Find the Longest Equal Subarray
<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> and an integer <code>k</code>.</p>

<p>A subarray is called <strong>equal</strong> if all of its elements are equal. Note that the empty subarray is an <strong>equal</strong> subarray.</p>

<p>Return <em>the length of the <strong>longest</strong> possible equal subarray after deleting <strong>at most</strong> </em><code>k</code><em> elements from </em><code>nums</code>.</p>

<p>A <b>subarray</b> is a contiguous, possibly empty sequence of elements within an array.</p>

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

<pre>
<strong>Input:</strong> nums = [1,3,2,3,1,3], k = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> It&#39;s optimal to delete the elements at index 2 and index 4.
After deleting them, nums becomes equal to [1, 3, 3, 3].
The longest equal subarray starts at i = 1 and ends at j = 3 with length equal to 3.
It can be proven that no longer equal subarrays can be created.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,1,2,2,1,1], k = 2
<strong>Output:</strong> 4
<strong>Explanation:</strong> It&#39;s optimal to delete the elements at index 2 and index 3.
After deleting them, nums becomes equal to [1, 1, 1, 1].
The array itself is an equal subarray, so the answer is 4.
It can be proven that no longer equal subarrays can be created.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= nums.length</code></li>
	<li><code>0 &lt;= k &lt;= nums.length</code></li>
</ul>
Array; Hash Table; Binary Search; Sliding Window
function longestEqualSubarray(nums: number[], k: number): number {
    const cnt: Map<number, number> = new Map();
    let mx = 0;
    let l = 0;
    for (let r = 0; r < nums.length; ++r) {
        cnt.set(nums[r], (cnt.get(nums[r]) ?? 0) + 1);
        mx = Math.max(mx, cnt.get(nums[r])!);
        if (r - l + 1 - mx > k) {
            cnt.set(nums[l], cnt.get(nums[l])! - 1);
            ++l;
        }
    }
    return mx;
}
