#3134

Hard

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Find the Median of the Uniqueness Array

Given a nums array, find the median of its uniqueness array by considering all subarrays and their distinct element counts.

array hash table binary search sliding window

Problem Statement

You are given an integer array nums. The uniqueness array of nums is a sorted array containing the number of distinct elements in every possible subarray of nums. In other words, it is the sorted array consisting of distinct elements for every subarray nums[i..j], for all 0 <= i <= j < nums.length.

Your task is to return the median of this uniqueness array. The median is the middle value of the sorted uniqueness array, or the average of the two middle values if the size is even.

Examples

Example 1

Input: nums = [1,2,3]

Output: 1

The uniqueness array of nums is [distinct(nums[0..0]), distinct(nums[1..1]), distinct(nums[2..2]), distinct(nums[0..1]), distinct(nums[1..2]), distinct(nums[0..2])] which is equal to [1, 1, 1, 2, 2, 3] . The uniqueness array has a median of 1. Therefore, the answer is 1.

Example 2

Input: nums = [3,4,3,4,5]

Output: 2

The uniqueness array of nums is [1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3] . The uniqueness array has a median of 2. Therefore, the answer is 2.

Example 3

Input: nums = [4,3,5,4]

Output: 2

The uniqueness array of nums is [1, 1, 1, 1, 2, 2, 2, 3, 3, 3] . The uniqueness array has a median of 2. Therefore, the answer is 2.

Constraints

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

Solution Approach

Array Scanning and Hash Lookup

You can iterate through each subarray and use a hash table to count the distinct elements. This helps avoid re-computing the distinct count for overlapping parts of the subarrays, improving efficiency.

Binary Search Over the Answer

Given the large input constraints, binary search over the possible values of the median in the uniqueness array is a viable approach. This reduces the problem's complexity by focusing on narrowing down the potential answers based on counts of distinct elements.

Efficient Memory Management

Since the problem involves potentially large arrays, careful management of memory and using algorithms that don't require holding all subarrays in memory at once is crucial. Techniques such as sliding windows and caching distinct elements can help.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the chosen approach, but binary search combined with efficient hashing or sliding windows can bring the time complexity down significantly from brute force. The space complexity is influenced by the need to store distinct elements and intermediate results for subarrays.

What Interviewers Usually Probe

The candidate should demonstrate knowledge of efficient subarray processing, especially using hash tables for distinct element counting.
Look for understanding of binary search and its application to this problem for narrowing down the potential median.
Candidates should be able to explain memory management strategies for handling large inputs efficiently.

Common Pitfalls or Variants

Common pitfalls

Misunderstanding the problem's complexity by using brute-force approaches, leading to time-limit exceeded errors for large inputs.
Confusing the median of the uniqueness array with other statistical measures, such as the mean or mode.
Not managing memory properly, causing excessive memory usage due to storing all subarrays.

Follow-up variants

What if the input array nums is sorted? Does it change the approach?
How can this problem be adapted to find the mode of the uniqueness array instead of the median?
Can this problem be optimized further with parallel processing or advanced hashing techniques?

FAQ

What is the median of the uniqueness array?

The median is the middle element in the sorted uniqueness array, or the average of the two middle elements if the array has an even length.

How does binary search help in solving the problem?

Binary search helps narrow down potential median values by focusing on counts of distinct elements within the uniqueness array, improving efficiency.

What pattern does this problem follow?

This problem follows the 'array scanning plus hash lookup' pattern, with an emphasis on efficiently counting distinct elements in subarrays.

Can I use brute force to solve this problem?

Brute force would be too slow for large inputs, so it's recommended to use more efficient methods like hash tables and binary search.

How can I optimize memory usage for this problem?

You can optimize memory usage by using techniques like sliding windows and caching distinct elements, rather than storing all subarrays at once.

terminal

Solution

Solution 1: Binary Search + Two Pointers

Let the length of the array $\textit{nums}$ be $n$. The length of the uniqueness array is $m = \frac{(1 + n) \times n}{2}$, and the median of the uniqueness array is the $\frac{m + 1}{2}$-th smallest number among these $m$ numbers.

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class Solution:
    def medianOfUniquenessArray(self, nums: List[int]) -> int:
        def check(mx: int) -> bool:
            cnt = defaultdict(int)
            k = l = 0
            for r, x in enumerate(nums):
                cnt[x] += 1
                while len(cnt) > mx:
                    y = nums[l]
                    cnt[y] -= 1
                    if cnt[y] == 0:
                        cnt.pop(y)
                    l += 1
                k += r - l + 1
                if k >= (m + 1) // 2:
                    return True
            return False

        n = len(nums)
        m = (1 + n) * n // 2
        return bisect_left(range(n), True, key=check)

Continue Practicing

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