Given a matrix arr[][] of size M*N, where M is the number of rows and N is the number of columns. The task is to flip the matrix by both diagonals. Flipping the matrix means rotating all elements of the matrix in a clockwise direction, along the diagonal.
Examples:
Input: arr[][] = { {1, 2, 3}, {4, 5, 6}, {7, 8, 9} }
Output: { {9, 8, 7}, {6, 5, 4}, {3, 2, 1} }
Explanation: Resultant matrix after flipping the matrix along the main diagonal: { {1, 4, 7}, {2, 5, 8}, {3, 6, 9} }
Resultant matrix after flipping the matrix along the second diagonal: { {9, 8, 7}, {6, 5, 4}, {3, 2, 1} }Input: arr[][] = { {1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}, {13, 14, 15, 16} }
Output: { {16, 15, 14, 13}, {12, 11, 10, 9}, {8, 7, 6, 5}, {4, 3, 2, 1} }
Approach: The task can easily be solved using observations. One can observe that the resultant matrix would contain reversed rows in reverse order. Follow the below steps to solve the problem:
- Iterate over the rows of the matrix, and swap elements of the first row & last row in reverse order, and similarly second row & second last row, and so on.
Below is the implementation of the above code:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;const int m = 4, n = 4;// Function to flip the matrix along // both the diagonalsvoid flip(int matrix[m][n]){ int i, j, temp; // Swapping elements for (i = 0; i < m / 2; i++) { for (j = 0; j < n; j++) { temp = matrix[i][j]; matrix[i][j] = matrix[m - 1 - i][n - 1 - j]; matrix[m - 1 - i][n - 1 - j] = temp; } }}// Function to print the matrixvoid show(int matrix[m][n]){ int i, j; for (i = 0; i < m; i++) { for (j = 0; j < n; j++) { cout << matrix[i][j] << " "; } cout << endl; }}// Driver Codeint main(){ int matrix[4][4] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; flip(matrix); show(matrix);} |
Java
// Java program for the above approachimport java.util.*;class GFG{static int m = 4, n = 4;// Function to flip the matrix along // both the diagonalsstatic void flip(int matrix[][]){ int i, j, temp; // Swapping elements for (i = 0; i < m / 2; i++) { for (j = 0; j < n; j++) { temp = matrix[i][j]; matrix[i][j] = matrix[m - 1 - i][n - 1 - j]; matrix[m - 1 - i][n - 1 - j] = temp; } }}// Function to print the matrixstatic void show(int matrix[][]){ int i, j; for (i = 0; i < m; i++) { for (j = 0; j < n; j++) { System.out.print(matrix[i][j]+ " "); } System.out.println(); }}// Driver Codepublic static void main(String[] args){ int matrix[][] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; flip(matrix); show(matrix);}}// This code is contributed by Rajput-Ji |
Python3
# Python3 program for the above approachm = 4n = 4# Function to flip the matrix along# both the diagonalsdef flip(matrix): i = None j = None temp = None # Swapping elements for i in range(m // 2): for j in range(n): temp = matrix[i][j] matrix[i][j] = matrix[m - 1 - i][n - 1 - j] matrix[m - 1 - i][n - 1 - j] = temp# Function to print the matrixdef show(matrix): i = None j = None for i in range(m): for j in range(n): print(matrix[i][j], end = " ") print("")# Driver Codematrix = [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 10, 11, 12 ], [ 13, 14, 15, 16 ] ]flip(matrix)show(matrix)# This code is contributed by gfgking |
C#
// C# program for the above approachusing System;class GFG{ const int m = 4, n = 4; // Function to flip the matrix along // both the diagonals static void flip(int[, ] matrix) { int i, j, temp; // Swapping elements for (i = 0; i < m / 2; i++) { for (j = 0; j < n; j++) { temp = matrix[i, j]; matrix[i, j] = matrix[m - 1 - i, n - 1 - j]; matrix[m - 1 - i, n - 1 - j] = temp; } } } // Function to print the matrix static void show(int[, ] matrix) { int i, j; for (i = 0; i < m; i++) { for (j = 0; j < n; j++) { Console.Write(matrix[i, j] + " "); } Console.WriteLine(); } } // Driver Code public static void Main() { int[, ] matrix = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; flip(matrix); show(matrix); }}// This code is contributed by ukasp. |
Javascript
<script> // JavaScript code for the above approach let m = 4, n = 4; // Function to flip the matrix along // both the diagonals function flip(matrix) { let i, j, temp; // Swapping elements for (i = 0; i < Math.floor(m / 2); i++) { for (j = 0; j < n; j++) { temp = matrix[i][j]; matrix[i][j] = matrix[m - 1 - i][n - 1 - j]; matrix[m - 1 - i][n - 1 - j] = temp; } } } // Function to print the matrix function show(matrix) { let i, j; for (i = 0; i < m; i++) { for (j = 0; j < n; j++) { document.write(matrix[i][j] + " "); } document.write('<br>') } } // Driver Code let matrix = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]; flip(matrix); show(matrix); // This code is contributed by Potta Lokesh </script> |
Output
16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
Time Complexity: O(M*N)
Auxiliary Space: O(1)
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