<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<int>ans;
        int m=mat.size();
        int n=mat[0].size();
        if(m==1 or n==1){
            for(auto x:mat){
                for(auto k:x)
                    ans.push_back(k);
            }
            return ans;
            
        }
        ans.push_back(mat[0][0]);
        int i=0,j=1;
        while(ans.size()!=m*n){
            while(i<m and j>=0){
                ans.push_back(mat[i][j]);
                i++;
                j--;
            }

            if(i<m)
                j=0;
            else{
                i=m-1;
                j=j+2;
            }
            while(i>=0 and j<n){
                ans.push_back(mat[i][j]);
                i--;
                j++;
            }
            if(j<n)
                i=0;
            else{
                i=i+2;
                j=n-1;
            }      
        }
        return ans;  
    }
};
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