<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();
        int prefix[n];
        int suffix[n];
        int total=0;
        prefix[0]=height[0];
        suffix[n-1]=height[n-1];
        for(int i=1;i<n;i++){
            prefix[i]=max(prefix[i-1],height[i]);
        }
        for(int i=n-2;i>=0;i--){
            suffix[i]=max(suffix[i+1],height[i]);
        }
        for(int i=0;i<n;i++){
            int leftmax=prefix[i],rightmax=suffix[i];
            if(height[i]<leftmax and height[i]<rightmax){
                total+=min(leftmax,rightmax)-height[i];
            }
        }
        return total;
    }
};
25000