Non-decreasing Array
<p>Given an array <code>nums</code> with <code>n</code> integers, your task is to check if it could become non-decreasing by modifying <strong>at most one element</strong>.</p>

<p>We define an array is non-decreasing if <code>nums[i] &lt;= nums[i + 1]</code> holds for every <code>i</code> (<strong>0-based</strong>) such that (<code>0 &lt;= i &lt;= n - 2</code>).</p>

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

<pre>
<strong>Input:</strong> nums = [4,2,3]
<strong>Output:</strong> true
<strong>Explanation:</strong> You could modify the first 4 to 1 to get a non-decreasing array.
</pre>

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

<pre>
<strong>Input:</strong> nums = [4,2,1]
<strong>Output:</strong> false
<strong>Explanation:</strong> You cannot get a non-decreasing array by modifying at most one element.
</pre>

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

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>5</sup> &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>
Array
func checkPossibility(nums []int) bool {
	isSorted := func(nums []int) bool {
		for i, b := range nums[1:] {
			if nums[i] > b {
				return false
			}
		}
		return true
	}
	for i := 0; i < len(nums)-1; i++ {
		a, b := nums[i], nums[i+1]
		if a > b {
			nums[i] = b
			if isSorted(nums) {
				return true
			}
			nums[i] = a
			nums[i+1] = a
			return isSorted(nums)
		}
	}
	return true
}