How do you solve Count the Number of Good Subarrays on LeetCode?

Count the Number of Good Subarrays (LeetCode #2537) 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: Understanding how pairs are formed is crucial for counting efficiently.

What is the time complexity of Count the Number of Good Subarrays?

The optimal solution for Count the Number of Good Subarrays runs in O(n) time and uses O(n) space. This complexity is achieved because we only traverse the array once with two pointers, maintaining a frequency map.

What is the brute force solution for Count the Number of Good Subarrays?

The brute force approach for Count the Number of Good Subarrays is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible subarray to count the number of good subarrays. This is straightforward but inefficient, as it examines all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count the Number of Good Subarrays?

Count the Number of Good Subarrays uses the following concepts: Array, Hash Table, Sliding Window. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count the Number of Good Subarrays?

Always consider edge cases, such as very small arrays or arrays with all unique elements. Explain your thought process clearly; interviewers value understanding over just getting the right answer. Practice writing clean and efficient code; readability is as important as functionality.

What are common mistakes when solving Count the Number of Good Subarrays?

Not maintaining the frequency map correctly, leading to incorrect pair counts. Failing to realize that once pairs_count reaches k, all remaining elements can form valid subarrays.

#2537

Count the Number of Good Subarrays

Medium

ArrayHash TableSliding WindowHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal solution uses a sliding window approach to efficiently count good subarrays without needing to check every possible subarray. This reduces the time complexity significantly.

⚙️

Algorithm

4 steps

1Step 1: Use two pointers (left and right) to represent the current subarray.
2Step 2: Maintain a frequency map to count occurrences of each number in the current subarray.
3Step 3: Expand the right pointer to include new elements until the number of pairs is at least k.
4Step 4: Move the left pointer to shrink the window while still maintaining at least k pairs, counting valid subarrays as you go.

solution.py14 lines

1def countGoodSubarrays(nums, k):
2    count = 0
3    left = 0
4    pairs_count = 0
5    freq = {}
6    for right in range(len(nums)):
7        freq[nums[right]] = freq.get(nums[right], 0) + 1
8        pairs_count += freq[nums[right]] - 1
9        while pairs_count >= k:
10            count += len(nums) - right
11            freq[nums[left]] -= 1
12            pairs_count -= freq[nums[left]]
13            left += 1
14    return count

ℹ

Complexity note: This complexity is achieved because we only traverse the array once with two pointers, maintaining a frequency map.

1Understanding how pairs are formed is crucial for counting efficiently.
2Using a frequency map allows us to quickly determine how many pairs exist as we expand or contract our window.

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