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:4 6Explanation: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 best period for investment returns.

Potential pitfalls include misunderstanding the requirement for the subarray to be contiguous and exactly of length **k**.

To solve this problem, we can use a sliding window approach, which is optimal for problems involving contiguous subarrays.

**Naive Solution:** A naive solution would involve checking the sum of every possible subarray of length **k**. This would have a time complexity of O(n*k), which is not efficient for large arrays.

**Optimized Solution:** The sliding window technique allows us to achieve a time complexity of O(n). The idea is to maintain a window of size **k** and slide it across the array while keeping track of the maximum sum.

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

2. Slide the window across the array by adding the next element and removing the first element of the previous window.

3. Keep track of the maximum sum encountered during this process.

```
public class MaximumSumSubarray {
public static int maxSumSubarray(int[] nums, int k) {
// Initialize the sum of the first window
int maxSum = 0;
for (int i = 0; i < k; i++) {
maxSum += nums[i];
}
// Current sum of the window
int windowSum = maxSum;
// Slide the window from start to end of the array
for (int i = k; i < nums.length; i++) {
// Add the next element and remove the first element of the previous window
windowSum += nums[i] - nums[i - k];
// Update the maximum sum if the current window sum is greater
maxSum = Math.max(maxSum, windowSum);
}
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 traverse the array only once. The space complexity is O(1) as we are using a constant amount of extra space.

1. If the array length is less than **k**, the problem constraints should ensure this does not happen.

2. If all elements are negative, the algorithm still works correctly by finding the least negative sum.

To test the solution comprehensively, consider the following cases:

- Array with all positive numbers
- Array with all negative numbers
- Array with mixed positive and negative numbers
- Array with length exactly
**k**

When approaching such problems, always consider if a sliding window or two-pointer technique can be applied. Practice similar problems to get a better understanding of these techniques.

Understanding and solving the maximum sum subarray problem using the sliding window technique is crucial for optimizing performance in real-world applications. Practice and familiarity with such problems will enhance your problem-solving skills.