Formatted question description: https://leetcode.ca/all/760.html

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:

A, B have equal lengths in range [1, 100].
A[i], B[i] are integers in range [0, 10^5].

Companies:
Google

Related Topics:
Hash Table

Solution 1.

// OJ: https://leetcode.com/problems/find-anagram-mappings/

// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<int> anagramMappings(vector<int>& A, vector<int>& B) {
        unordered_map<int, int> m;
        for (int i = 0; i < B.size(); ++i) m[B[i]] = i;
        vector<int> ans;
        for (int n : A) ans.push_back(m[n]);
        return ans;
    }
};

Java

class Solution {
    public int[] anagramMappings(int[] A, int[] B) {
        int length = A.length;
        List<Integer> listB = new ArrayList<Integer>();
        for (int i = 0; i < length; i++)
            listB.add(B[i]);
        int[] anagram = new int[length];
        for (int i = 0; i < length; i++) {
            int num = A[i];
            int index = listB.indexOf(num);
            anagram[i] = index;
            listB.set(index, -1);
        }
        return anagram;
    }
}

// OJ: https://leetcode.com/problems/find-anagram-mappings/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<int> anagramMappings(vector<int>& A, vector<int>& B) {
        unordered_map<int, int> m;
        for (int i = 0; i < B.size(); ++i) m[B[i]] = i;
        vector<int> ans;
        for (int n : A) ans.push_back(m[n]);
        return ans;
    }
};

class Solution:
    def anagramMappings(self, A, B):
        """
        :type A: List[int]
        :type B: List[int]
        :rtype: List[int]
        """
        answer = []
        for a in A:
            for i,b in enumerate(B):
                if a == b:
                    answer.append(i)
                    break
        return answer

760 - Find Anagram Mappings

760. Find Anagram Mappings (Easy)

Solution 1.

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