<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>& h) {
    //     int n=h.size();
    //     vector<int> t(n);
    //     t[n-1]=h[n-1];
    //     for(int i = n-2; i>=0;i--){
    //         t[i]=min(t[i+1],h[i]);
    //     }
    //     int sum=0;
    //     int maxi=h[0];
    //     for(int i=1;i<n-1;i++){
    //         sum+= max(0,min(maxi,t[i+1])-h[i]);
    //         maxi = max(maxi, h[i]);
    //     }
    // }

    int trap(vector<int>& A){
        int n=A.size();
        vector<int> dp(n);
        int left_max=0;
        for(int i=0;i<n;i++){
            dp[i]=left_max-A[i]>=0 ? left_max-A[i]: 0 ;
            left_max = max(left_max, A[i]);
        }

        int right_max=0;
        for(int i=n-1;i>=0;i--){
            dp[i]=min(dp[i], right_max-A[i]>=0 ? right_max-A[i]: 0);
            right_max = max(right_max, A[i]);
        }

        return accumulate(dp.begin(), dp.end(), 0);
    }
};
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