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].
If the mapping index cannot be re-used (as there might be duplicates in the arrays), we need to maintain a vector of indices. The C++ code below illustrates the mapping algorithm
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 | class Solution { public: vector<int> anagramMappings(vector<int>& A, vector<int>& B) { unordered_map<int, vector<int>> table; for (int i = 0; i < B.size(); ++ i) { if (table.find(B[i]) == table.end()) { table[B[i]] = {i}; } else { table[B[i]].push_back(i); // add duplicate index } } vector<int> r; for (int i = 0; i < A.size(); ++ i) { r.push_back(table[A[i]].back()); // pick a mapping index table[A[i]].pop_back(); // remove it from the queue } return r; } }; |
class Solution { public: vector<int> anagramMappings(vector<int>& A, vector<int>& B) { unordered_map<int, vector<int>> table; for (int i = 0; i < B.size(); ++ i) { if (table.find(B[i]) == table.end()) { table[B[i]] = {i}; } else { table[B[i]].push_back(i); // add duplicate index } } vector<int> r; for (int i = 0; i < A.size(); ++ i) { r.push_back(table[A[i]].back()); // pick a mapping index table[A[i]].pop_back(); // remove it from the queue } return r; } };
The complexity is O(N) and the space complexity is O(N) as well. If the same index can be used to map duplicate numbers, the above can be simplified by using a hash map i.e. unordered_map in C++.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 | class Solution { public: vector<int> anagramMappings(vector<int>& A, vector<int>& B) { unordered_map<int, int> table; for (int i = 0; i < B.size(); ++ i) { table[B[i]] = {i}; } vector<int> r; for (int i = 0; i < A.size(); ++ i) { r.push_back(table[A[i]]); } return r; } }; |
class Solution { public: vector<int> anagramMappings(vector<int>& A, vector<int>& B) { unordered_map<int, int> table; for (int i = 0; i < B.size(); ++ i) { table[B[i]] = {i}; } vector<int> r; for (int i = 0; i < A.size(); ++ i) { r.push_back(table[A[i]]); } return r; } };
The same anagram mapping algorithm can be implemented by Java using the HashMap i.e. O(N) both time and space complexity.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 | class Solution { public int[] anagramMappings(int[] A, int[] B) { Map<Integer, Integer> D = new HashMap(); for (int i = 0; i < B.length; ++i) { D.put(B[i], i); } int[] ans = new int[A.length]; int t = 0; for (int x: A) { ans[t++] = D.get(x); } return ans; } } |
class Solution { public int[] anagramMappings(int[] A, int[] B) { Map<Integer, Integer> D = new HashMap(); for (int i = 0; i < B.length; ++i) { D.put(B[i], i); } int[] ans = new int[A.length]; int t = 0; for (int x: A) { ans[t++] = D.get(x); } return ans; } }
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