Skip to content
Related Articles

Related Articles

C++ Program to Rotate matrix by 45 degrees

View Discussion
Improve Article
Save Article
Like Article
  • Last Updated : 27 Jan, 2022

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 <bits/stdc++.h>
using namespace std;
  
// Function to rotate matrix by 45 degree
void matrix(int n, int m, vector<vector<int>> 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<vector<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);
  
    return 0;
}
  
// This code is contributed by divyeshrabadiya07

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)

Please refer complete article on Rotate matrix by 45 degrees for more details!


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!