760 Find Anagram Mappings
Given two lists
A
andB
, andB
is an anagram ofA
.B
is an anagram ofA
meansB
is made by randomizing the order of the elements inA
.We want to find anindex mapping
P
, fromA
toB
. A mappingP[i] = j
means thei
th element inA
appears inB
at indexj
.These lists
A
andB
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.
Note:
- 1.A, B have equal lengths in range [1, 100].
- 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
Last modified 4yr ago