Skip to content
Related Articles
Rotate matrix by 45 degrees
• Difficulty Level : Hard
• Last Updated : 28 Sep, 2020

Given a matrix mat[][] of size N*N, the task is to rotate the matrix by 45 degrees and print the matrix.

Examples:

Input: N = 6,
mat[][] = {{3, 4, 5, 1, 5, 9, 5},
{6, 9, 8, 7, 2, 5, 2},
{1, 5, 9, 7, 5, 3, 2},
{4, 7, 8, 9, 3, 5, 2},
{4, 5, 2, 9, 5, 6, 2},
{4, 5, 7, 2, 9, 8, 3}}
Output:
3
6 4
1 9 5
4 5 8 1
4 7 9 7 5
4 5 8 7 2 9
5 2 9 5 5
7 9 3 3
2 5 5
9 6
8

Input: N = 4,
mat[][] = {{2, 5, 7, 2},
{9, 1, 4, 3},
{5, 8, 2, 3},
{6, 4, 6, 3}}

Output:
2
9 5
5 1 7
6 8 4 2
4 2 3
6 3
3

Approach: Follow the steps given below in order to solve the problem:

1. Store the diagonal elements in a list using a counter variable.
2. Print the number of spaces required to make the output look like the desired pattern.
3. Print the list elements after reversing the list.
4. Traverse through only diagonal elements to optimize the time taken by the operation.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to rotate matrix by 45 degree``void` `matrix(``int` `n, ``int` `m, vector> li)``{``    ` `    ``// Counter Variable``    ``int` `ctr = 0;``    ` `    ``while` `(ctr < 2 * n - 1)``    ``{``        ``for``(``int` `i = 0;``                ``i < ``abs``(n - ctr - 1);``                ``i++)``        ``{``            ``cout << ``" "``;``        ``}``        ` `        ``vector<``int``> lst;` `        ``// Iterate [0, m]``        ``for``(``int` `i = 0; i < m; i++)``        ``{``            ` `            ``// Iterate [0, n]``            ``for``(``int` `j = 0; j < n; j++)``            ``{``                ` `                ``// Diagonal Elements``                ``// Condition``                ``if` `(i + j == ctr)``                ``{``                    ` `                    ``// Appending the``                    ``// Diagonal Elements``                    ``lst.push_back(li[i][j]);``                ``}``            ``}``        ``}``            ` `        ``// Printing reversed Diagonal``        ``// Elements``        ``for``(``int` `i = lst.size() - 1; i >= 0; i--)``        ``{``            ``cout << lst[i] << ``" "``;``        ``}``        ``cout << endl;``        ``ctr += 1;``    ``}``}` `// Driver code   ``int` `main()``{``    ` `    ``// Dimensions of Matrix``    ``int` `n = 8;``    ``int` `m = n;``    ` `    ``// Given matrix``    ``vector> li{``        ``{ 4, 5, 6, 9, 8, 7, 1, 4 },``        ``{ 1, 5, 9, 7, 5, 3, 1, 6 },``        ``{ 7, 5, 3, 1, 5, 9, 8, 0 },``        ``{ 6, 5, 4, 7, 8, 9, 3, 7 },``        ``{ 3, 5, 6, 4, 8, 9, 2, 1 },``        ``{ 3, 1, 6, 4, 7, 9, 5, 0 },``        ``{ 8, 0, 7, 2, 3, 1, 0, 8 },``        ``{ 7, 5, 3, 1, 5, 9, 8, 5 } };``    ` `    ``// Function call``    ``matrix(n, m, li);` `    ``return` `0;``}` `// This code is contributed by divyeshrabadiya07`

## Java

 `// Java program for``// the above approach``import` `java.util.*;``class` `GFG{` `// Function to rotate``// matrix by 45 degree``static` `void` `matrix(``int` `n, ``int` `m,``                   ``int` `[][]li)``{``  ``// Counter Variable``  ``int` `ctr = ``0``;` `  ``while` `(ctr < ``2` `* n - ``1``)``  ``{``    ``for``(``int` `i = ``0``; i < Math.abs(n - ctr - ``1``);``            ``i++)``    ``{``      ``System.out.print(``" "``);``    ``}` `    ``Vector lst = ``new` `Vector();` `    ``// Iterate [0, m]``    ``for``(``int` `i = ``0``; i < m; i++)``    ``{``      ``// Iterate [0, n]``      ``for``(``int` `j = ``0``; j < n; j++)``      ``{``        ``// Diagonal Elements``        ``// Condition``        ``if` `(i + j == ctr)``        ``{``          ``// Appending the``          ``// Diagonal Elements``          ``lst.add(li[i][j]);``        ``}``      ``}``    ``}` `    ``// Printing reversed Diagonal``    ``// Elements``    ``for``(``int` `i = lst.size() - ``1``; i >= ``0``; i--)``    ``{``      ``System.out.print(lst.get(i) + ``" "``);``    ``}``    ` `    ``System.out.println();``    ``ctr += ``1``;``  ``}``}` `// Driver code   ``public` `static` `void` `main(String[] args)``{   ``  ``// Dimensions of Matrix``  ``int` `n = ``8``;``  ``int` `m = n;` `  ``// Given matrix``  ``int``[][] li = {{``4``, ``5``, ``6``, ``9``, ``8``, ``7``, ``1``, ``4``},``                ``{``1``, ``5``, ``9``, ``7``, ``5``, ``3``, ``1``, ``6``},``                ``{``7``, ``5``, ``3``, ``1``, ``5``, ``9``, ``8``, ``0``},``                ``{``6``, ``5``, ``4``, ``7``, ``8``, ``9``, ``3``, ``7``},``                ``{``3``, ``5``, ``6``, ``4``, ``8``, ``9``, ``2``, ``1``},``                ``{``3``, ``1``, ``6``, ``4``, ``7``, ``9``, ``5``, ``0``},``                ``{``8``, ``0``, ``7``, ``2``, ``3``, ``1``, ``0``, ``8``},``                ``{``7``, ``5``, ``3``, ``1``, ``5``, ``9``, ``8``, ``5``}};` `  ``// Function call``  ``matrix(n, m, li);``}``}` `// This code is contributed by Princi Singh`

