Given a square matrix mat[][] of dimension N and an integer K, the task is to rotate the matrix by 90 degrees K times without changing the position of the diagonal elements.
Examples:
Input: mat[][] = {{1, 2, 3, 4, 5}, {6, 7, 8, 9, 10}, {11, 12, 13, 14, 15}, {16, 17, 18, 19, 20}, {21, 22, 23, 24, 25}}, K = 1
Output:
1 16 11 6 5
22 7 12 9 2
23 18 13 8 3
24 17 14 19 4
21 20 15 10 25Input: mat[][] = {{10, 11}, {12, 13}}, K = 2
Output:
10 11
12 13
Approach: The given problem can be solved by using the idea discussed in this article and the fact that the matrix restores after performing clockwise rotation 4 times. Follow the below steps to solve the given problem:
- Update the value of K as K % 4.
- Iterate until K is a positive and perform the following steps:
- Traverse the matrix, for i over the range [0, N / 2) and j over the range[0, N – i – 1) and perform the following steps:
- If the value of i != j and (i + j) != (N – 1), then perform the following steps:
- Store the value of mat[i][j] in a temporary variable temp.
- Update the value of mat[i][j] as mat[N – 1 – j][i].
- Update the value of mat[N – 1 – j][i] as mat[N – 1 -i][N – 1 – j].
- Update the value of mat[N – 1 – i][N – 1 – j] as mat[j][N – 1 – i].
- Update the value of mat[j][N – 1 – i] as temp.
- After completing the above steps, print the updated matrix obtained.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to print the matrixvoid print(vector<vector<int> >& mat){ // Iterate over the rows for (int i = 0; i < mat.size(); i++) { // Iterate over the columns for (int j = 0; j < mat[0].size(); j++) // Print the value cout << setw(3) << mat[i][j]; cout << "\n"; }}// Function to perform the swapping of// matrix elements in clockwise mannervoid performSwap(vector<vector<int> >& mat, int i, int j){ int N = mat.size(); // Stores the last row int ei = N - 1 - i; // Stores the last column int ej = N - 1 - j; // Perform the swaps int temp = mat[i][j]; mat[i][j] = mat[ej][i]; mat[ej][i] = mat[ei][ej]; mat[ei][ej] = mat[j][ei]; mat[j][ei] = temp;}// Function to rotate non - diagonal// elements of the matrix K times in// clockwise directionvoid rotate(vector<vector<int> >& mat, int N, int K){ // Update K to K % 4 K = K % 4; // Iterate until K is positive while (K--) { // Iterate each up to N/2-th row for (int i = 0; i < N / 2; i++) { // Iterate each column // from i to N - i - 1 for (int j = i; j < N - i - 1; j++) { // Check if the element // at i, j is not a // diagonal element if (i != j && (i + j) != N - 1) { // Perform the swapping performSwap(mat, i, j); } } } } // Print the matrix print(mat);}// Driver Codeint main(){ int K = 5; vector<vector<int> > mat = { { 1, 2, 3, 4 }, { 6, 7, 8, 9 }, { 11, 12, 13, 14 }, { 16, 17, 18, 19 }, }; int N = mat.size(); rotate(mat, N, K); return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.lang.*;import java.util.*;public class GFG { // Function to print the matrix static void print(int mat[][]) { // Iterate over the rows for (int i = 0; i < mat.length; i++) { // Iterate over the columns for (int j = 0; j < mat[0].length; j++) // Print the value System.out.print(mat[i][j] + " "); System.out.println(); } } // Function to perform the swapping of // matrix elements in clockwise manner static void performSwap(int mat[][], int i, int j) { int N = mat.length; // Stores the last row int ei = N - 1 - i; // Stores the last column int ej = N - 1 - j; // Perform the swaps int temp = mat[i][j]; mat[i][j] = mat[ej][i]; mat[ej][i] = mat[ei][ej]; mat[ei][ej] = mat[j][ei]; mat[j][ei] = temp; } // Function to rotate non - diagonal // elements of the matrix K times in // clockwise direction static void rotate(int mat[][], int N, int K) { // Update K to K % 4 K = K % 4; // Iterate until K is positive while (K-- > 0) { // Iterate each up to N/2-th row for (int i = 0; i < N / 2; i++) { // Iterate each column // from i to N - i - 1 for (int j = i; j < N - i - 1; j++) { // Check if the element // at i, j is not a // diagonal element if (i != j && (i + j) != N - 1) { // Perform the swapping performSwap(mat, i, j); } } } } // Print the matrix print(mat); } // Driver Code public static void main(String[] args) { int K = 5; int mat[][] = { { 1, 2, 3, 4 }, { 6, 7, 8, 9 }, { 11, 12, 13, 14 }, { 16, 17, 18, 19 }, }; int N = mat.length; rotate(mat, N, K); }}// This code is contributed by Kingash. |
Python3
