Maximum Subarray Sum With Length Divisible by K
<p>You are given an array of integers <code>nums</code> and an integer <code>k</code>.</p>

<p>Return the <strong>maximum</strong> sum of a <span data-keyword="subarray-nonempty">subarray</span> of <code>nums</code>, such that the size of the subarray is <strong>divisible</strong> by <code>k</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1,2], k = 1</span></p>

<p><strong>Output:</strong> <span class="example-io">3</span></p>

<p><strong>Explanation:</strong></p>

<p>The subarray <code>[1, 2]</code> with sum 3 has length equal to 2 which is divisible by 1.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [-1,-2,-3,-4,-5], k = 4</span></p>

<p><strong>Output:</strong> <span class="example-io">-10</span></p>

<p><strong>Explanation:</strong></p>

<p>The maximum sum subarray is <code>[-1, -2, -3, -4]</code> which has length equal to 4 which is divisible by 4.</p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [-5,1,2,-3,4], k = 2</span></p>

<p><strong>Output:</strong> <span class="example-io">4</span></p>

<p><strong>Explanation:</strong></p>

<p>The maximum sum subarray is <code>[1, 2, -3, 4]</code> which has length equal to 4 which is divisible by 2.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= nums.length &lt;= 2 * 10<sup>5</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>
Array; Hash Table; Prefix Sum
function maxSubarraySum(nums: number[], k: number): number {
    const f: number[] = Array(k).fill(Infinity);
    f[k - 1] = 0;
    let ans = -Infinity;
    let s = 0;
    for (let i = 0; i < nums.length; ++i) {
        s += nums[i];
        ans = Math.max(ans, s - f[i % k]);
        f[i % k] = Math.min(f[i % k], s);
    }
    return ans;
}
