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:
- Traverse the given matrix in row – wise manner and for every ith 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:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to rotate every i-th// row of the matrix i timesvoid rotateMatrix(vector<vector<int> >& mat){ int i = 0; // Traverse the matrix row-wise for (auto& it : mat) { // Reverse the current row reverse(it.begin(), it.end()); // Reverse the first i elements reverse(it.begin(), it.begin() + i); // Reverse the last (N - i) elements reverse(it.begin() + i, it.end()); // Increment count i++; } // Print final matrix for (auto rows : mat) { for (auto cols : rows) { cout << cols << " "; } cout << "\n"; }}// Driver Codeint main(){ vector<vector<int> > mat = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } }; rotateMatrix(mat); return 0;} |
Java
// java program for the above approachimport java.io.*;import java.lang.*;import java.util.*;class GFG { // Function to reverse arr[] from start to end 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--; } } // Function to rotate every i-th // row of the matrix i times static void rotateMatrix(int mat[][]) { int i = 0; // Traverse the matrix row-wise for (int rows[] : mat) { // Reverse the current row reverse(rows, 0, rows.length - 1); // Reverse the first i elements reverse(rows, 0, i - 1); // Reverse the last (N - i) elements reverse(rows, i, rows.length - 1); // Increment count i++; } // Print final matrix for (int rows[] : mat) { for (int cols : rows) { System.out.print(cols + " "); } System.out.println(); } } // Driver Code public static void main(String[] args) { int mat[][] = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } }; rotateMatrix(mat); }}// This code is contributed by Kingash. |
Python3
# Python3 program for the above approach# Function to rotate every i-th# row of the matrix i timesdef 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 Codeif __name__ == "__main__": mat = [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] rotateMatrix(mat)# This code is contributed by ukasp |
C#
// C# program to Modify a matrix // by rotating ith row exactly // i times in clockwise directionusing System;class GFG{ // Reverse each row of matrix static void reverse(int N, int[, ] mat, int start, int end) { // Till start < end, swap the element // at start and end index while (start < end) { // Swap the element int temp = mat[N,start]; mat[N, start] = mat[N, end]; mat[N, end] = temp; // Increment start and decrement // end for next pair of swapping start++; end--; } } // An Inplace function to // rotate a N x N matrix // by 90 degrees in anti- // clockwise direction static void rotateMatrix(int N, int[, ] mat) { int i = 0; // Performing Transpose for(int j = 0; j < N; j++) { // Reverse the current row reverse(j, mat, 0, N - 1); // Reverse the first i elements reverse(j, mat, 0, i - 1); // Reverse the last (N - i) elements reverse(j, mat, i, N - 1); // Increment count i++; } for (i = 0; i < N; i++) { for (int j = 0; j < N; j++) Console.Write(mat[i, j] + " "); Console.Write("\n"); } } // Driver Code static public void Main() { int N = 3; // Test Case 1 int[, ] mat = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 }}; rotateMatrix(N, mat); }}// This code is contributed by Aarti_Rathi |
Javascript
<script>// javascript program for the above approach// Function to reverse arr[] from start to endfunction reverse(arr,start,end){ while (start < end) { let temp = arr[start]; arr[start] = arr[end]; arr[end] = temp; start++; end--; }} // Function to rotate every i-th // row of the matrix i timesfunction rotateMatrix(mat){ let i = 0; // Traverse the matrix row-wise for (let rows=0;rows<mat.length;rows++) { // Reverse the current row reverse(mat[rows], 0, mat[rows].length - 1); // Reverse the first i elements reverse(mat[rows], 0, i - 1); // Reverse the last (N - i) elements reverse(mat[rows], i, mat[rows].length - 1); // Increment count i++; } // Print final matrix for (let rows=0;rows< mat.length;rows++) { for (let cols=0;cols< mat[rows].length;cols++) { document.write(mat[rows][cols] + " "); } document.write("<br>"); }}// Driver Codelet mat=[[ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ]];rotateMatrix(mat);// This code is contributed by avanitrachhadiya2155</script> |
Output:
1 2 3 6 4 5 8 9 7
Time Complexity: O(M * N)
Auxiliary Space: O(1)
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