How do you solve Find the Median of the Uniqueness Array on LeetCode?

Find the Median of the Uniqueness Array (LeetCode #3134) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log(max(nums))) time complexity and O(n) space complexity. Key insight: Understanding how to count distinct elements efficiently is crucial.

What is the brute force solution for Find the Median of the Uniqueness Array?

The brute force approach for Find the Median of the Uniqueness Array is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute-force approach involves generating all possible subarrays of the input array and counting the distinct elements in each. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n log(max(nums))).

What algorithm or data structure is used to solve Find the Median of the Uniqueness Array?

Find the Median of the Uniqueness Array uses the following concepts: Array, Hash Table, Binary Search, Sliding Window. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find the Median of the Uniqueness Array?

Always clarify the problem statement and constraints before diving into coding. Discuss your thought process and any assumptions you make while solving the problem. Practice counting distinct elements in subarrays as it is a common operation in many problems.

What are common mistakes when solving Find the Median of the Uniqueness Array?

Not considering edge cases with small arrays. Misunderstanding the definition of median in the context of sorted arrays.

#3134

Find the Median of the Uniqueness Array

Hard

ArrayHash TableBinary SearchSliding WindowHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log(max(nums)))
Space	O(n)	O(n)

💡

Intuition

Time O(n log(max(nums)))Space O(n)

The optimal approach uses binary search to find the median directly by counting how many subarrays have distinct counts less than or equal to a given value. This reduces the need to generate all subarrays explicitly.

⚙️

Algorithm

5 steps

1Step 1: Define a function to count the number of subarrays with distinct elements less than or equal to x.
2Step 2: Use binary search on the range of possible distinct counts (from 1 to the maximum element in nums).
3Step 3: For each mid value in the binary search, use the counting function to determine if it is the median.
4Step 4: Adjust the binary search range based on the count of subarrays compared to the desired median position.
5Step 5: Return the median value once found.

solution.py28 lines

1# Full working Python code
2
3def countSubarraysWithDistinctLessEqual(nums, x):
4    count = 0
5    left = 0
6    freq = {}
7    for right in range(len(nums)):
8        freq[nums[right]] = freq.get(nums[right], 0) + 1
9        while len(freq) > x:
10            freq[nums[left]] -= 1
11            if freq[nums[left]] == 0:
12                del freq[nums[left]]
13            left += 1
14        count += right - left + 1
15    return count
16
17
18def findMedian(nums):
19    low, high = 1, max(nums)
20    n = len(nums)
21    total_subarrays = n * (n + 1) // 2
22    while low < high:
23        mid = (low + high) // 2
24        if countSubarraysWithDistinctLessEqual(nums, mid) < (total_subarrays + 1) // 2:
25            low = mid + 1
26        else:
27            high = mid
28    return low

ℹ

Complexity note: The time complexity is O(n log(max(nums))) due to the binary search over the range of possible distinct counts and the linear time complexity of counting subarrays with distinct elements.

1Understanding how to count distinct elements efficiently is crucial.
2Binary search can significantly reduce the time complexity when looking for a median.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.