Find the Maximum Length of Valid Subsequence I
You are given an integer array <code>nums</code>.

<p>A <span data-keyword="subsequence-array">subsequence</span> <code>sub</code> of <code>nums</code> with length <code>x</code> is called <strong>valid</strong> if it satisfies:</p>

<ul>
	<li><code>(sub[0] + sub[1]) % 2 == (sub[1] + sub[2]) % 2 == ... == (sub[x - 2] + sub[x - 1]) % 2.</code></li>
</ul>

<p>Return the length of the <strong>longest</strong> <strong>valid</strong> subsequence of <code>nums</code>.</p>

<p>A <strong>subsequence</strong> is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.</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 = [1,2,3,4]</span></p>

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

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

<p>The longest valid subsequence is <code>[1, 2, 3, 4]</code>.</p>
</div>

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

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

<p><strong>Output:</strong> 6</p>

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

<p>The longest valid subsequence is <code>[1, 2, 1, 2, 1, 2]</code>.</p>
</div>

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

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

<p><strong>Output:</strong> <span class="example-io">2</span></p>

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

<p>The longest valid subsequence is <code>[1, 3]</code>.</p>
</div>

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

<ul>
	<li><code>2 &lt;= nums.length &lt;= 2 * 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>7</sup></code></li>
</ul>
Array; Dynamic Programming
func maximumLength(nums []int) (ans int) {
	k := 2
	f := make([][]int, k)
	for i := range f {
		f[i] = make([]int, k)
	}
	for _, x := range nums {
		x %= k
		for j := 0; j < k; j++ {
			y := (j - x + k) % k
			f[x][y] = f[y][x] + 1
			ans = max(ans, f[x][y])
		}
	}
	return
}