Given an array of integers, find the maximum sum of any subarray of length k.
Question Analysis
The problem is asking us to find the maximum sum of any subarray with a fixed length ( k ) from a given array of integers. A subarray is a contiguous part of the array, and we need to consider all possible subarrays of length ( k ) to find the one with the maximum sum.
To solve this problem efficiently, we can utilize the sliding window technique. This method allows us to compute the sum of the subarray in constant time by reusing the sum of the previous subarray, hence optimizing the process instead of recalculating the sum from scratch each time.
Answer
To solve the problem, follow these steps using the sliding window technique:
- Calculate the sum of the first ( k ) elements, as this will be our initial window.
- Initialize a variable to track the maximum sum, setting it initially to the sum of the first window.
- Slide the window across the array by adding the next element and subtracting the first element of the previous window.
- Update the maximum sum if the current window's sum is greater.
- Continue this process until you've processed all possible windows of length ( k ).
Here is a Python implementation of this approach:
def max_sum_subarray(arr, k):
# Check if the array is smaller than k
if len(arr) < k:
return "Array length is smaller than k"
# Initial sum of the first window
window_sum = sum(arr[:k])
max_sum = window_sum
# Slide the window through the array
for i in range(k, len(arr)):
# Slide the window to the right by adding the next element and removing the first element of the previous window
window_sum += arr[i] - arr[i - k]
# Update max_sum if the current window's sum is greater
max_sum = max(max_sum, window_sum)
return max_sum
# Example usage:
arr = [2, 1, 5, 1, 3, 2]
k = 3
print(max_sum_subarray(arr, k)) # Output: 9 (subarray [5, 1, 3])
This solution efficiently computes the maximum sum of any subarray of length ( k ) with a time complexity of ( O(n) ), where ( n ) is the number of elements in the array.