760. Find Anagram Mappings
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 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[0] = 1
because the 0
th element of A
appears at B[1]
, and P[1] = 4
because the 1
st element of A
appears at B[4]
, and so on.Note:
A, B
have equal lengths in range[1, 100]
.A[i], B[i]
are integers in range[0, 10^5]
.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 | class Solution { public int[] anagramMappings(int[] A, int[] B) { int n = A.length; Map<Integer, Set<Integer>> mapB = new HashMap<>(); for(int i=0; i<n; ++i){ mapB.computeIfAbsent(B[i], k->new HashSet<>()).add(i); } int[] ret = new int[n]; for(int i=0; i<n; ++i){ int v = A[i]; int index = mapB.get(v).iterator().next(); mapB.get(v).remove(index); ret[i] = index; } return ret; } } |
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