Given a matrix, clockwise rotate elements in it.
Examples:
Input 1 2 3 4 5 6 7 8 9 Output: 4 1 2 7 5 3 8 9 6 For 4*4 matrix Input: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Output: 5 1 2 3 9 10 6 4 13 11 7 8 14 15 16 12
The idea is to use loops similar to the program for printing a matrix in spiral form. One by one rotate all rings of elements, starting from the outermost. To rotate a ring, we need to do following.
- Move elements of top row.
- Move elements of last column.
- Move elements of bottom row.
- Move elements of first column.
Repeat above steps for inner ring while there is an inner ring.
Below is the implementation of above idea. Thanks to Gaurav Ahirwar for suggesting below solution.
C++
// C++ program to rotate a matrix #include <bits/stdc++.h> #define R 4 #define C 4 using namespace std; // A function to rotate a matrix mat[][] of size R x C. // Initially, m = R and n = C void rotatematrix(int m, int n, int mat[R][C]) { int row = 0, col = 0; int prev, curr; /* row - Starting row index m - ending row index col - starting column index n - ending column index i - iterator */ while (row < m && col < n) { if (row + 1 == m || col + 1 == n) break; // Store the first element of next row, this // element will replace first element of current // row prev = mat[row + 1][col]; /* Move elements of first row from the remaining rows */ for (int i = col; i < n; i++) { curr = mat[row][i]; mat[row][i] = prev; prev = curr; } row++; /* Move elements of last column from the remaining columns */ for (int i = row; i < m; i++) { curr = mat[i][n-1]; mat[i][n-1] = prev; prev = curr; } n--; /* Move elements of last row from the remaining rows */ if (row < m) { for (int i = n-1; i >= col; i--) { curr = mat[m-1][i]; mat[m-1][i] = prev; prev = curr; } } m--; /* Move elements of first column from the remaining rows */ if (col < n) { for (int i = m-1; i >= row; i--) { curr = mat[i][col]; mat[i][col] = prev; prev = curr; } } col++; } // Print rotated matrix for (int i=0; i<R; i++) { for (int j=0; j<C; j++) cout << mat[i][j] << " "; cout << endl; } } /* Driver program to test above functions */int main() { // Test Case 1 int a[R][C] = { {1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}, {13, 14, 15, 16} }; // Test Case 2 /* int a[R][C] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9} }; */ rotatematrix(R, C, a); return 0; } |
Java
// Java program to rotate a matriximport java.lang.*;import java.util.*;class GFG{ static int R = 4; static int C = 4; // A function to rotate a matrix // mat[][] of size R x C. // Initially, m = R and n = C static void rotatematrix(int m, int n, int mat[][]) { int row = 0, col = 0; int prev, curr; /* row - Starting row index m - ending row index col - starting column index n - ending column index i - iterator */ while (row < m && col < n) { if (row + 1 == m || col + 1 == n) break; // Store the first element of next // row, this element will replace // first element of current row prev = mat[row + 1][col]; // Move elements of first row // from the remaining rows for (int i = col; i < n; i++) { curr = mat[row][i]; mat[row][i] = prev; prev = curr; } row++; // Move elements of last column // from the remaining columns for (int i = row; i < m; i++) { curr = mat[i][n-1]; mat[i][n-1] = prev; prev = curr; } n--; // Move elements of last row // from the remaining rows if (row < m) { for (int i = n-1; i >= col; i--) { curr = mat[m-1][i]; mat[m-1][i] = prev; prev = curr; } } m--; // Move elements of first column // from the remaining rows if (col < n) { for (int i = m-1; i >= row; i--) { curr = mat[i][col]; mat[i][col] = prev; prev = curr; } } col++; } // Print rotated matrix for (int i = 0; i < R; i++) { for (int j = 0; j < C; j++) System.out.print( mat[i][j] + " "); System.out.print("\n"); } }/* Driver program to test above functions */ public static void main(String[] args) { // Test Case 1 int a[][] = { {1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}, {13, 14, 15, 16} }; // Test Case 2 /* int a[][] = new int {{1, 2, 3}, {4, 5, 6}, {7, 8, 9} };*/ rotatematrix(R, C, a); }}// This code is contributed by Sahil_Bansall |
