algorithm notes: leetcode#760 find anagram mappings

Problem

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 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].

Soultion

Basic idea

P[i] = j, A[i] = B[j]. Traverse list A, map each value of A to its index in B. And before that, we need to traverse B, and find a hash table that can map value of B to its index.

Python implementation

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class :
    def anagramMappings(self, A, B):
        """
        :type A: List[int]
        :type B: List[int]
        :rtype: List[int]
        """
        lookup = { val: i for i, val in enumerate(B)}
        return [lookup[val] for val in A]

Java implementation

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class  {
    public int[] anagramMappings(int[] A, int[] B) {
        Map<Integer, Integer> lookup = new HashMap();
        
        for(int i = 0; i < B.length; i++)
            lookup.put(B[i], i);
        
        int[] ans = new int[A.length];
        for(int j = 0; j < ans.length; j++)
            ans[j] = lookup.get(A[j]);
        
        return ans;
    }
}

Time complexity analysis

Traverse list A and B. O(n).

Space complexity analysis

Use a hash table to store the mapping from value of B to its index. O(n).

Link

760. Find Anagram Mappings
(中文版) 算法笔记: Leetcode#760 Find Anagram Mappings