<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()) return {};
  vector<int> ret;

  unordered_map<int, vector<int>> mp; // key = (i + j)                                                                                                                                        
  int m = mat.size();
  int n = mat[0].size();
  for (int i = 0;i < m; ++i) {
    for (int j = 0; j < n; ++j) {
      mp[i+j].push_back(mat[i][j]);
    }
  }
  for (int i = 0; i <= m + n -2; i++) {
    if (i % 2 == 1) {
      ret.insert(ret.end(), mp[i].begin(), mp[i].end());
    } else {
      ret.insert(ret.end(), mp[i].rbegin(), mp[i].rend());
    }
  }

  return ret;
}

};
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