<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 m = mat.size();

    int n = mat[0].size();

    vector<int> ans(m * n);

    int k = 0;

    int count = 0;

    

    int i = 0, j = 0;



    while (count <= m + n - 2) {

        if (count % 2 == 0) {

            // Traverse upwards

            while (i >= 0 && j < n) {

                ans[k++] = mat[i--][j++];

            }

            if (j < n) { 

                i++;

            } else { 

                i += 2;

                j--;

            }

        } else {

            // Traverse downwards

            while (i < m && j >= 0) {

                ans[k++] = mat[i++][j--];

            }

            if (i < m) { 

                j++;

            } else {

                j += 2;

                i--;

            }

        }

        count++;

    }



    return ans;
    }
};
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