760__easy__find-anagram-mappings.js

/**

    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]

 */
/**
 *
 * O(n) time
 * O(n) space
 *
 * Runtime: 52 ms, faster than 100.00%
 */
/**
 * @param {number[]} A
 * @param {number[]} B
 * @return {number[]}
 */
var anagramMappings = function(A, B) {
  if (!A || !B) return null;

  let map = {}; // {} is faster than Map()
  for (let i = 0; i < B.length; i++) {
    if (!map[B[i]]) map[B[i]] = i;
  }

  let res = new Array(A.length);
  for (let i = 0; i < A.length; i++) {
    res[i] = map[A[i]];
  }

  return res;
};