LeetCode [760] Find Anagram Mappings

 760. Find Anagram Mappings

Easy

Given two lists Aand 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.

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].

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; } }