Rotate Matrix Clockwise by 1 (original) (raw)
Last Updated : 17 Oct, 2024
Given a square matrix, the task is to rotate its elements clockwise by one step.
**Examples:
**Input
1 2 3
4 5 6
7 8 9
**Output:
4 1 2
7 5 3
8 9 6**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 nested loops to move elements in four directions (right, down, left, and up) one step at a time for each layer starting from the outermost layer. This simulates the clockwise rotation by rotating each "ring" or layer of the matrix.
1. First move the elements of the outermost layer.
a) Move the top row elements one position ahead (except the last element)
b) Move the rightmost column elements one position down (except the last element)
c) Move the bottom row elements one position left (Except the leftmost element)
d) Move the first column elements one position up (Except the top elements)
2. Repeat the same process for inner layers.
C++ `
#include <bits/stdc++.h> using namespace std;
// Function to rotate a matrix represented by a vector of vectors void rotateMatrix(vector<vector>& mat) {
int m = mat.size();
int n = mat[0].size();
int row = 0, col = 0;
int prev, curr;
// Rotate the matrix in layers
while (row < m && col < n) {
if (row + 1 == m || col + 1 == n)
break;
// Store the first element of the next row
prev = mat[row + 1][col];
// Move elements of the first row
for (int i = col; i < n; i++) {
curr = mat[row][i];
mat[row][i] = prev;
prev = curr;
}
row++;
// Move elements of the last column
for (int i = row; i < m; i++) {
curr = mat[i][n - 1];
mat[i][n - 1] = prev;
prev = curr;
}
n--;
// Move elements of the last row
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 the first column
if (col < n) {
for (int i = m - 1; i >= row; i--) {
curr = mat[i][col];
mat[i][col] = prev;
prev = curr;
}
}
col++;
}}
int main() { vector<vector> mat = { {1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}, {13, 14, 15, 16} };
rotateMatrix(mat);
// Print the rotated matrix
for (auto& r : mat) {
for (int val : r)
cout << val << " ";
cout << endl;
}
return 0;}
Java
// Java program to rotate a matrix import 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)
{
int a[][] = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 } };
rotatematrix(R, C, a);
}}
Python
Python program to rotate a matrix
Function to rotate a matrix
def 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 matUtility Function
def printMatrix(mat): for row in mat: print row
matrix = [ [1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16] ]
matrix = rotateMatrix(matrix)
Print modified matrix
printMatrix(matrix)
C#
// C# program to rotate a matrix
using 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");
}
}
static void Main() {
int[, ] a = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 } };
rotatematrix(R, C, a);
}}
JavaScript
// Function to rotate the matrix in a clockwise direction function rotateMatrix(matrix) { let m = matrix.length; let n = matrix[0].length; let row = 0, col = 0; let prev, curr;
/*
row - Starting row index
m - ending row index
col - starting column index
n - ending column index
*/
while (row < m && col < n) {
if (row + 1 === m || col + 1 === n)
break;
// Store the first element of the next row, this
// element will replace the first element of the current row
prev = matrix[row + 1][col];
// Move elements of the first row from the remaining rows
for (let i = col; i < n; i++) {
curr = matrix[row][i];
matrix[row][i] = prev;
prev = curr;
}
row++;
// Move elements of the last column from the remaining columns
for (let i = row; i < m; i++) {
curr = matrix[i][n - 1];
matrix[i][n - 1] = prev;
prev = curr;
}
n--;
// Move elements of the last row from the remaining rows
if (row < m) {
for (let i = n - 1; i >= col; i--) {
curr = matrix[m - 1][i];
matrix[m - 1][i] = prev;
prev = curr;
}
}
m--;
// Move elements of the first column from the remaining rows
if (col < n) {
for (let i = m - 1; i >= row; i--) {
curr = matrix[i][col];
matrix[i][col] = prev;
prev = curr;
}
}
col++;
}}
// Function to print the matrix function printMatrix(matrix) { for (let i = 0; i < matrix.length; i++) { console.log(matrix[i].join(' ')); } }
// Main function to test the rotation function main() { let matrix = [ [1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16] ];
rotateMatrix(matrix);
printMatrix(matrix);}
// Run the main function main();
`
Output
5 1 2 3 9 10 6 4 13 11 7 8 14 15 16 12
**Time Complexity: O(m*n) where m is the number of rows & n is the number of columns.
**Auxiliary Space: O(1).