Relative sort array

Time: O(NLogN); Space: O(N); easy

Given two arrays arr1 and arr2, the elements of arr2 are distinct, and all elements in arr2 are also in arr1.

Sort the elements of arr1 such that the relative ordering of items in arr1 are the same as in arr2.

Elements that don’t appear in arr2 should be placed at the end of arr1 in ascending order.

Example 1:

Input: arr1 = [2,3,1,3,2,4,6,7,9,2,19], arr2 = [2,1,4,3,9,6]

Output: [2,2,2,1,4,3,3,9,6,7,19]

Constraints:

  • len(arr1), len(arr2) <= 1000

  • 0 <= arr1[i], arr2[i] <= 1000

  • Each arr2[i] is distinct.

  • Each arr2[i] is in arr1.

Hints:

  1. Using a hashmap, we can map the values of arr2 to their position in arr2.

  2. After, we can use a custom sorting function.

[1]:

class Solution1(object):
    """
    Time:  O(NLogN)
    Space: O(N)
    """
    def relativeSortArray(self, arr1, arr2):
        """
        :type arr1: List[int]
        :type arr2: List[int]
        :rtype: List[int]
        """
        lookup = {v: i for i, v in enumerate(arr2)}

        return sorted(arr1, key=lambda i: lookup.get(i, len(arr2) + i))

[2]:

s = Solution1()
arr1 = [2,3,1,3,2,4,6,7,9,2,19]
arr2 = [2,1,4,3,9,6]
assert s.relativeSortArray(arr1, arr2) == [2,2,2,1,4,3,3,9,6,7,19]