Is Lexicographically Maximum MEX Array easy or hard?

Lexicographically Maximum MEX Array is generally classified as a hard problem because it requires understanding MEX behavior, greedy decision making, and maintaining dynamic state efficiently. The logic is simple once the invariant is recognized, but identifying that greedy structure is the challenging part.

Lexicographically Maximum MEX Array Python/Java solution

Python and Java implementations typically maintain an integer array for frequencies and a variable storing the current MEX. Each step updates the frequency of the removed element and advances the MEX pointer when necessary, producing an O(n) solution.

How to solve Lexicographically Maximum MEX Array in O(n)?

Count frequencies of all numbers first, then iterate while simulating element removals. Maintain a pointer representing the current MEX and advance it whenever all smaller values are present. Because each value is processed a constant number of times, the total runtime remains linear.

What is the best approach for Lexicographically Maximum MEX Array?

The best approach uses a greedy strategy with a frequency array and a dynamically maintained MEX pointer. By tracking how many times each value appears and updating the smallest missing number as elements are removed, you can build the lexicographically maximum sequence in O(n) time and O(n) space.

Is Lexicographically Maximum MEX Array asked at Google/Amazon/Meta?

MEX-based greedy and frequency tracking problems appear in interviews at companies like Google, Amazon, and Meta. While the exact problem may vary, understanding how to maintain a dynamic MEX and build lexicographically optimal results is a common interview theme.

What data structure is used in Lexicographically Maximum MEX Array?

Typical implementations use a frequency array or hash map to count occurrences and track removals. Some versions also use a set or priority queue to maintain the smallest missing number, although the optimal approach uses only arrays and a moving pointer.

What is the time complexity of Lexicographically Maximum MEX Array?

The optimal solution runs in O(n) time with O(n) space. It maintains frequencies of numbers and updates the MEX pointer only when needed. Simpler implementations using ordered sets take O(n log n) due to insertion and deletion operations.

Description Hints 4 Video 16:49

3948. Lexicographically Maximum MEX Array

Hard53.6% acceptance2 companies

You are given an integer array nums.

You want to construct an array result by repeatedly performing the following operation until nums becomes empty:

Choose an integer k such that 1 <= k <= len(nums).
Compute the MEX of the first k elements of nums.
Append this MEX to result.
Remove the first k elements from nums.

Return the lexicographically maximum array result that can be obtained after performing the operations.

The MEX of an array is the smallest non-negative integer not present in the array.

An array a is lexicographically greater than an array b if in the first position where a and b differ, array a has an element that is greater than the corresponding element in b. If the first min(a.length, b.length) elements do not differ, then the longer array is the lexicographically greater one.

Example 1:

Input: nums = [0,1,0]

Output: [2,1]

Explanation:

Take the first k = 2 elements [0, 1] which has MEX = 2. Current result = [2].
Remaining array [0] has MEX = 1. Thus, the final result = [2, 1].

Example 2:

Input: nums = [1,0,2]

Output: [3]

Explanation:

Take the first k = 3 elements [1, 0, 2] which has MEX = 3.
nums is now empty. Thus, the final result = [3].

Example 3:

Input: nums = [3,1]

Output: [0,0]

Explanation:

Take k = 1, first element [3] has MEX = 0. Current result = [0].
Remaining array [1] has MEX = 0. Thus, the final result = [0, 0].

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁵

Asked at 2 companies

Tekion Amazon

Topics

Hints open one at a time

Hint 1

At any step, the best next value is the MEX of the entire remaining suffix. You cannot make the first result element larger than that.

Hint 2

If the current suffix MEX is x, you should take the shortest prefix that contains all values 0, 1, ..., x - 1. This appends x while leaving as many elements as possible for later.

Hint 3

If the current suffix MEX is 0, then no remaining element is 0, so taking one element at a time is optimal and appends many 0s.

Hint 4

Maintain remaining frequencies to know the current suffix MEX, and use a sliding process to find the shortest prefix containing all numbers below it.

LanguageLoading editor…

1class Solution:
2    def maximumMEX(self, nums: List[int]) -> List[int]:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

[0,1,0]

Input

[1,0,2]

Input

[3,1]

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