How do you solve Count Subarrays With Median K on LeetCode?

Count Subarrays With Median K (LeetCode #2488) can be solved using Brute Force. The optimal approach is Brute Force with O(n²) time complexity and O(1) space complexity. Key insight: The brute force approach checks all possible subarrays to see if their median equals k. This is straightforward but inefficient for large arrays since it examines every combination.

What algorithm or data structure is used to solve Count Subarrays With Median K?

Count Subarrays With Median K uses the following concepts: Array, Hash Table, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Subarrays With Median K?

Always clarify the definition of median in your own words. Discuss your thought process out loud to show your understanding. Consider edge cases and how they might affect your solution.

What are common mistakes when solving Count Subarrays With Median K?

Forgetting to handle edge cases like single-element subarrays. Not considering the transformation of the array properly.

Count Subarrays With Median K — LeetCode #2488 (Hard)

Tags: Array, Hash Table, Prefix Sum

Related patterns: Hash Map, Array

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute force approach checks all possible subarrays to see if their median equals k. This is straightforward but inefficient for large arrays since it examines every combination.

This complexity arises because we check every possible subarray, which results in a nested loop. Each subarray requires sorting, leading to O(n log n) for each check.

Step 1: Iterate through all possible starting points of subarrays.
Step 2: For each starting point, iterate through all possible ending points to form subarrays.
Step 3: For each subarray, sort it and check if the median equals k.

For nums = [3,2,1,4,5], k = 4: 1. i=0, j=0: subarray [3], median=3 (not k) 2. i=0, j=1: subarray [3,2], median=3 (not k) 3. i=0, j=2: subarray [3,2,1], median=2 (not k) 4. i=0, j=3: subarray [3,2,1,4], median=2 (not k) 5. i=0, j=4: subarray [3,2,1,4,5], median=3 (not k) 6. i=1, j=3: subarray [2,1,4], median=2 (not k) 7. i=3, j=3: subarray [4], median=4 (is k) => count=1 8. i=3, j=4: subarray [4,5], median=4 (is k) => count=2 9. i=2, j=4: subarray [1,4,5], median=4 (is k) => count=3.

Common Mistakes

Forgetting to handle edge cases like single-element subarrays.
Not considering the transformation of the array properly.

Interview Tips

Always clarify the definition of median in your own words.
Discuss your thought process out loud to show your understanding.
Consider edge cases and how they might affect your solution.

#2488

Count Subarrays With Median K

Hard

Array↗Hash Table↗Prefix Sum↗Hash Map↗Array↗

LeetCode ↗

Approaches

💡

Intuition

Time O(n²)Space O(1)

The brute force approach checks all possible subarrays to see if their median equals k. This is straightforward but inefficient for large arrays since it examines every combination.

⚙️

Algorithm

3 steps

1Step 1: Iterate through all possible starting points of subarrays.
2Step 2: For each starting point, iterate through all possible ending points to form subarrays.
3Step 3: For each subarray, sort it and check if the median equals k.

solution.py13 lines

1def countSubarrays(nums, k):
2    count = 0
3    for i in range(len(nums)):
4        for j in range(i, len(nums)):
5            subarray = nums[i:j+1]
6            if median(subarray) == k:
7                count += 1
8    return count
9
10def median(arr):
11    arr.sort()
12    n = len(arr)
13    return arr[n // 2] if n % 2 != 0 else arr[n // 2 - 1]

ℹ

Complexity note: This complexity arises because we check every possible subarray, which results in a nested loop. Each subarray requires sorting, leading to O(n log n) for each check.

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