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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:

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

Companies:
Google

Related Topics:
Hash Table

Solution 1.

• Java
• C++
• Python

• 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;
      }
  }
  
  ############
  
  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;
      }
  }
  

• // 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, 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]
  
  ############
  
  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
  

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