<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) {
        int N = mat.size();
        int M = mat[0].size();
        vector<int> answer;

        int x = 0; 
        int sum = 0; 
        int dir = -1;

        auto addToAnswer = [&]() {
            int y = sum - x;
            answer.push_back(mat[x][y]);
        };
        auto ok = [&](int x) {
            int y = sum - x;
            return x >= 0 && y >= 0 && x < N && y < M;
        };
        addToAnswer();
        while (sum != N + M - 2) {
            sum++;
            if ((dir == -1 && !ok(x)) || (dir == 1 && (ok(x + 1)))) {
                x++;
            }
            addToAnswer();
            dir = -dir;

            while (ok(x + dir)) {
                x += dir;
                addToAnswer();
            }
        }
        return answer;   
    }
};
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