<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(),m=mat[0].size();

        vector<int>ans(n*m);
        int i=0;
map<int,vector<int> >mp;
for(int i=0;i<n;i++){
    for(int j=0;j<m;j++){
        mp[i+j].push_back(mat[i][j]);
    }
}
bool up=1;
for(auto & it:mp){


    if(up){
        reverse(it.second.begin(),it.second.end());
    }
    for(int & num:it.second)ans[i++]=num;

    up=!up;
}
return ans;






       }
};
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