<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) {
        auto result = 0;
        
        if(height.size() < 1) {
            return result;
        }

        size_t left = 0, right = height.size() -1, current = 0;
        int max_left_value = height.at(left), max_right_value = height.at(right);

        while(left != right) {
            auto min_height = std::min(max_left_value, max_right_value);
            auto temp_result = min_height - height.at(current);
            if(temp_result > 0) {
                result += temp_result;
            }

            if(max_left_value < max_right_value) {
                left++;
                max_left_value = std::max(max_left_value, height.at(left));
                current = left;
            } else {
                right--;
                max_right_value = std::max(max_right_value, height.at(right));
                current = right;
            }
        }
        return result;
        
    }
};
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