<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) {
        if (mat.empty() || mat[0].empty()) return {};
        
        int m = mat.size();
        int n = mat[0].size();
        vector<int> result(m * n);
        int row = 0, col = 0;
        
        for (int i = 0; i < m * n; i++) {
            result[i] = mat[row][col];
            
            if ((row + col) % 2 == 0) {
                // Moving up
                if (col == n - 1) {
                    row++;
                } else if (row == 0) {
                    col++;
                } else {
                    row--;
                    col++;
                }
            } else {
                // Moving down
                if (row == m - 1) {
                    col++;
                } else if (col == 0) {
                    row++;
                } else {
                    row++;
                    col--;
                }
            }
        }
        
        return result;        
    }
};
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