760. Find Anagram Mappings
Tags: ‘Hash Table’
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]
.
Solution
public int[] anagramMappings(int[] A, int[] B) {
Map<Integer, Integer> map = new HashMap<>();
int[] result = new int[A.length];
for (int i = 0; i < B.length; i++) {
map.put(B[i], i);
}
int i = 0;
for (int a: A) {
result[i++] = map.get(a);
}
return result;
}