Maximum Size Subarray Sum Equals k
<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <em>the maximum length of a </em><span data-keyword="subarray"><em>subarray</em></span><em> that sums to</em> <code>k</code>. If there is not one, return <code>0</code> instead.</p>

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

<pre>
<strong>Input:</strong> nums = [1,-1,5,-2,3], k = 3
<strong>Output:</strong> 4
<strong>Explanation:</strong> The subarray [1, -1, 5, -2] sums to 3 and is the longest.
</pre>

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

<pre>
<strong>Input:</strong> nums = [-2,-1,2,1], k = 1
<strong>Output:</strong> 2
<strong>Explanation:</strong> The subarray [-1, 2] sums to 1 and is the longest.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>9</sup>&nbsp;&lt;= k &lt;= 10<sup>9</sup></code></li>
</ul>
Array; Hash Table; Prefix Sum
function maxSubArrayLen(nums: number[], k: number): number {
    const d: Map<number, number> = new Map();
    d.set(0, -1);
    let ans = 0;
    let s = 0;
    for (let i = 0; i < nums.length; ++i) {
        s += nums[i];
        if (d.has(s - k)) {
            ans = Math.max(ans, i - d.get(s - k)!);
        }
        if (!d.has(s)) {
            d.set(s, i);
        }
    }
    return ans;
}
