<p>Given an <code>m x n</code> matrix <code>mat</code>, return <em>an array of all the elements of the array in a diagonal order</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/10/diag1-grid.jpg" style="width: 334px; height: 334px;" />
<pre>
<strong>Input:</strong> mat = [[1,2,3],[4,5,6],[7,8,9]]
<strong>Output:</strong> [1,2,4,7,5,3,6,8,9]
</pre>

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

<pre>
<strong>Input:</strong> mat = [[1,2],[3,4]]
<strong>Output:</strong> [1,2,3,4]
</pre>

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

<ul>
	<li><code>m == mat.length</code></li>
	<li><code>n == mat[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>5</sup> &lt;= mat[i][j] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    vector<int> findDiagonalOrder(vector<vector<int>>& mat) {
        vector<vector<int>> fin;
        int i,j;
        vector<int> tmp;
        fin.push_back(tmp);
        for(i=0;i<mat.size();i++)
        {
            for(j=0;j<mat[0].size();j++)
            {
                if(i+j>fin.size() - 1)
                {
                    vector<int> tmp;
                    fin.push_back(tmp);
                }
                //cout<<i<<' '<<j<<'\n';

                fin[i+j].push_back(mat[i][j]);

            }
        }

        vector<int> ans;

        for(i=0;i<fin.size();i++)
        {
            if(i%2==0)
            {
                for(j=fin[i].size()-1;j>=0;j--)
                {
                    ans.push_back(fin[i][j]);
                }
            }
            else
            {
                for(j=0;j<fin[i].size();j++)
                {
                    ans.push_back(fin[i][j]);
                }
            }
        }

        

        return ans;
    }
};
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