<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 pref[n],suff[n];
         pref[0]=height[0];
         for(int i=1; i<n; i++){
            pref[i]= max(pref[i-1], height[i]);
         }
         suff[n-1]=height[n-1];
         for(int i=n-2; i>=0; i--)
         { 
             suff[i]=max(suff[i+1], height[i]);
         }
          int ans=0;
          for(int i=0; i<n; i++){
            int lmax= pref[i];
            int rmax= suff[i];
             if( lmax>=height[i] && rmax>= height[i]){
                ans+= min(lmax, rmax)-height[i];
             }
          }
          return ans;

    }
};
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