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:

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// 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
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// 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<Integer> lst = new Vector<Integer>();
 
    // 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
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# 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)
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// 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
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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)

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