Python3 Program to Modify a matrix by rotating ith row exactly i times in clockwise direction

Given a matrix mat[][] of dimensions M * N, the task is to print the matrix obtained after rotating every i^th 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 0^th row is rotated 0 times. Therefore, the 0^th row remains the same as {1, 2, 3}.
The 1^st row is rotated 1 times. Therefore, the 1^st row modifies to {6, 4, 5}.
The 2^nd row is rotated 2 times. Therefore, the 2^nd 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:

• Traverse the given matrix in row – wise manner and for every i^th row, perform the following steps:
  • Reverse the current row of the matrix.
  • Reverse the first i elements of the current row.
  • Reverse the last (N – i) elements of the current row, where N is the current size of the row.
• After completing the above steps, print the matrix mat[][].

Below is the implementation of the above approach:

Python3

# Python3 program for the above approach

# Function to rotate every i-th
# row of the matrix i times
def rotateMatrix(mat):

    i = 0
    mat1 = []

    # Traverse the matrix row-wise
    for it in mat:

        # Reverse the current row
        it.reverse()

        # Reverse the first i elements
        it1 = it[:i]
        it1.reverse()

        # Reverse the last (N - i) elements
        it2 = it[i:]
        it2.reverse()

        # Increment count
        i += 1
        mat1.append(it1 + it2)

    # Print final matrix
    for rows in mat1:
        for cols in rows:
            print(cols, end = " ")

        print()

# Driver Code
if __name__ == "__main__":

    mat = [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ]

    rotateMatrix(mat)

# This code is contributed by ukasp

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.

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