README_EN.md

760. Find Anagram Mappings

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Description

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].

Solutions

Python3

class Solution:
    def anagramMappings(self, nums1: List[int], nums2: List[int]) -> List[int]:
        mapper = collections.defaultdict(set)
        for i, num in enumerate(nums2):
            mapper[num].add(i)
        return [mapper[num].pop() for num in nums1]

Java

class Solution {
    public int[] anagramMappings(int[] nums1, int[] nums2) {
        Map<Integer, Set<Integer>> map = new HashMap<>();
        for (int i = 0; i < nums2.length; ++i) {
            Set<Integer> s = map.getOrDefault(nums2[i], new HashSet<>());
            s.add(i);
            map.put(nums2[i], s);
        }
        int[] res = new int[nums1.length];
        for (int i = 0; i < nums1.length; ++i) {
            int idx = map.get(nums1[i]).iterator().next();
            res[i] = idx;
            map.get(nums1[i]).remove(idx);
        }
        return res;
    }
}

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