Python
# Python program to rotate a matrix# Function to rotate a matrixdef rotateMatrix(mat): if not len(mat): return """ top : starting row index bottom : ending row index left : starting column index right : ending column index """ top = 0 bottom = len(mat)-1 left = 0 right = len(mat[0])-1 while left < right and top < bottom: # Store the first element of next row, # this element will replace first element of # current row prev = mat[top+1][left] # Move elements of top row one step right for i in range(left, right+1): curr = mat[top][i] mat[top][i] = prev prev = curr top += 1 # Move elements of rightmost column one step downwards for i in range(top, bottom+1): curr = mat[i][right] mat[i][right] = prev prev = curr right -= 1 # Move elements of bottom row one step left for i in range(right, left-1, -1): curr = mat[bottom][i] mat[bottom][i] = prev prev = curr bottom -= 1 # Move elements of leftmost column one step upwards for i in range(bottom, top-1, -1): curr = mat[i][left] mat[i][left] = prev prev = curr left += 1 return mat# Utility Functiondef printMatrix(mat): for row in mat: print row# Test case 1matrix =[ [1, 2, 3, 4 ], [5, 6, 7, 8 ], [9, 10, 11, 12 ], [13, 14, 15, 16 ] ]# Test case 2"""matrix =[ [1, 2, 3], [4, 5, 6], [7, 8, 9] ]"""matrix = rotateMatrix(matrix)# Print modified matrixprintMatrix(matrix) |
C#
// C# program to rotate a matrixusing System;class GFG { static int R = 4; static int C = 4; // A function to rotate a matrix // mat[][] of size R x C. // Initially, m = R and n = C static void rotatematrix(int m, int n, int [,]mat) { int row = 0, col = 0; int prev, curr; /* row - Starting row index m - ending row index col - starting column index n - ending column index i - iterator */ while (row < m && col < n) { if (row + 1 == m || col + 1 == n) break; // Store the first element of next // row, this element will replace // first element of current row prev = mat[row + 1, col]; // Move elements of first row // from the remaining rows for (int i = col; i < n; i++) { curr = mat[row,i]; mat[row, i] = prev; prev = curr; } row++; // Move elements of last column // from the remaining columns for (int i = row; i < m; i++) { curr = mat[i,n-1]; mat[i, n-1] = prev; prev = curr; } n--; // Move elements of last row // from the remaining rows if (row < m) { for (int i = n-1; i >= col; i--) { curr = mat[m-1,i]; mat[m-1,i] = prev; prev = curr; } } m--; // Move elements of first column // from the remaining rows if (col < n) { for (int i = m-1; i >= row; i--) { curr = mat[i,col]; mat[i,col] = prev; prev = curr; } } col++; } // Print rotated matrix for (int i = 0; i < R; i++) { for (int j = 0; j < C; j++) Console.Write( mat[i,j] + " "); Console.Write("\n"); } } /* Driver program to test above functions */ public static void Main() { // Test Case 1 int [,]a = { {1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}, {13, 14, 15, 16} }; // Test Case 2 /* int a[][] = new int {{1, 2, 3}, {4, 5, 6}, {7, 8, 9} };*/ rotatematrix(R, C, a); }}// This code is contributed by nitin mittal. |
PHP
<?php// PHP program to rotate a matrix$R = 4;$C = 4;// A function to rotate a matrix // mat[][] of size R x C. Initially,// m = R and n = Cfunction rotatematrix($m, $n, $mat){ global $R, $C; $row = 0; $col = 0; $prev = 0; $curr = 0; /* row - Starting row index m - ending row index col - starting column index n - ending column index i - iterator */ while ($row < $m && $col < $n) { if ($row + 1 == $m || $col + 1 == $n) break; // Store the first element // of next row, this element // will replace first element // of current row $prev = $mat[$row + 1][$col]; /* Move elements of first row from the remaining rows */ for ($i = $col; $i < $n; $i++) { $curr = $mat[$row][$i]; $mat[$row][$i] = $prev; $prev = $curr; } $row++; /* Move elements of last column from the remaining columns */ for ($i = $row; $i < $m; $i++) { $curr = $mat[$i][$n - 1]; $mat[$i][$n - 1] = $prev; $prev = $curr; } $n--; /* Move elements of