Maximum Sum of Subarrays of Size K

nums = [2, 1, 5, 1, 3, 2]
k = 3

9

public class Solution {
    public Integer maxSum(int[] nums, Integer k) {
        // Your code goes here
    }
}

Explanation

This problem uses a fixed-size sliding window to efficiently find the maximum sum among all subarrays of length k. Instead of recalculating the sum for each subarray from scratch, we slide the window across the array and update the sum incrementally.

This approach is efficient because we calculate each window's sum in constant time by:

• Adding the new element entering the window (nums[end])
• Subtracting the old element leaving the window (nums[start])

Instead of summing k elements for each window (which would be O(n*k)), we do constant work per window, giving us O(n) time complexity.

Solution

public class Solution {
    public int maxSubarraySum(int[] nums, int k) {
        int maxSum = Integer.MIN_VALUE;
        int windowSum = 0;
        int start = 0;
        for (int end = 0; end < nums.length; end++) {
            windowSum += nums[end];
            if (end - start + 1 == k) {
                maxSum = Math.max(maxSum, windowSum);
                windowSum -= nums[start];
                start++;
            }
        }
        return maxSum;
    }
}