C++ Program to Rotate matrix by 45 degrees

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(N) since using auxiliary space for vector

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