# 30. Sort the Matrix Diagonally

A **matrix diagonal** is a diagonal line of cells starting from some cell in either the topmost row or leftmost column and going in the bottom-right direction until reaching the matrix's end. For example, the **matrix diagonal** starting from `mat[2][0]`, where `mat` is a `6 x 3` matrix, includes cells `mat[2][0]`, `mat[3][1]`, and `mat[4][2]`.

Given an `m x n` matrix `mat` of integers, sort each **matrix diagonal** in ascending order and return *the resulting matrix*.

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/01/21/1482_example_1_2.png)

```
Input: mat = [[3,3,1,1],[2,2,1,2],[1,1,1,2]]
Output: [[1,1,1,1],[1,2,2,2],[1,2,3,3]]
```

**Example 2:**

```
Input: mat = [[11,25,66,1,69,7],[23,55,17,45,15,52],[75,31,36,44,58,8],[22,27,33,25,68,4],[84,28,14,11,5,50]]
Output: [[5,17,4,1,52,7],[11,11,25,45,8,69],[14,23,25,44,58,15],[22,27,31,36,50,66],[84,28,75,33,55,68]]
```

## **Solution:**

**Approach**\
**Using index(i-j) as the key and storing all diagonal elements**\
**Sorting all diagonal**\
**Coping values to original matrix**

```cpp
class Solution
{
public:
    vector<vector<int>> diagonalSort(vector<vector<int>> &mat)
    {

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

        unordered_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]);
            }
        }

        for (auto itr = mp.begin(); itr != mp.end(); itr++)
        {
            sort(itr->second.begin(), itr->second.end());
        }

        for (int i = n - 1; i >= 0; i--)
        {
            for (int j = m - 1; j >= 0; j--)
            {
                int idx = i - j;
                mat[i][j] = mp[idx].back();
                mp[idx].pop_back();
            }
        }

        return mat;
    }
};
```

**Time Complexity: O(n \*m \* dlogd )  d=min(n,m)**