last row from the remaining rows */ if ($row < $m) { for ($i = $n - 1; $i >= $col; $i--) { $curr = $mat[$m - 1][$i]; $mat[$m - 1][$i] = $prev; $prev = $curr; } } $m--; /* Move elements of first column from the remaining rows */ if ($col < $n) { for ($i = $m - 1; $i >= $row; $i--) { $curr = $mat[$i][$col]; $mat[$i][$col] = $prev; $prev = $curr; } } $col++; } // Print rotated matrix for ($i = 0; $i < $R; $i++) { for ($j = 0; $j < $C; $j++) echo $mat[$i][$j] . " "; echo "\n"; }}// Driver code// Test Case 1$a = array(array(1, 2, 3, 4), array(5, 6, 7, 8), array(9, 10, 11, 12), array(13, 14, 15, 16));// Test Case 2/* int $a = array(array(1, 2, 3), array(4, 5, 6), array(7, 8, 9));*/ rotatematrix($R, $C, $a); return 0; // This code is contributed// by ChitraNayal?> |
Javascript
<script>// Javascript program to rotate a matrix let R = 4;let C = 4;// A function to rotate a matrix // mat[][] of size R x C.// Initially, m = R and n = Cfunction rotatematrix(m, n, mat){ let row = 0, col = 0; let prev, curr; /* row - Starting row index m - ending row index col - starting column index n - ending column index i - iterator */ while (row < m && col < n) { if (row + 1 == m || col + 1 == n) break; // Store the first element of next // row, this element will replace // first element of current row prev = mat[row + 1][col]; // Move elements of first row // from the remaining rows for(let i = col; i < n; i++) { curr = mat[row][i]; mat[row][i] = prev; prev = curr; } row++; // Move elements of last column // from the remaining columns for(let i = row; i < m; i++) { curr = mat[i][n - 1]; mat[i][n - 1] = prev; prev = curr; } n--; // Move elements of last row // from the remaining rows if (row < m) { for(let i = n - 1; i >= col; i--) { curr = mat[m - 1][i]; mat[m - 1][i] = prev; prev = curr; } } m--; // Move elements of first column // from the remaining rows if (col < n) { for(let i = m - 1; i >= row; i--) { curr = mat[i][col]; mat[i][col] = prev; prev = curr; } } col++; } // Print rotated matrix for(let i = 0; i < R; i++) { for(let j = 0; j < C; j++) document.write( mat[i][j] + " "); document.write("<br>"); }}// Driver code// Test Case 1let a = [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 10, 11, 12 ], [ 13, 14, 15, 16 ] ]; rotatematrix(R, C, a);// This code is contributed by avanitrachhadiya2155</script> |
Output
5 1 2 3 9 10 6 4 13 11 7 8 14 15 16 12
Complexity Analysis:
- Time Complexity: O(m*n) where m is the number of rows & n is the number of columns.
- Auxiliary Space: O(1).
Example: (Rotate anticlockwise – By using vectors in c++)
C++
#include <iostream>#include <vector>using namespace std;// Function to rotate the matrix in a clockwise directionvoid rotateMatrix(vector<vector<int>> &matrix) { int n = matrix.size(); // Transpose the matrix for (int i = 0; i < n; i++) { for (int j = i; j < n; j++) { swap(matrix[i][j], matrix[j][i]); } } // Reverse the columns for (int i = 0; i < n; i++) { for (int j = 0, k = n - 1; j < k; j++, k--) { swap(matrix[j][i], matrix[k][i]); } }}// Function to print the matrixvoid printMatrix(vector<vector<int>> &matrix) { for (int i = 0; i < matrix.size(); i++) { for (int j = 0; j < matrix[i].size(); j++) { cout << matrix[i][j] << " "; } cout << endl; }}int main() { vector<vector<int>> matrix = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}; cout << "Original matrix:" << endl; printMatrix(matrix); rotateMatrix(matrix); cout << "Rotated matrix:" << endl; printMatrix(matrix); return 0;} |
Java
