<p>Given a <code>triangle</code> array, return <em>the minimum path sum from top to bottom</em>.</p>

<p>For each step, you may move to an adjacent number of the row below. More formally, if you are on index <code>i</code> on the current row, you may move to either index <code>i</code> or index <code>i + 1</code> on the next row.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> triangle = [[2],[3,4],[6,5,7],[4,1,8,3]]
<strong>Output:</strong> 11
<strong>Explanation:</strong> The triangle looks like:
   <u>2</u>
  <u>3</u> 4
 6 <u>5</u> 7
4 <u>1</u> 8 3
The minimum path sum from top to bottom is 2 + 3 + 5 + 1 = 11 (underlined above).
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> triangle = [[-10]]
<strong>Output:</strong> -10
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= triangle.length &lt;= 200</code></li>
	<li><code>triangle[0].length == 1</code></li>
	<li><code>triangle[i].length == triangle[i - 1].length + 1</code></li>
	<li><code>-10<sup>4</sup> &lt;= triangle[i][j] &lt;= 10<sup>4</sup></code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Could you&nbsp;do this using only <code>O(n)</code> extra space, where <code>n</code> is the total number of rows in the triangle?
class Solution {
public:
    
    //  vector<vector<int>> dp;
    //  Solution() : dp(201,vector<int>(201,-1)) {}
    // int solve(vector<vector<int>>& triangle,int i,int j){
    //     if(j==triangle.size()-1) return triangle[j][i];
    //     if(dp[j][i]!=-1){
    //         return dp[j][i];
    //     }

    //     int down=solve(triangle,i,j+1);
    //     int dia=solve(triangle,i+1,j+1);

    //     return dp[j][i]=min(down,dia)+triangle[j][i];

    // }

    int minimumTotal(vector<vector<int>>& triangle) {

        vector<vector<int>> dp(triangle.size()+1,vector<int> (triangle.size()+1,0));
        
        for(int i=0;i<triangle.size();i++){
            dp[triangle.size()-1][i]=triangle[triangle.size()-1][i];
        }

        for(int i=triangle.size()-2;i>=0;i--){
            for(int j=0;j<=i;j++){
                int down=dp[i+1][j];
                int dia=dp[i+1][j+1];

               dp[i][j]=min(down,dia)+triangle[i][j];

            }
        }
        
        return dp[0][0];
        
    }
};
11800