2766 Find The Prefix Common Array Of Two Arrays

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.

use std::collections::HashMap;
impl Solution {
    pub fn find_the_prefix_common_array(a: Vec<i32>, b: Vec<i32>) -> Vec<i32> {
        
        let mut mp = HashMap::new();
        let mut res = Vec::new();

        let mut i = 0;
        let mut j = 0;
        let mut temp = 0;
        while(i < a.len() && i < b.len()) {
            *mp.entry(a[i]).or_insert(0) += 1;
            *mp.entry(b[i]).or_insert(0) += 1;
            //print!("{:?}", mp);
            if mp[&a[i]] % 2 == 0 {
                temp += 1;
            }
            if a[i] != b[i] && mp[&b[i]] % 2 == 0 {
                temp += 1;
            }
            res.push(temp);
            i += 1;
        }
        return res;



    }
}