Rotate matrix by 45 degrees - GeeksforGeeks

05 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 54 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 76 8 4 2 4 2 3  6 3   3 
Follow the steps given below in order to solve the problem:
Store the diagonal elements in a list using a counter variable.Print the number of spaces required to make the output look like the desired pattern.Print the list elements after reversing the list.Traverse through only diagonal elements to optimize the time taken by the operation.
Store the diagonal elements in a list using a counter variable.
Print the number of spaces required to make the output look like the desired pattern.
Print the list elements after reversing the list.
Traverse through only diagonal elements to optimize the time taken by the operation.
Below is the implementation of the above approach:
C++
Java
Python3
C#
// C++ program for the above approach#include <bits/stdc++.h>using namespace std; // Function to rotate matrix by 45 degreevoid 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
// Java program for// the above approachimport java.util.*;class GFG{ // Function to rotate// matrix by 45 degreestatic 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
# 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 Matrixn = 8m = n # Given matrixli = [[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 Callmatrix(n, m, li)
// C# program for// the above approachusing System;using System.Collections;class GFG{  // Function to rotate// matrix by 45 degreestatic 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
       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(N3)Auxiliary Space: O(N)
 (by rythmrana2)
Follow the given steps to print the matrix rotated by 45 degree:
print the spaces required.print the element this way –
print the spaces required.
print the element this way –
          traverse the matrix the way you want to print it – 
We will print the matrix by dividing it into two parts using two for loops , the first loop will print the first triangle of the matrix and the second loop will print the second triangle of the matrix
for the first half of matrix, take a loop that goes from the first row to the last row of the matrix.
at each row , take a while loop which prints the first element of the row and the second element of the row above the current row and the third element of the row which is twice above the current row, going this way we will print the first half triangle of the matrix.
take two counters, c for counting how many more elements the current diagonal have and counter inc for accessing those elements.
similarly do this for the second half of the matrix.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>using namespace std; // function to print the matrix rotated at 45 degree.void rotate(int m, int n, int* a){    // for printing the first half of matrix    for (int i = 0; i < m; i++) {         // print spaces        for (int k = 0; k < m - i; k++) {            cout << " ";        }        // counter to keep a track of the number of elements        // left for the current diagonal        int c = min(m - 1, min(i, n - 1));         // to access the diagonal elements        int inc = 0;         // while loop prints the diagonal elements of the        // current loop.        while (i <= m - 1 && c != -1) {            cout << *(a + (i - inc) * n + inc) << " ";             // increasing this variable to reach the            // remaining elements of the diagonal            inc++;             // decreasing the counter to as an element is            // printed            c--;        }         cout << endl;    }     // for printing second half of the matrix    for (int j = 1; j < n; j++) {         // print spaces        if (m < n) {            for (int k = 0; k <= j - 1; k++) {                cout << " ";            }        }        else{            for (int k = 0; k <= j; k++) {                cout << " ";            }        }                  // counter to keep a track of the number of elements        // left for the current diagonal        int c2 = min(m - 1, n - j - 1);         // to access the diagonal elements        int inc2 = 0;         // while loop prints the diagonal elements of the        // current loop.        while (j <= n - 1 && c2 != -1) {            cout << *((a + (m - 1 - inc2) * n) + j + inc2)                 << " ";            inc2++;            c2--;        }         cout << endl;    }}// Driver codeint main(){    int m, n;    cin >> m >> n;    int a[m][n];    for (int i = 0; i < m; i++) {        for (int j = 0; j < n; j++) {            cin >> a[i][j];        }    }    rotate(m, n, (int*)a);    return 0;} // contributed by rythmrana2
      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 
Time Complexity: O(N2)
Auxiliary Space: O(1)
divyeshrabadiya07
princi singh
rutvik_56
rythmrana2
rotation
school-programming
Matrix
Matrix
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