How do you solve Count Subarrays With Score Less Than K on LeetCode?

Count Subarrays With Score Less Than K (LeetCode #2302) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding how the score is calculated helps in optimizing the solution.

What is the brute force solution for Count Subarrays With Score Less Than K?

The brute force approach for Count Subarrays With Score Less Than K is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible subarrays and calculating their scores. This is straightforward but inefficient for large arrays, as it requires examining each subarray individually. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Subarrays With Score Less Than K?

Count Subarrays With Score Less Than K uses the following concepts: Array, Binary Search, Sliding Window, Prefix Sum. The recommended patterns to study are: Sliding Window, Prefix Sum.

What are the interview tips for Count Subarrays With Score Less Than K?

Always clarify the problem statement and constraints before diving into coding. Think about edge cases, such as arrays with a single element or all elements being the same. Explain your thought process and approach as you code, as this shows your problem-solving skills.

What are common mistakes when solving Count Subarrays With Score Less Than K?

Failing to account for the length of the subarray when calculating the score. Not efficiently managing the window size, leading to unnecessary recalculations.

#2302

Count Subarrays With Score Less Than K

Hard

ArrayBinary SearchSliding WindowPrefix SumSliding WindowPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses a sliding window technique combined with prefix sums to efficiently count valid subarrays. By maintaining a window of elements, we can dynamically calculate scores without recalculating sums from scratch.

⚙️

Algorithm

5 steps

1Step 1: Initialize variables for the current sum, left pointer, and a counter for valid subarrays.
2Step 2: Iterate through the array with a right pointer to expand the window.
3Step 3: For each new element added, calculate the new score and check if it's less than k.
4Step 4: If the score exceeds k, increment the left pointer to shrink the window until the score is valid again.
5Step 5: For each valid configuration, add the number of valid subarrays ending at the right pointer to the counter.

solution.py13 lines

1# Full working Python code
2
3def countSubarrays(nums, k):
4    count = 0
5    current_sum = 0
6    left = 0
7    for right in range(len(nums)):
8        current_sum += nums[right]
9        while (current_sum * (right - left + 1)) >= k:
10            current_sum -= nums[left]
11            left += 1
12        count += (right - left + 1)
13    return count

ℹ

Complexity note: The time complexity is O(n) because each element is processed at most twice (once added and once removed). The space complexity is O(1) since we only use a few extra variables.

1Understanding how the score is calculated helps in optimizing the solution.
2Using a sliding window allows us to efficiently manage the sum and length of subarrays.

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