import java.util.ArrayList;import java.util.List;public class Main { // Function to rotate the matrix in a clockwise // direction public static void rotateMatrix(List<List<Integer> > matrix) { int n = matrix.size(); // Transpose the matrix for (int i = 0; i < n; i++) { for (int j = i; j < n; j++) { int temp = matrix.get(i).get(j); matrix.get(i).set(j, matrix.get(j).get(i)); matrix.get(j).set(i, temp); } } // Reverse the columns for (int i = 0; i < n; i++) { for (int j = 0, k = n - 1; j < k; j++, k--) { int temp = matrix.get(j).get(i); matrix.get(j).set(i, matrix.get(k).get(i)); matrix.get(k).set(i, temp); } } } // Function to print the matrix public static void printMatrix(List<List<Integer> > matrix) { for (int i = 0; i < matrix.size(); i++) { for (int j = 0; j < matrix.get(i).size(); j++) { System.out.print(matrix.get(i).get(j) + " "); } System.out.println(); } } public static void main(String[] args) { List<List<Integer> > matrix = new ArrayList<>(); matrix.add(new ArrayList<Integer>() { { add(1); add(2); add(3); } }); matrix.add(new ArrayList<Integer>() { { add(4); add(5); add(6); } }); matrix.add(new ArrayList<Integer>() { { add(7); add(8); add(9); } }); System.out.println("Original matrix:"); printMatrix(matrix); rotateMatrix(matrix); System.out.println("Rotated matrix:"); printMatrix(matrix); }} |
Python3
def rotate_matrix(matrix): n = len(matrix) # Transpose the matrix for i in range(n): for j in range(i, n): matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j] # Reverse the columns for i in range(n): for j, k in zip(range(n//2), range(n-1, n//2-1, -1)): matrix[j][i], matrix[k][i] = matrix[k][i], matrix[j][i]def print_matrix(matrix): for row in matrix: print(' '.join(str(elem) for elem in row))matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]print("Original matrix:")print_matrix(matrix)rotate_matrix(matrix)print("Rotated matrix:")print_matrix(matrix) |
C#
using System;using System.Collections.Generic;class Program {// Function to rotate the matrix in a clockwise// directionpublic static void RotateMatrix(List<List<int>> matrix) {int n = matrix.Count; // Transpose the matrix for (int i = 0; i < n; i++) { for (int j = i; j < n; j++) { int temp = matrix[i][j]; matrix[i][j] = matrix[j][i]; matrix[j][i] = temp; } } // Reverse the columns for (int i = 0; i < n; i++) { for (int j = 0, k = n - 1; j < k; j++, k--) { int temp = matrix[j][i]; matrix[j][i] = matrix[k][i]; matrix[k][i] = temp; } }}// Function to print the matrixpublic static void PrintMatrix(List<List<int>> matrix) { for (int i = 0; i < matrix.Count; i++) { for (int j = 0; j < matrix[i].Count; j++) { Console.Write(matrix[i][j] + " "); } Console.WriteLine(); }}static void Main(string[] args) { List<List<int>> matrix = new List<List<int>>(); matrix.Add(new List<int>() { 1, 2, 3 }); matrix.Add(new List<int>() { 4, 5, 6 }); matrix.Add(new List<int>() { 7, 8, 9 }); Console.WriteLine("Original matrix:"); PrintMatrix(matrix); RotateMatrix(matrix); Console.WriteLine("Rotated matrix:"); PrintMatrix(matrix);}} |
Javascript
function rotateMatrix(grid) { const n = grid.length; // Transpose the matrix for (let i = 0; i < n; i++) { for (let j = i; j < n; j++) { [grid[i][j], matrix[j][i]] = [grid[j][i], grid[i][j]]; } } // Reverse the columns for (let i = 0; i < n; i++) { for (let j = 0, k = n - 1; j < k; j++, k--) { [grid[j][i], matrix[k][i]] = [grid[k][i], matrix[j][i]]; } }}function printMatrix(matrix) { for (let i = 0; i < grid.length; i++) { let row = ""; for (let j = 0; j < grid[i].length; j++) { row += grid[i][j] + " "; } console.log(row); }}let matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]];console.log("Original matrix:");printMatrix(matrix);rotateMatrix(matrix);console.log("Rotated matrix:");printMatrix(matrix); |
Output
Original matrix: 1 2 3 4 5 6 7 8 9 Rotated matrix: 3 6 9 2 5 8 1 4 7
Complexity Analysis:
Time Complexity:
The time complexity of the given implementation is O(n^2), where n is the size of the matrix. This is because we need to traverse through all the elements of the matrix twice (once for transposing and once for reversing the columns). Therefore, the time complexity of this algorithm is quadratic.
Auxiliary Space:
The auxiliary space complexity of this implementation is O(1), which means that the amount of extra memory required for the algorithm is constant and does not depend on the input size. In this implementation, we are modifying the matrix in-place without using any additional data structure. Therefore, the space required for this algorithm is constant.
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