How do you solve Count Subarrays With K Distinct Integers on LeetCode?

Count Subarrays With K Distinct Integers (LeetCode #3859) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a sliding window optimizes the search for valid subarrays.

What is the time complexity of Count Subarrays With K Distinct Integers?

The optimal solution for Count Subarrays With K Distinct Integers runs in O(n) time and uses O(n) space. Sliding window allows us to process each element in linear time, maintaining a frequency map.

What is the brute force solution for Count Subarrays With K Distinct Integers?

The brute force approach for Count Subarrays With K Distinct Integers is Brute Force, with O(n²) time complexity and O(1) space complexity. Generate all possible subarrays and check each for the required conditions. This is straightforward but inefficient for large arrays. The optimal Optimal Solution approach improves this to O(n).

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

Count Subarrays With K Distinct Integers uses the following concepts: Array, Hash Table, Sliding Window, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Subarrays With K Distinct Integers?

Explain your thought process clearly while coding. Consider edge cases and constraints before finalizing your approach. Practice similar problems to build confidence.

What are common mistakes when solving Count Subarrays With K Distinct Integers?

Failing to check frequency conditions correctly. Not handling edge cases with distinct counts properly.

#3859

Count Subarrays With K Distinct Integers

Hard

ArrayHash TableSliding WindowCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Use a sliding window to efficiently count subarrays with at most k distinct integers, then subtract those with at most k-1 distinct integers.

⚙️

Algorithm

3 steps

1Step 1: Create a helper function to count subarrays with at most k distinct integers, ensuring each appears at least m times.
2Step 2: Use the helper function to get counts for k and k-1.
3Step 3: Return the difference between the two counts.

solution.py14 lines

1def at_most(nums, k, m):
2    count, left, freq = 0, 0, {}
3    for right in range(len(nums)):
4        freq[nums[right]] = freq.get(nums[right], 0) + 1
5        while len(freq) > k or any(v < m for v in freq.values()):
6            freq[nums[left]] -= 1
7            if freq[nums[left]] == 0:
8                del freq[nums[left]]
9            left += 1
10        count += right - left + 1
11    return count
12
13def countSubarrays(nums, k, m):
14    return at_most(nums, k, m) - at_most(nums, k - 1, m)

ℹ

Complexity note: Sliding window allows us to process each element in linear time, maintaining a frequency map.

1Using a sliding window optimizes the search for valid subarrays.
2Counting distinct integers efficiently is crucial for performance.

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