Find Anagram Mappings

• LeetCode: Find Anagram Mapping

• Problem:

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.

• Note:

1. A, B have equal lengths in range [1, 100].

2. A[i], B[i] are integers in range [0, 10^5].

• Examples:

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.

• Analysis:

Use a hash map to store <value, index> of array B. 

Then, use value in A to get index in the hash map.

• Code - Java:

class Solution {
    public int[] anagramMappings(int[] A, int[] B) {
        int len = A.length;
        int[] result = new int[len];
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        for (int i = 0; i < len; i++) {
            map.put(B[i], i);
        }
        for (int i = 0; i < len; i++) {
            result[i] = map.get(A[i]);
        }
        return result;
    }
}

• Code - C++:

class Solution {
public:
    vector<int> anagramMappings(vector<int>& A, vector<int>& B) {
        map<int, int> index;
        int size = B.size();
        for (int i = 0; i < size; i++) {
            index.insert(pair<int, int>(B[i], i));
        }
        vector<int> result;
        for (int i = 0; i < size; i++) {
            result.push_back(index[A[i]]);
        }
        return result;
    }
};

• Code - Python:

class Solution(object):
    def anagramMappings(self, A, B):
        """
        :type A: List[int]
        :type B: List[int]
        :rtype: List[int]
        """

        index = {}
        BSize = len(B)
        for i in range(BSize):
            index[B[i]] = i
        result = []
        for value in A:
            result.append(index[value])
        return result

• Time Complexity: O(N), N is the length of array B

• Space Complexity: O(N), N is the length of array B