<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>>& g) {
    int m = g.size(), n = g[0].size();
    vector<int> res;
    res.reserve(m * n);
    auto in_bound = [&](int x, int y) {
      return 0 <= x && x < m && 0 <= y && y < n;
    };
    for (int sum = 0; sum < m * n; ++sum) {
      if (sum & 1) {
        for (int i = 0; i <= sum; ++i)
          if (in_bound(i, sum - i)) res.push_back(g[i][sum - i]);
      } else {
        for (int i = sum; i >= 0; --i)
          if (in_bound(i, sum - i)) res.push_back(g[i][sum - i]);
      }
    }
    return res;
  }
};
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