algoadvance

Given an integer array nums consisting of n elements and an integer k, return the maximum average value of a subarray of length k. You need to compute the result as a floating-point number.

Clarifying Questions

1. Bounds on the size of nums and k:
   • What is the upper limit for n (the length of nums) and k?
   • Is it guaranteed that k <= n?
2. Element Range:
   • What is the range of element values in the nums array? (e.g., are they all positive, can they be negative?)
3. Return Type:
   • Should the result be returned as a floating-point number with any specific precision?

Strategy

We need an efficient way to find the maximum average subarray of length k. A brute-force solution would involve checking all possible subarrays of length k which would be very inefficient with time complexity of O(n*k). So, we will use a sliding window approach to achieve a time complexity of O(n).

1. Sliding Window Approach:
   • First, compute the sum of the first k elements.
   • Then slide the window one element at a time from the beginning to the end of the array.
   • Update the sum by subtracting the element that is left out of the window and adding the new element that enters the window.
   • Keep track of the maximum sum encountered during this process.
   • Finally, return the maximum sum divided by k to get the maximum average.

Code

def findMaxAverage(nums, k):
    # Compute the sum of the first window of size `k`
    current_sum = sum(nums[:k])
    max_sum = current_sum
    
    # Slide the window from the start to the end of the array
    for i in range(k, len(nums)):
        # Update the sum for the new window
        current_sum += nums[i] - nums[i - k]
        # Update max_sum if we find a new maximum
        if current_sum > max_sum:
            max_sum = current_sum
    
    # The maximum average is the maximum sum divided by `k`
    return max_sum / k

# Example usage
nums = [1, 12, -5, -6, 50, 3]
k = 4
print(findMaxAverage(nums, k))  # Output should be 12.75

Time Complexity

• Time Complexity: O(n)
  • We traverse the list once for the first sum computation, and then again sliding the window across the array, which both are linear operations.
• Space Complexity: O(1)
  • We are using only a few extra variables to keep track of the sums and the maximum, thus having constant space complexity.

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