<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:
    def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
        N, M = len(mat), len(mat[0])
        def check(i, j):
            if i >= 0 and j >= 0 and i < N and j < M:
                return True
            return False

        ans = []
        for i in range(N):
            row = []
            for j in range(N):
                if check(i-j, j):
                    row.append(mat[i-j][j])
            if i%2 == 0:
                ans.extend(row)
            else:
                ans.extend(row[::-1])

        for j in range(1, M):
            row = []
            for i in range(N-1, -1, -1):
                if check(i, j-i+N-1):
                    row.append(mat[i][j-i+N-1])
            if (j+N-1)%2 == 0:
                ans.extend(row)
            else:
                ans.extend(row[::-1])
        
        return ans

    # 2 5 8
    # 4 0 -1


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