Given an image represented by m x n matrix, rotate the image by 90 degrees in counterclockwise direction. Please note the dimensions of the result matrix are going to n x m for an m x n input matrix.
Input:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Output:
4 8 12 16
3 7 11 15
2 6 10 14
1 5 9 13Input:
1 2 3 4
5 6 7 8
9 10 11 12
Output:
4 8 12
3 7 11
2 6 10
1 5 9
[Best Approach] – O(m x n) Time
We mainly need to move first row elements to first column in revers order, second row elements to second column in reverse order.
Let us first try to find out a pattern to solve the problem for n = 4 (second example matrix above)
mat[0][0] goes to mat[3][0]
mat[0][1] goes to mat[2][0]
………………………………………
mat[1][0] goes to mat[3][1]
……………………………………..
mat[3][3] goes to mat[0][3]
Do you see a pattern? Mainly we need to move mat[i][j] to mat[n-j-1][i].
#include <bits/stdc++.h>
using namespace std;
vector<vector<int>> rotateMatrix(vector<vector<int>> &mat)
{
int m = mat.size();
int n = mat[0].size();
// Create a result matrix of size n x m
vector<vector<int>> res(n, vector<int>(m));
// Move mat[i][j] to res[j][m - i- 1]
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
{
res[j][m - i - 1] = mat[i][j];
}
}
return res;
}
int main()
{
vector<vector<int>> mat{
{1, 2, 3, 4},
{5, 6, 7, 8},
{9, 10, 11, 12},
};
vector<vector<int>> res = rotateMatrix(mat);
for (auto &row : res)
{
for (auto &x : row)
cout << x << " ";
cout << endl;
}
return 0;
}
import java.util.*;
class RotateMatrix {
static int[][] rotateMatrix(int[][] mat) {
int m = mat.length;
int n = mat[0].length;
// Create a result matrix of size n x m
int[][] res = new int[n][m];
// Move mat[i][j] to res[j][m - i - 1]
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
res[j][m - i - 1] = mat[i][j];
}
}
return res;
}
public static void main(String[] args) {
int[][] mat = {
{1, 2, 3, 4},
{5, 6, 7, 8},
{9, 10, 11, 12}
};
int[][] res = rotateMatrix(mat);
for (int[] row : res) {
for (int x : row) {
System.out.print(x + " ");
}
System.out.println();
}
}
}
def rotate_matrix(mat):
m = len(mat)
# Create a result matrix of size n x m
res = [[0] * m for _ in range(len(mat[0]))]
# Move mat[i][j] to res[j][m - i - 1]
for i in range(m):
for j in range(len(mat[0])):
res[j][m - i - 1] = mat[i][j]
return res
if __name__ == "__main__":
mat = [
[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12]
]
res = rotate_matrix(mat)
for row in res:
print(" ".join(map(str, row)))
Output
9 5 1 10 6 2 11 7 3 12 8 4
[Alternate Approach] – O(m x n) Time
When you think about rotating a matrix 90 degrees counterclockwise, each element moves to a new position. The top row becomes the leftmost column, the second row becomes the second-left column, and so forth. If we first transpose the matrix, and then reverse individual columns, we get the desired result.
Follow the given steps to solve the problem:
- Perform Transpose of the matrix
- Reverse Individual Columns
#include <iostream>
#include <vector>
using namespace std;
vector<vector<int>> rotateMatrix(vector<vector<int>> &mat)
{
int m = mat.size(), n = mat[0].size();
vector<vector<int>> res(n, vector<int>(m));
// Transpose the matrix
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
{
res[j][i] = mat[i][j];
}
}
// Reverse each column of result which is
// a n x m matrix
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
{
swap(res[i][j], res[n - i - 1][j]);
}
}
return res;
}
int main()
{
vector<vector<int>> mat = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}};
vector<vector<int>> res = rotateMatrix(mat);
for (auto &row : res)
{
for (int i = 0; i < row.size(); i++)
{
cout << row[i] << " ";
}
cout << endl;
}
return 0;
}
import java.util.*;
class RotateMatrix {
static int[][] rotateMatrix(int[][] mat) {
int m = mat.length, n = mat[0].length;
int[][] res = new int[n][m];
// Transpose the matrix
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
res[j][i] = mat[i][j];
}
}
// Reverse each column (not row) of the result which is a n x m matrix
for (int j = 0; j < m; j++) {
for (int i = 0; i < n / 2; i++) {
int temp = res[i][j];
res[i][j] = res[n - i - 1][j];
res[n - i - 1][j] = temp;
}
}
return res;
}
public static void main(String[] args) {
int[][] mat = {
{1, 2, 3, 4},
{5, 6, 7, 8},
{9, 10, 11, 12}
};
int[][] res = rotateMatrix(mat);
for (int[] row : res) {
for (int val : row) {
System.out.print(val + " ");
}
System.out.println();
}
}
}
def rotate_matrix(mat):
m, n = len(mat), len(mat[0])
res = [[0] * m for _ in range(n)]
# Transpose the matrix
for i in range(m):
for j in range(n):
res[j][i] = mat[i][j]
# Reverse each column (not row) of the result which is a n x m matrix
for j in range(m):
for i in range(n // 2):
res[i][j], res[n - i - 1][j] = res[n - i - 1][j], res[i][j]
return res
if __name__ == "__main__":
mat = [
[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12]
]
res = rotate_matrix(mat)
for row in res:
print(" ".join(map(str, row)))
Output
4 8 12 2 6 10 3 7 11 1 5 9
Note : We can alternatively do
- Reverse individual rows
- Transpose Matrix
