# 665-non-decreasing-array


Try it on <a href='https://leetcode.com/problems/665-non-decreasing-array'>leetcode</a>

## Description
<div class="description">
<div><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>Example 1:</strong></p>

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

<p><strong>Example 2:</strong></p>

<pre><strong>Input:</strong> nums = [4,2,1]
<strong>Output:</strong> false
<strong>Explanation:</strong> You can't get a non-decreasing array by modify 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>
</div>
</div>

## Solution(Python)
```Python
class Solution:
    def checkPossibility(self, N: List[int]) -> bool:
        err = 0
        for i in range(1, len(N)):
            if N[i] < N[i-1]:
                if err or (i > 1 and i < len(N) - 1 and N[i-2] > N[i] and N[i+1] < N[i-1]):
                    return False
                err = 1
        return True
```