Find anagram mappings

Time: O(N); Space: O(N); easy

Given two lists A and 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.

Example 1:

Input: A = [12, 28, 46, 32, 50], B = [50, 12, 32, 46, 28]

Output: [1, 4, 3, 2, 0]

Explanation:

  • 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.

Example 2:

Input: A = [1, 2, 3, 4, 5], B = [5, 4, 3, 2, 1]

Output: [4, 3, 2, 1, 0]

Explanation:

  • As P[0] = 4 because the 0th element of A appears at B[4], and P[1] = 3 because the 1st element of A appears at B[3], and so on.

Constraints:

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

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

[1]:

import collections
class Solution1(object):
    """
    Time: O(N)
    Space: O(N)
    """
    def anagramMappings(self, A, B):
        """
        :type A: List[int]
        :type B: List[int]
        :rtype: List[int]
        """
        lookup = collections.defaultdict(collections.deque)

        for i, n in enumerate(B):
            lookup[n].append(i)
        result = []

        for n in A:
            result.append(lookup[n].popleft())

        return result

[2]:

s = Solution1()

A = [12, 28, 46, 32, 50]
B = [50, 12, 32, 46, 28]
assert s.anagramMappings(A, B) == [1, 4, 3, 2, 0]

A = [1, 2, 3, 4, 5]
B = [5, 4, 3, 2, 1]
assert s.anagramMappings(A, B) == [4, 3, 2, 1, 0]