C++ Program to Rotate all Matrix elements except the diagonal K times by 90 degrees in clockwise direction
Given a square matrix mat[][] of dimension N and an integer K, the task is to rotate the matrix by 90 degrees K times without changing the position of the diagonal elements.
Examples:
Input: mat[][] = {{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}}, K = 1
Output:
1 16 11 6 5
22 7 12 9 2
23 18 13 8 3
24 17 14 19 4
21 20 15 10 25Input: mat[][] = {{10, 11}, {12, 13}}, K = 2
Output:
10 11
12 13
Approach: The given problem can be solved by using the idea discussed in this article and the fact that the matrix restores after performing clockwise rotation 4 times. Follow the below steps to solve the given problem:
- Update the value of K as K % 4.
- Iterate until K is a positive and perform the following steps:
- Traverse the matrix, for i over the range [0, N / 2) and j over the range[0, N - i - 1) and perform the following steps:
- If the value of i != j and (i + j) != (N - 1), then perform the following steps:
- Store the value of mat[i][j] in a temporary variable temp.
- Update the value of mat[i][j] as mat[N - 1 - j][i].
- Update the value of mat[N - 1 - j][i] as mat[N - 1 -i][N - 1 - j].
- Update the value of mat[N - 1 - i][N - 1 - j] as mat[j][N - 1 - i].
- Update the value of mat[j][N - 1 - i] as temp.
- After completing the above steps, print the updated matrix obtained.
Below is the implementation of the above approach:
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the matrix
void print(vector<vector<int> >& mat)
{
// Iterate over the rows
for (int i = 0; i < mat.size(); i++) {
// Iterate over the columns
for (int j = 0; j < mat[0].size(); j++)
// Print the value
cout << setw(3) << mat[i][j];
cout << "
";
}
}
// Function to perform the swapping of
// matrix elements in clockwise manner
void performSwap(vector<vector<int> >& mat,
int i, int j)
{
int N = mat.size();
// Stores the last row
int ei = N - 1 - i;
// Stores the last column
int ej = N - 1 - j;
// Perform the swaps
int temp = mat[i][j];
mat[i][j] = mat[ej][i];
mat[ej][i] = mat[ei][ej];
mat[ei][ej] = mat[j][ei];
mat[j][ei] = temp;
}
// Function to rotate non - diagonal
// elements of the matrix K times in
// clockwise direction
void rotate(vector<vector<int> >& mat,
int N, int K)
{
// Update K to K % 4
K = K % 4;
// Iterate until K is positive
while (K--) {
// Iterate each up to N/2-th row
for (int i = 0; i < N / 2; i++) {
// Iterate each column
// from i to N - i - 1
for (int j = i;
j < N - i - 1; j++) {
// Check if the element
// at i, j is not a
// diagonal element
if (i != j
&& (i + j) != N - 1) {
// Perform the swapping
performSwap(mat, i, j);
}
}
}
}
// Print the matrix
print(mat);
}
// Driver Code
int main()
{
int K = 5;
vector<vector<int> > mat = {
{ 1, 2, 3, 4 },
{ 6, 7, 8, 9 },
{ 11, 12, 13, 14 },
{ 16, 17, 18, 19 },
};
int N = mat.size();
rotate(mat, N, K);
return 0;
}
Output:
1 11 6 4 17 7 8 2 18 12 13 3 16 14 9 19
Time Complexity: O(N2)
Auxiliary Space: O(1)
