<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>& h) {
        
        int leftmax[h.size()];
        leftmax[0]=0;
        for(int i=1;i<h.size();i++){
              leftmax[i]=max(leftmax[i-1],h[i-1]);
        }
        int rightmax[h.size()];
        rightmax[h.size()-1]=0;
        for(int j=h.size()-2;j>=0;j--){
               rightmax[j]=max(rightmax[j+1],h[j+1]);
        }
        int totalarea=0;
        for(int i=0;i<h.size();i++){
            int area=0;
             if(min(leftmax[i],rightmax[i])<=h[i]){
              
             }
             else{
                 area=min(leftmax[i],rightmax[i])-h[i];
                 totalarea+=area;

             }
        }
    return totalarea;    }
    
    
};
25100