<p>Given an integer array <code>nums</code>, return <code>true</code><em> if there exists a triple of indices </em><code>(i, j, k)</code><em> such that </em><code>i &lt; j &lt; k</code><em> and </em><code>nums[i] &lt; nums[j] &lt; nums[k]</code>. If no such indices exists, return <code>false</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5]
<strong>Output:</strong> true
<strong>Explanation:</strong> Any triplet where i &lt; j &lt; k is valid.
</pre>

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

<pre>
<strong>Input:</strong> nums = [5,4,3,2,1]
<strong>Output:</strong> false
<strong>Explanation:</strong> No triplet exists.
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,1,5,0,4,6]
<strong>Output:</strong> true
<strong>Explanation:</strong> The triplet (3, 4, 5) is valid because nums[3] == 0 &lt; nums[4] == 4 &lt; nums[5] == 6.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>5</sup></code></li>
	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Could you implement a solution that runs in <code>O(n)</code> time complexity and <code>O(1)</code> space complexity?
class Solution {
public:
    bool increasingTriplet(vector<int>& nums) {
        if(nums.size()<3){
            return false;
        }
        if(nums.size() == 3){
            return nums[0]<nums[1] && nums[1] < nums[2];
        }
        // arr -> 1 0 2 0 3 0 4 0 5 
        // brute force
        // select k 
        // select j that satisfies 
        // find i that satisfies


        // for(int k = nums.size() - 1; i>1 ; i--){
        //     for (int j = k-1 ; j>0 ;j--){
        //         for (int i = j-1 ; i>=0 ; i--){
        //             if(nums[k]>nums[j] && nums[j] > nums[i]) return true;
        //         }
        //     }
        // }

        // need to optimise this code

        // think in nlogn

        // for each element store the min before it and the max after it

        vector<int> min_arr = vector<int> (nums.size(),INT_MAX);
        int max_num = nums[nums.size() -1];
        // min[0] = nums[0];
        for (int i =1 ; i < nums.size() ; i++){
            min_arr[i] = min(min_arr[i-1] , nums[i-1]);
        }
        for (int i = nums.size() -2 ; i >0 ; i--){
            if(nums[i]< max_num &&  nums[i]> min_arr[i]) return true;
            max_num = max(max_num ,nums[i]);
        }

        return false;
    }
};
119100