Maximum Sum Subarray of length K II in Python (Time Complexity: O(n))

Given an input array of integers and a positive integer k, find out the maximum sum of a contiguous subarray of length exactly k

Example

Input: nums = [5, 6, 1, 2, 6, 6, 4, 3], k = 3
Output: 4 6
Explanation: The subarray nums[4...6] has the maximum sum of 16

Understanding the Problem

The core challenge of this problem is to find the maximum sum of any contiguous subarray of length exactly k. This problem is significant in various applications such as financial analysis, where one might want to find the maximum profit over a fixed number of days.

Potential pitfalls include misunderstanding the requirement for the subarray to be contiguous and not considering edge cases such as when the array length is less than k.

Approach

To solve this problem, we can use a sliding window approach, which is optimal for this type of problem. The naive approach would involve calculating the sum of every possible subarray of length k and then finding the maximum, but this would be inefficient with a time complexity of O(n*k).

Instead, the sliding window approach allows us to maintain a running sum of the current window of size k and slide the window across the array, updating the sum by subtracting the element that is left behind and adding the new element that comes into the window. This approach has a time complexity of O(n), which is much more efficient.

Algorithm

Here is a step-by-step breakdown of the sliding window algorithm:

Initialize the sum of the first window of size k.
Set this initial sum as the maximum sum.
Slide the window one element at a time to the right, updating the sum by subtracting the element that is left behind and adding the new element.
Update the maximum sum if the current window's sum is greater.
Continue this process until the end of the array is reached.

Code Implementation

def max_sum_subarray(nums, k):
    # Initialize the sum of the first window
    window_sum = sum(nums[:k])
    max_sum = window_sum
    
    # Slide the window from start to end of the array
    for i in range(len(nums) - k):
        # Update the window sum by subtracting the element that is left behind
        # and adding the new element that comes into the window
        window_sum = window_sum - nums[i] + nums[i + k]
        # Update the maximum sum if the current window's sum is greater
        max_sum = max(max_sum, window_sum)
    
    return max_sum

# Example usage
nums = [5, 6, 1, 2, 6, 6, 4, 3]
k = 3
print(max_sum_subarray(nums, k))  # Output: 16

Complexity Analysis

The time complexity of the sliding window approach is O(n) because we only pass through the array once. The space complexity is O(1) since we are using a fixed amount of extra space.

Edge Cases

Potential edge cases include:

Array length less than k: In this case, the problem is invalid as we cannot form a subarray of length k.
Array with all negative numbers: The algorithm should still correctly find the maximum sum subarray.
Array with all zeros: The maximum sum will be zero.

Testing

To test the solution comprehensively, consider the following test cases:

Simple case: nums = [1, 2, 3, 4, 5], k = 2
All negative numbers: nums = [-1, -2, -3, -4, -5], k = 3
Array with zeros: nums = [0, 0, 0, 0, 0], k = 1
Array length less than k: nums = [1, 2], k = 3

Thinking and Problem-Solving Tips

When approaching such problems, it is crucial to:

Understand the problem requirements and constraints thoroughly.
Consider edge cases and how they might affect your solution.
Think about efficient algorithms and data structures that can help optimize your solution.
Practice similar problems to improve your problem-solving skills.

Conclusion

In this blog post, we discussed how to find the maximum sum of a contiguous subarray of length k using a sliding window approach. This method is efficient with a time complexity of O(n) and handles various edge cases effectively. Understanding and solving such problems is crucial for developing strong problem-solving skills in programming.

Additional Resources

For further reading and practice, consider the following resources: