Navie way
        cnt = 0
        ans = []
        n = len(A)
        for i in range(n):
            cnt = 0
            for j in range(i,-1,-1):
                for k in range(i,-1,-1):
                    if B[j]==A[k] : cnt+=1
            ans.append(cnt)
        return ans

 

2. C

# C
/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* findThePrefixCommonArray(int* A, int ASize, int* B, int BSize, int* returnSize) {
    *returnSize = ASize;
    int cnt = 0;
    int* ans = (int*)malloc(ASize*sizeof(int));
    int check[51] = {0};

    for(int i=0;i<ASize;i++){
        check[A[i]]++;
        if(check[A[i]]==2) cnt++;
        check[B[i]]++;
        if(check[B[i]]==2) cnt++;
        ans[i] = cnt;
    }
    return ans;
}

 

 

3. C++

# C++
class Solution {
public:
    vector<int> findThePrefixCommonArray(vector<int>& A, vector<int>& B) {
        int cnt = 0;
        vector<int> vec;
        int n = A.size();
        vector<int> check(n+1,0);
        for(int i=0;i<n;i++){
            check[A[i]]++;
            if(check[A[i]]==2) cnt++;
            check[B[i]]++;
            if(check[B[i]]==2) cnt++;
            vec.push_back(cnt);
        }
        return vec;
    }
};