<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) {
        unordered_map<int,vector<pair<int,int>>> map;
        int n = mat.size(), m = mat[0].size();
        vector<int> res;

        for(int r=0;r<n;r++) {
            for(int c=0;c<m;c++) {
                map[r+c].push_back({r,mat[r][c]});
            }
        }
        
        for(int diag_value=0;diag_value<=m+n-2;diag_value++) {
            bool odd = diag_value%2;
            if(odd) {
                // sort(map[diag_value].begin(),map[diag_value].end());
            } else {
                reverse(map[diag_value].begin(),map[diag_value].end());
            }
            for(pair<int,int>& cell:map[diag_value]) res.push_back(cell.second);
        }

        return res;

    }
};
/**
map<diagonal_value, vector<>>
diag_value = x+y / r+c for each cell

if(diag_value&1) -> sort vector in order of increasing rows
else sort desc by row
*/
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