<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) {
        
        int N = nums.size();
        int l_r_min[N]; 
        int r_l_max[N];
        int maxx_num = INT_MIN;
        int minn_num = INT_MAX;
        for(int i =0; i < nums.size(); i++) {

            maxx_num = max(maxx_num, nums[N-i-1]);
            minn_num = min(minn_num, nums[i]);
            l_r_min[i] = minn_num;
            r_l_max[N-i-1] = maxx_num;
            
        }
        

        for(int i =1; i < N-1; i++) {
            
            if(nums[i] > l_r_min[i-1] and nums[i] < r_l_max[i]) {
                return true;
            }
        }
        return false;
    }
};
118900