Given a square matrix, the task is that we turn it by 180 degrees in an anti-clockwise direction without using any extra space.
Examples :
Input : 1 2 3
4 5 6
7 8 9
Output : 9 8 7
6 5 4
3 2 1
Input : 1 2 3 4
5 6 7 8
9 0 1 2
3 4 5 6
Output : 6 5 4 3
2 1 0 9
8 7 6 5
4 3 2 1
Method: 1 (Only prints rotated matrix)
The solution of this problem is that to rotate a matrix by 180 degrees we can easily follow that step
Matrix = a00 a01 a02
a10 a11 a12
a20 a21 a22
when we rotate it by 90 degree
then matrix is
Matrix = a02 a12 a22
a01 a11 a21
a00 a10 a20
when we rotate it by again 90
degree then matrix is
Matrix = a22 a21 a20
a12 a11 a10
a02 a01 a00
From the above illustration, we get that simply to rotate the matrix by 180 degrees then we will have to print the given matrix in a reverse manner.
C++
// C++ program to rotate a matrix by 180 degrees #include <bits/stdc++.h> #define N 3 using namespace std; // Function to Rotate the matrix by 180 degree void rotateMatrix(int mat[][N]) { // Simply print from last cell to first cell. for (int i = N - 1; i >= 0; i--) { for (int j = N - 1; j >= 0; j--) printf("%d ", mat[i][j]); printf(" "); } } // Driven code int main() { int mat[N][N] = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } }; rotateMatrix(mat); return 0; } |
Output :
9 8 7 6 5 4 3 2 1
Time complexity: O(N*N)
Auxiliary Space: O(1)
Method : 2(In-place rotation)
There are four steps :
1- Find transpose of a matrix.
2- Reverse columns of the transpose.
3- Find transpose of a matrix.
4- Reverse columns of the transpose
Let the given matrix be 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 First we find transpose. 1 5 9 13 2 6 10 14 3 7 11 15 4 8 12 16 Then we reverse elements of every column. 4 8 12 16 3 7 11 15 2 6 10 14 1 5 9 13 then transpose again 4 3 2 1 8 7 6 5 12 11 10 9 16 15 14 13 Then we reverse elements of every column again 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
C++
// C++ program for left rotation of matrix by 180 #include <bits/stdc++.h> using namespace std; #define R 4 #define C 4 // Function to rotate the matrix by 180 degree void reverseColumns(int arr[R][C]) { for (int i = 0; i < C; i++) for (int j = 0, k = C - 1; j < k; j++, k--) swap(arr[j][i], arr[k][i]); } // Function for transpose of matrix void transpose(int arr[R][C]) { for (int i = 0; i < R; i++) for (int j = i; j < C; j++) swap(arr[i][j], arr[j][i]); } // Function for display the matrix void printMatrix(int arr[R][C]) { for (int i = 0; i < R; i++) { for (int j = 0; j < C; j++) cout << arr[i][j] << " "; cout << ' '; } } // Function to anticlockwise rotate matrix // by 180 degree void rotate180(int arr[R][C]) { transpose(arr); reverseColumns(arr); transpose(arr); reverseColumns(arr); } // Driven code int main() { int arr[R][C] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; rotate180(arr); printMatrix(arr); return 0; } |
Output :
16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
Time complexity : O(R*C)
Auxiliary Space : O(1)
In the code above, the transpose of the matrix has to be found twice, and also, columns have to be reversed twice.
So, we can have a better solution.
Method : 3 (Position swapping)
Here, we swap the values in the respective positions.
C++
#include <bits/stdc++.h> using namespace std; /** * Reverse Row at specified index in the matrix * @param data matrix * @param index row index */void reverseRow(vector<vector<int>>& data, int index) { int cols = data[index].size(); for(int i = 0; i < cols / 2; i++) { int temp = data[index][i]; data[index][i] = data[index][cols - i - 1]; data[index][cols - i - 1] = temp; } } /** * Print Matrix data * @param data matrix */void printMatrix(vector<vector<int>>& data) { for(int i = 0; i < data.size(); i++) { for(int j = 0; j < data[i].size(); j++) { cout << data[i][j] << " "; } cout << endl; } } /** * Rotate Matrix by 180 degrees * @param data matrix */void rotateMatrix180(vector<vector<int>>& data) { int rows = data.size(); int cols = data[0].size(); if (rows % 2 != 0) { // If N is odd reverse the middle // row in the matrix reverseRow(data, data.size() / 2); } // Swap the value of matrix [i][j] with // [rows - i - 1][cols - j - 1] for half // the rows size. for(int i = 0; i <= (rows/2) - 1; i++) { for(int j = 0; j < cols; j++) { int temp = data[i][j]; data[i][j] = data[rows - i - 1][cols - j - 1]; data[rows - i - 1][cols - j - 1] = temp; } } } // Driver code int main() { vector<vector<int>> data{ { 1, 2, 3, 4, 5 }, { 6, 7, 8, 9, 10 }, { 11, 12, 13, 14, 15 }, { 16, 17, 18, 19, 20 }, { 21, 22, 23, 24, 25 } }; // Rotate Matrix rotateMatrix180(data); // Print Matrix printMatrix(data); return 0; } // This code is contributed by divyeshrabadiya07 |
Output :
25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
Time complexity : O(R*C)
Auxiliary Space : O(1)
Please refer complete article on Rotate a Matrix by 180 degree for more details!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