## Python3

 `# Python3 program for the above approach` `# Function to rotate matrix by 45 degree`  `def` `matrix(n, m, li):` `    ``# Counter Variable``    ``ctr ``=` `0``    ``while``(ctr < ``2` `*` `n``-``1``):``        ``print``(``" "``*``abs``(n``-``ctr``-``1``), end ``=``"")``        ``lst ``=` `[]` `        ``# Iterate [0, m]``        ``for` `i ``in` `range``(m):` `                ``# Iterate [0, n]``            ``for` `j ``in` `range``(n):` `                ``# Diagonal Elements``                ``# Condition``                ``if` `i ``+` `j ``=``=` `ctr:` `                    ``# Appending the``                    ``# Diagonal Elements``                    ``lst.append(li[i][j])` `        ``# Printing reversed Diagonal``        ``# Elements``        ``lst.reverse()``        ``print``(``*``lst)``        ``ctr ``+``=` `1`  `# Driver Code` `# Dimensions of Matrix``n ``=` `8``m ``=` `n` `# Given matrix``li ``=` `[[``4``, ``5``, ``6``, ``9``, ``8``, ``7``, ``1``, ``4``],``      ``[``1``, ``5``, ``9``, ``7``, ``5``, ``3``, ``1``, ``6``],``      ``[``7``, ``5``, ``3``, ``1``, ``5``, ``9``, ``8``, ``0``],``      ``[``6``, ``5``, ``4``, ``7``, ``8``, ``9``, ``3``, ``7``],``      ``[``3``, ``5``, ``6``, ``4``, ``8``, ``9``, ``2``, ``1``],``      ``[``3``, ``1``, ``6``, ``4``, ``7``, ``9``, ``5``, ``0``],``      ``[``8``, ``0``, ``7``, ``2``, ``3``, ``1``, ``0``, ``8``],``      ``[``7``, ``5``, ``3``, ``1``, ``5``, ``9``, ``8``, ``5``]]` `# Function Call``matrix(n, m, li)`

## C#

 `// C# program for``// the above approach``using` `System;``using` `System.Collections;``class` `GFG{`` ` `// Function to rotate``// matrix by 45 degree``static` `void` `matrix(``int` `n, ``int` `m,``                   ``int` `[,]li)``{``  ``// Counter Variable``  ``int` `ctr = 0;` `  ``while` `(ctr < 2 * n - 1)``  ``{``    ``for``(``int` `i = 0;``            ``i < Math.Abs(n - ctr - 1);``            ``i++)``    ``{``      ``Console.Write(``" "``);``    ``}` `    ``ArrayList lst = ``new` `ArrayList();` `    ``// Iterate [0, m]``    ``for``(``int` `i = 0; i < m; i++)``    ``{``      ``// Iterate [0, n]``      ``for``(``int` `j = 0; j < n; j++)``      ``{``        ``// Diagonal Elements``        ``// Condition``        ``if` `(i + j == ctr)``        ``{``          ``// Appending the``          ``// Diagonal Elements``          ``lst.Add(li[i, j]);``        ``}``      ``}``    ``}` `    ``// Printing reversed Diagonal``    ``// Elements``    ``for``(``int` `i = lst.Count - 1;``            ``i >= 0; i--)``    ``{``      ``Console.Write((``int``)lst[i] + ``" "``);``    ``}` `    ``Console.Write(``"\n"``);``    ``ctr += 1;``  ``}``}`` ` `// Driver code   ``public` `static` `void` `Main(``string``[] args)``{   ``  ``// Dimensions of Matrix``  ``int` `n = 8;``  ``int` `m = n;`` ` `  ``// Given matrix``  ``int``[,] li = {{4, 5, 6, 9, 8, 7, 1, 4},``               ``{1, 5, 9, 7, 5, 3, 1, 6},``               ``{7, 5, 3, 1, 5, 9, 8, 0},``               ``{6, 5, 4, 7, 8, 9, 3, 7},``               ``{3, 5, 6, 4, 8, 9, 2, 1},``               ``{3, 1, 6, 4, 7, 9, 5, 0},``               ``{8, 0, 7, 2, 3, 1, 0, 8},``               ``{7, 5, 3, 1, 5, 9, 8, 5}};`` ` `  ``// Function call``  ``matrix(n, m, li);``}``}` `// This code is contributed by Rutvik_56`
Output:
```       4
1 5
7 5 6
6 5 9 9
3 5 3 7 8
3 5 4 1 5 7
8 1 6 7 5 3 1
7 0 6 4 8 9 1 4
5 7 4 8 9 8 6
3 2 7 9 3 0
1 3 9 2 7
5 1 5 1
9 0 0
8 8
5

```

Time Complexity: O(N2)
Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live

My Personal Notes arrow_drop_up