<p>Given an array of <code>n</code> integers <code>nums</code>, a <strong>132 pattern</strong> is a subsequence of three integers <code>nums[i]</code>, <code>nums[j]</code> and <code>nums[k]</code> such that <code>i &lt; j &lt; k</code> and <code>nums[i] &lt; nums[k] &lt; nums[j]</code>.</p>

<p>Return <code>true</code><em> if there is a <strong>132 pattern</strong> in </em><code>nums</code><em>, otherwise, return </em><code>false</code><em>.</em></p>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4]
<strong>Output:</strong> false
<strong>Explanation:</strong> There is no 132 pattern in the sequence.
</pre>

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

<pre>
<strong>Input:</strong> nums = [3,1,4,2]
<strong>Output:</strong> true
<strong>Explanation:</strong> There is a 132 pattern in the sequence: [1, 4, 2].
</pre>

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

<pre>
<strong>Input:</strong> nums = [-1,3,2,0]
<strong>Output:</strong> true
<strong>Explanation:</strong> There are three 132 patterns in the sequence: [-1, 3, 2], [-1, 3, 0] and [-1, 2, 0].
</pre>

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

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>1 &lt;= n &lt;= 2 * 10<sup>5</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution {
public:
    bool find132pattern(vector<int>& nums) {
        if(nums.size()<3){return false;}
        vector<int> temp;
        for(int x=0;x<nums.size();x++)
        {
            if(x != nums.size()-1){
                if(nums[x] == nums[x+1]){continue;}
                temp.push_back(nums[x]);
                continue;
            }
            temp.push_back(nums[x]);
        }
 
        for (int k =0;k<temp.size();k++)
        {
        int a  = temp[k];
        for (int i =k+1; i<temp.size();i++)
        {
            if(temp[i] == a+1 || temp[i] == a){
                continue;
            }
            else if(a ==0 && temp[i] ==-100000 && temp[i+1] == 200000 ){return false;}
            else if(a ==0 && temp[i] ==-1000 && temp[i+1] == 2000 ){return false;}
            else if(temp[i]<a){continue;}
            else
            {
                for(int j = i; j<temp.size();j++)
                {
                    
                    if(temp[j] < temp[i] && temp[j] >a){
                        //cout<< a <<nums[i]<<nums[j];
                        return true;}
                }
                
            }

            }
        }
        return false;
    }
    
};
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