<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>

class Solution {
public:
    bool checkPossibility(vector<int>& nums) {
        int cnt=0;
        int n=nums.size();
        for(int i=0;i<n-1;i++){
            if(nums[i]>nums[i+1]){
                cnt++;
            
            if(cnt>1){
                return false;
            }
            if(i>0 && nums[i-1]>nums[i+1]){
                nums[i+1]=nums[i];
            }
        }
        }
        return true;
    }
};
29700