<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 lastindex=height.size()-1;
        int L=1;
        int R=lastindex-1;
        int Lmax=height[0];
        int Rmax=height[lastindex];
        vector<int> w(height.size(),0);
        
        while(L<=R)
        {
          
          if(Lmax <= Rmax)
          {
           // index=i;
            w[L]=Lmax-height[L];
            Lmax=max(Lmax,height[L]);
            L++;
          }
          
          if(Rmax <= Lmax)
          {
            w[R]=Rmax-height[R];
            Rmax=max(Rmax,height[R]);
            R--;
          }
          
          
        }
        int water=0;
        for(int i=0;i<=lastindex;i++)
        {
         // cout << " i "<< i<<" "<<w[i]<<endl;
         if(w[i]>0)
         water=water+w[i];
        }
        return water;
        
        
    }
};
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