find-the-longest-equal-subarray

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Algorithms

You are given a 0-indexed integer array <code>nums</code> and an integer <code>k</code>.

A subarray is called equal if all of its elements are equal. Note that the empty subarray is an equal subarray.

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

A subarray is a contiguous, possibly empty sequence of elements within an array.

&nbsp;
Example 1:

<pre>
Input: nums = [1,3,2,3,1,3], k = 3
Output: 3
Explanation: 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>

Example 2:

<pre>
Input: nums = [1,1,2,2,1,1], k = 2
Output: 4
Explanation: 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>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= nums.length &lt;= 105</code></li>
	<li><code>1 &lt;= nums[i] &lt;= nums.length</code></li>
	<li><code>0 &lt;= k &lt;= nums.length</code></li>
</ul>

longestEqualSubarray

Find the Longest Equal Subarray

Array

Hash Table

Binary Search

Sliding Window

给你一个下标从 0 开始的整数数组 <code>nums</code> 和一个整数 <code>k</code> 。

如果子数组中所有元素都相等，则认为子数组是一个 等值子数组 。注意，空数组是 等值子数组 。

从 <code>nums</code> 中删除最多 <code>k</code> 个元素后，返回可能的最长等值子数组的长度。

子数组 是数组中一个连续且可能为空的元素序列。

&nbsp;

示例 1：

<pre>
输入：nums = [1,3,2,3,1,3], k = 3
输出：3
解释：最优的方案是删除下标 2 和下标 4 的元素。
删除后，nums 等于 [1, 3, 3, 3] 。
最长等值子数组从 i = 1 开始到 j = 3 结束，长度等于 3 。
可以证明无法创建更长的等值子数组。
</pre>

示例 2：

<pre>
输入：nums = [1,1,2,2,1,1], k = 2
输出：4
解释：最优的方案是删除下标 2 和下标 3 的元素。 
删除后，nums 等于 [1, 1, 1, 1] 。 
数组自身就是等值子数组，长度等于 4 。 
可以证明无法创建更长的等值子数组。
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= nums.length &lt;= 105</code></li>
	<li><code>1 &lt;= nums[i] &lt;= nums.length</code></li>
	<li><code>0 &lt;= k &lt;= nums.length</code></li>
</ul>