# Python3 program for the above approach# Function to print the matrixdef printMat(mat): # Iterate over the rows for i in range(len(mat)): # Iterate over the columns for j in range(len(mat[0])): # Print the value print(mat[i][j], end = " ") print()# Function to perform the swapping of# matrix elements in clockwise mannerdef performSwap(mat, i, j): N = len(mat) # Stores the last row ei = N - 1 - i # Stores the last column ej = N - 1 - j # Perform the swaps temp = mat[i][j] mat[i][j] = mat[ej][i] mat[ej][i] = mat[ei][ej] mat[ei][ej] = mat[j][ei] mat[j][ei] = temp# Function to rotate non - diagonal# elements of the matrix K times in# clockwise directiondef rotate(mat, N, K): # Update K to K % 4 K = K % 4 # Iterate until K is positive while (K > 0): # Iterate each up to N/2-th row for i in range(int(N / 2)): # Iterate each column # from i to N - i - 1 for j in range(i, N - i - 1): # Check if the element # at i, j is not a # diagonal element if (i != j and (i + j) != N - 1): # Perform the swapping performSwap(mat, i, j) K -= 1 # Print the matrix printMat(mat)# Driver CodeK = 5mat = [ [ 1, 2, 3, 4 ], [ 6, 7, 8, 9 ], [ 11, 12, 13, 14 ], [ 16, 17, 18, 19 ] ]N = len(mat)rotate(mat, N, K)# This code is contributed by Dharanendra L V. |
C#
// C# program for the above approachusing System;public class GFG { // Function to print the matrix static void print(int[, ] mat) { // Iterate over the rows for (int i = 0; i < mat.GetLength(0); i++) { // Iterate over the columns for (int j = 0; j < mat.GetLength(1); j++) // Print the value Console.Write(mat[i, j] + " "); Console.WriteLine(); } } // Function to perform the swapping of // matrix elements in clockwise manner static void performSwap(int[, ] mat, int i, int j) { int N = mat.GetLength(0); // Stores the last row int ei = N - 1 - i; // Stores the last column int ej = N - 1 - j; // Perform the swaps int temp = mat[i, j]; mat[i, j] = mat[ej, i]; mat[ej, i] = mat[ei, ej]; mat[ei, ej] = mat[j, ei]; mat[j, ei] = temp; } // Function to rotate non - diagonal // elements of the matrix K times in // clockwise direction static void rotate(int[, ] mat, int N, int K) { // Update K to K % 4 K = K % 4; // Iterate until K is positive while (K-- > 0) { // Iterate each up to N/2-th row for (int i = 0; i < N / 2; i++) { // Iterate each column // from i to N - i - 1 for (int j = i; j < N - i - 1; j++) { // Check if the element // at i, j is not a // diagonal element if (i != j && (i + j) != N - 1) { // Perform the swapping performSwap(mat, i, j); } } } } // Print the matrix print(mat); } // Driver Code public static void Main(string[] args) { int K = 5; int[, ] mat = { { 1, 2, 3, 4 }, { 6, 7, 8, 9 }, { 11, 12, 13, 14 }, { 16, 17, 18, 19 }, }; int N = mat.GetLength(0); rotate(mat, N, K); }}// This code is contributed by ukasp. |
Javascript
<script>// Javascript implementation of the above approach // Function to print the matrix function print(mat) { // Iterate over the rows for (let i = 0; i < mat.length; i++) { // Iterate over the columns for (let j = 0; j < mat[0].length; j++) // Print the value document.write(mat[i][j] + " "); document.write("<br/>"); } } // Function to perform the swapping of // matrix elements in clockwise manner function performSwap(mat, i, j) { let N = mat.length; // Stores the last row let ei = N - 1 - i; // Stores the last column let ej = N - 1 - j; // Perform the swaps let temp = mat[i][j]; mat[i][j] = mat[ej][i]; mat[ej][i] = mat[ei][ej]; mat[ei][ej] = mat[j][ei]; mat[j][ei] = temp; } // Function to rotate non - diagonal // elements of the matrix K times in // clockwise direction function rotate(mat, N, K) { // Update K to K % 4 K = K % 4; // Iterate until K is positive while (K-- > 0) { // Iterate each up to N/2-th row for (let i = 0; i < N / 2; i++) { // Iterate each column // from i to N - i - 1 for (let j = i; j < N - i - 1; j++) { // Check if the element // at i, j is not a // diagonal element if (i != j && (i + j) != N - 1) { // Perform the swapping performSwap(mat, i, j); } } } } // Print the matrix print(mat); } // Driver Code let K = 5; let mat = [ [ 1, 2, 3, 4 ], [ 6, 7, 8, 9 ], [ 11, 12, 13, 14 ], [ 16, 17, 18, 19 ], ]; let N = mat.length; rotate(mat, N, K); </script> |
Output:
1 11 6 4 17 7 8 2 18 12 13 3 16 14 9 19
Time Complexity: O(N2)
Auxiliary Space: O(1), since no extra space has been taken.
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