760. Find Anagram Mappings

760. Find Anagram Mappings


Given two lists Aand B, and B is an anagram of A. B is an anagram of A means B is made by randomizing the order of the elements in A.
We want to find an index mapping P, from A to B. A mapping P[i] = j means the ith element in A appears in B at index j.
These lists A and B may contain duplicates. If there are multiple answers, output any of them.
For example, given
A = [12, 28, 46, 32, 50]
B = [50, 12, 32, 46, 28]

We should return
[1, 4, 3, 2, 0]
as P[0] = 1 because the 0th element of A appears at B[1], and P[1] = 4 because the 1st element of A appears at B[4], and so on.




Solution 1 : 

Use an int[] array to store the res 
Use a helper func to get the index of A[I] in B and put the index at res[I]
Return res


class Solution {
    public int[] anagramMappings(int[] A, int[] B) {
        int[] res = new int[A.length];
        for(int i = 0; i < A.length; i++){
            res[i] = helper(A[i], B);
        }
        return res;
    }
    private int helper(int target, int[] B){
        for(int i = 0; i < B.length; i++){
            if(B[i] == target) return i;
        }
        return -1;
    }
}


Solution 2 : use a hash map to optimize time 


class Solution {
    public int[] anagramMappings(int[] A, int[] B) {
        // use a hashmap to extract information from B, key is the element value in B, value is the index of this element in B 
        // loop thru every element in A and loop up its index in B 
        int[] res = new int[A.length];
        HashMap<Integer, Integer> map = new HashMap<>();
        for(int i = 0; i < B.length; i++){
            map.put(B[i], i);
        }
        
        for(int j = 0; j < A.length; j++){
            res[j] = map.get(A[j]);
        }
        return res;
        
    }
}




class Solution {
    public int[] anagramMappings(int[] A, int[] B) {
        // use a hashmap to extract information from B, key is the element value in B, value is the index of this element in B 
        // loop thru every element in A and loop up its index in B 
        int[] res = new int[A.length];
        HashMap<Integer, Integer> map = new HashMap<>();
        for(int i = 0; i < B.length; i++){
            map.put(B[i], i);
        }
        
        for(int i = 0; i < A.length; i++){ // also okay to use i , same var name as above, is it good practice tho? 
            res[i] = map.get(A[i]);
        }
        return res;
        
    }
}