Java Program to Modify a matrix by rotating ith row exactly i times in clockwise direction
Last Updated :
17 Aug, 2023
Given a matrix mat[][] of dimensions M * N, the task is to print the matrix obtained after rotating every ith row of the matrix i times in a clockwise direction.
Examples:
Input: mat[][] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}
Output:
1 2 3
6 4 5
8 9 7
Explanation:
The 0th row is rotated 0 times. Therefore, the 0th row remains the same as {1, 2, 3}.
The 1st row is rotated 1 times. Therefore, the 1st row modifies to {6, 4, 5}.
The 2nd row is rotated 2 times. Therefore, the 2nd row modifies to {8, 9, 7}.
After completing the above operations, the given matrix modifies to {{1, 2, 3}, {6, 4, 5}, {8, 9, 7}}.
Input: mat[][] = {{1, 2, 3, 4}, {4, 5, 6, 7}, {7, 8, 9, 8}, {7, 8, 9, 8}}
Output:
1 2 3 4
7 4 5 6
9 8 7 8
8 9 8 7
Approach: Follow the steps below to solve the problem:
Below is the implementation of the above approach:
Java
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {
static void reverse( int arr[], int start, int end)
{
while (start < end) {
int temp = arr[start];
arr[start] = arr[end];
arr[end] = temp;
start++;
end--;
}
}
static void rotateMatrix( int mat[][])
{
int i = 0 ;
for ( int rows[] : mat) {
reverse(rows, 0 , rows.length - 1 );
reverse(rows, 0 , i - 1 );
reverse(rows, i, rows.length - 1 );
i++;
}
for ( int rows[] : mat) {
for ( int cols : rows) {
System.out.print(cols + " " );
}
System.out.println();
}
}
public static void main(String[] args)
{
int mat[][] = { { 1 , 2 , 3 },
{ 4 , 5 , 6 },
{ 7 , 8 , 9 } };
rotateMatrix(mat);
}
}
|
Output:
1 2 3
6 4 5
8 9 7
Time Complexity: O(N*M), as we are using nested loops to traverse N*M times.
Auxiliary Space: O(1), as we are not using any extra space.
Please refer complete article on Modify a matrix by rotating ith row exactly i times in clockwise direction for more details!
Share your thoughts in the comments
Please Login to comment...