<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) {
        if(is_sorted(nums.begin(), nums.end())) {
            return true;
        }
        vector<int> ans;
        ans.push_back(nums[0]);
        for(int i=1;i<nums.size();i++) {
            if(nums[i] >= ans.back()) {
                ans.push_back(nums[i]);
            } else {
                int idx = upper_bound(ans.begin(), ans.end(), nums[i])-ans.begin();
                ans[idx] = nums[i];
            }
        }
        // cout<<ans.size()<<endl;
        return ans.size() == nums.size()-1;
    }
};
30500