Rotate Matrix Elements

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.

1. Move elements of top row. 
2. Move elements of last column. 
3. Move elements of bottom row. 
4. 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 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)
    {
    // 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 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 mat

# Utility Function
def printMatrix(mat):
    for row in mat:
        print row


# Test case 1
matrix =[
            [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 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");
            }
    }

    /* 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 = C
function 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 = C
function 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 1
let 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).