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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:

Related Topics:
Hash Table

## Solution 1.

• 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++)
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) {
}
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):
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]
"""