README_EN.md

760. Find Anagram Mappings

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Description

You are given two integer arrays nums1 and nums2 where nums2 is an anagram of nums1. Both arrays may contain duplicates.

Return an index mapping array mapping from nums1 to nums2 where mapping[i] = j means the ith element in nums1 appears in nums2 at index j. If there are multiple answers, return any of them.

An array a is an anagram of an array b means b is made by randomizing the order of the elements in a.

 

Example 1:

Input: nums1 = [12,28,46,32,50], nums2 = [50,12,32,46,28]
Output: [1,4,3,2,0]
Explanation: As mapping[0] = 1 because the 0th element of nums1 appears at nums2[1], and mapping[1] = 4 because the 1st element of nums1 appears at nums2[4], and so on.

Example 2:

Input: nums1 = [84,46], nums2 = [84,46]
Output: [0,1]

 

Constraints:

• 1 <= nums1.length <= 100
• nums2.length == nums1.length
• 0 <= nums1[i], nums2[i] <= 105
• nums2 is an anagram of nums1.

Solutions

Python3

class Solution:
    def anagramMappings(self, nums1: List[int], nums2: List[int]) -> List[int]:
        mapper = 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) {
            map.computeIfAbsent(nums2[i], k -> new HashSet<>()).add(i);
        }
        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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