<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 > & arr) {
  int n = arr.size();
  int prefix[n], suffix[n];
  prefix[0] = arr[0];
  for (int i = 1; i < n; i++) {
    prefix[i] = max(prefix[i - 1], arr[i]);
  }
  suffix[n - 1] = arr[n - 1];
  for (int i = n - 2; i >= 0; i--) {
    suffix[i] = max(suffix[i + 1], arr[i]);
  }
  int waterTrapped = 0;
  for (int i = 0; i < n; i++) {
    waterTrapped += min(prefix[i], suffix[i]) - arr[i];
  }
  return waterTrapped;
}
};
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