<p>Given <code>n</code> non-negative integers representing an elevation map where the width of each bar is <code>1</code>, compute how much water it can trap after raining.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img src="https://assets.leetcode.com/uploads/2018/10/22/rainwatertrap.png" style="width: 412px; height: 161px;" />
<pre>
<strong>Input:</strong> height = [0,1,0,2,1,0,1,3,2,1,2,1]
<strong>Output:</strong> 6
<strong>Explanation:</strong> The above elevation map (black section) is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped.
</pre>

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

<pre>
<strong>Input:</strong> height = [4,2,0,3,2,5]
<strong>Output:</strong> 9
</pre>

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

<ul>
	<li><code>n == height.length</code></li>
	<li><code>1 &lt;= n &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= height[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    
int trap(vector<int>& height) {
    int n=height.size();
    vector<int> maxleft(n,-1);
    vector<int> maxright(n,-1);
    int found=n;
    int curr_max_l=0,curr_max_r=0;
    int ans=0;
    for(int i=0;i<n;i++){
        maxleft[i]=curr_max_l;
        curr_max_l=max(curr_max_l,height[i]);
        maxright[n-1-i]=curr_max_r;
        curr_max_r=max(curr_max_r,height[n-1-i]);
        if(maxright[i]>=0 && maxleft[i]>=0){
        //if(i<found) found=i;
        int expected_val=(min(maxleft[i],maxright[i])-height[i]);
        ans+=(expected_val>0)?expected_val:0;  
        }
    }
    for(int i=0;i<n/2;i++){
        int expected_val=(min(maxleft[i],maxright[i])-height[i]);
        ans+=(expected_val>0)?expected_val:0;
    }
    return ans;
}
};
25800