<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> leftMax(n),rightMax(n);
       // to calculate leftMax array
       leftMax[0]=height[0];
       for(int i=1;i<n;i++){
          leftMax[i] = max(leftMax[i-1],height[i]);
       }
       rightMax[n-1]=height[n-1];
       for(int i=n-2;i>=0;i--){
        rightMax[i] = max(rightMax[i+1],height[i]);
       }
       int total = 0;
       for(int i=0;i<n;i++){
        total += min(leftMax[i],rightMax[i]) - height[i];
       }
       return total;
    }
};
25900