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= 3Output:16Explanation:The subarray nums[4...6] has the maximum sum of 16

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 period with the highest profit.

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**.

To solve this problem, we can use a sliding window approach which is more efficient than a naive solution.

The naive solution involves calculating the sum of every possible subarray of length **k** and keeping track of the maximum sum. This approach has a time complexity of O(n*k), which is not optimal for large arrays.

The sliding window approach involves maintaining a window of size **k** and sliding it across the array to find the maximum sum. This approach has a time complexity of O(n), which is much more efficient.

1. Calculate the sum of the first **k** elements.

2. Initialize the maximum sum as the sum of the first **k** elements.

3. Slide the window one element at a time by subtracting the element that is left behind and adding the new element.

4. Update the maximum sum if the new window sum is greater.

```
public class MaximumSumSubarray {
public static int maxSumSubarray(int[] nums, int k) {
// Edge case: if array length is less than k
if (nums.length < k) {
throw new IllegalArgumentException("Array length must be at least k");
}
// Calculate the sum of the first k elements
int maxSum = 0;
for (int i = 0; i < k; i++) {
maxSum += nums[i];
}
// Initialize the current sum as the maxSum
int currentSum = maxSum;
// Slide the window across the array
for (int i = k; i < nums.length; i++) {
currentSum += nums[i] - nums[i - k];
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}
public static void main(String[] args) {
int[] nums = {5, 6, 1, 2, 6, 6, 4, 3};
int k = 3;
System.out.println("Maximum sum of subarray of length " + k + " is: " + maxSumSubarray(nums, k));
}
}
```

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) as we are using a constant amount of extra space.

1. Array length less than **k**: The function should handle this by throwing an exception.

2. Array with all negative numbers: The algorithm still works as it finds the maximum sum, which could be negative.

3. Array with all positive numbers: The algorithm will find the maximum sum as expected.

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

- Simple case: [5, 6, 1, 2, 6, 6, 4, 3], k = 3
- Edge case: [1, 2], k = 3 (should throw an exception)
- All negative numbers: [-1, -2, -3, -4, -5], k = 2
- All positive numbers: [1, 2, 3, 4, 5], k = 2

When approaching such problems, it is crucial to understand the constraints and requirements clearly. Breaking down the problem and considering different approaches can help in finding an optimal solution. Practicing similar problems and studying algorithms like sliding window can improve problem-solving skills.

In this blog post, we discussed the problem of finding the maximum sum of a contiguous subarray of length **k**. We explored a naive solution and an optimized sliding window approach. Understanding and solving such problems is essential for improving algorithmic thinking and coding skills.