# 760 Find Anagram Mappings

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 anindex mapping`P`, from`A`to`B`. A mapping`P[i] = j`means the`i`th 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 = 1 because the 0th element of A appears at B, and P = 4 because the 1st element of A appears at B, and so on.
Note:
1. 1.
A, B have equal lengths in range [1, 100].
2. 2.
A[i], B[i] are integers in range [0, 10^5].
The Idea: Map the indices of `B` into a list. Then iterate through `A` and build a new list `P` that take the corresponding index of `B` that is mapped from a key of `A`. We solve the problem of duplicates by mapping to a list of elements, and simply taking the last one (as order does not pattern).
Complexity: O(N) time and space
import collections
class Solution:
def anagramMappings(self, A, B):
"""
:type A: List[int]
:type B: List[int]
:rtype: List[int]
"""
B_map = collections.defaultdict(list)
for i, b in enumerate(B):
B_map[b].append(i)
P = []
for a in A:
P.append(B_map[a].pop())
return P