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Maximum Sum Subarray Size K

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In Maximum Sum Subarray Size K, you are given an integer array and must find the largest sum among all contiguous windows of length k. This is a fixed-size sliding-window problem where recomputing every window from scratch is unnecessary and expensive.

The practical method is to compute the first window sum once, treat it as current best, then move the window one step at a time by adding the incoming right element and subtracting the outgoing left element. After each move, update the maximum if the new sum is bigger. This keeps the solution linear in array length.

The judge also checks invalid-window situations, such as k <= 0 or k > nums.length, where the expected return is 0 in this challenge. It includes positive-only, mixed-sign, and short arrays as well. Your function should return exactly one integer maximum sum and avoid extra output, because correctness is based on the numeric return value for each test case. Keep edge handling explicit so boundary behavior stays predictable.

Examples

Example 1
Input
nums = [2,1,5,1,3,2], k = 3
Output
9
Explanation

Window sums are 8, 7, 9, 6 so the maximum is 9.

Example 2
Input
nums = [2,3,4,1,5], k = 2
Output
7
Explanation

Window sums are 5, 7, 5, 6 so the maximum is 7.

Example 3
Input
nums = [1,1,1], k = 4
Output
0
Explanation

No valid window of size 4 exists.

Algorithm Flow

Recommendation Algorithm Flow for Maximum Sum Subarray Size K - Budibadu
Recommendation Algorithm Flow for Maximum Sum Subarray Size K - Budibadu

Best Answers

java
class Solution {
    public int maximum_sum_subarray_size_k(int[] nums, int k) {
        int n = nums.length;
        if (k <= 0 || k > n) return 0;
        int window = 0;
        for (int i = 0; i < k; i++) window += nums[i];
        int best = window;
        for (int i = k; i < n; i++) {
            window += nums[i] - nums[i-k];
            if (window > best) best = window;
        }
        return best;
    }
}