What is the time complexity of Number of Sub-arrays of Size K and Average Greater than or Equal to Threshold?

The optimal solution for Number of Sub-arrays of Size K and Average Greater than or Equal to Threshold runs in O(n) time and uses O(1) space. This complexity is achieved because we only traverse the array once, maintaining a running sum without needing nested loops.

What is the brute force solution for Number of Sub-arrays of Size K and Average Greater than or Equal to Threshold?

The brute force approach for Number of Sub-arrays of Size K and Average Greater than or Equal to Threshold is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible sub-array of size k in the array. For each sub-array, we calculate the average and compare it to the threshold. This method 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 Number of Sub-arrays of Size K and Average Greater than or Equal to Threshold?

Number of Sub-arrays of Size K and Average Greater than or Equal to Threshold uses the following concepts: Array, Sliding Window. The recommended patterns to study are: Sliding Window, Array.

What are the interview tips for Number of Sub-arrays of Size K and Average Greater than or Equal to Threshold?

Always consider edge cases and test your solution against them. Explain your thought process clearly while coding; it helps the interviewer understand your approach. Practice writing clean and efficient code, as readability is important in interviews.

What are common mistakes when solving Number of Sub-arrays of Size K and Average Greater than or Equal to Threshold?

Not maintaining the sum correctly while sliding the window, leading to incorrect averages. Forgetting to handle edge cases where k equals the length of the array.

#1343

Number of Sub-arrays of Size K and Average Greater than or Equal to Threshold

Medium

ArraySliding WindowSliding WindowArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal approach uses a sliding window technique to maintain the sum of the current sub-array of size k. This allows us to efficiently update the sum as we move the window, reducing the time complexity significantly.

⚙️

Algorithm

3 steps

1Step 1: Initialize a variable to hold the sum of the first k elements.
2Step 2: Loop through the array starting from index k, updating the sum by subtracting the element going out of the window and adding the new element coming into the window.
3Step 3: For each updated sum, check if the average (sum / k) is greater than or equal to the threshold and increment the count accordingly.

solution.py12 lines

1# Full working Python code
2
3def numOfSubarrays(arr, k, threshold):
4    count = 0
5    current_sum = sum(arr[:k])
6    if current_sum / k >= threshold:
7        count += 1
8    for i in range(k, len(arr)):
9        current_sum += arr[i] - arr[i - k]
10        if current_sum / k >= threshold:
11            count += 1
12    return count

ℹ

Complexity note: This complexity is achieved because we only traverse the array once, maintaining a running sum without needing nested loops.

1Using a sliding window allows us to efficiently calculate the sum without recalculating it from scratch.
2Understanding the average condition helps in optimizing the checks needed for counting valid sub-arrays.

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