You are given a square matrix of non-negative integers of size 'N x N'. Your task is to rotate that array by 90 degrees in an anti-clockwise direction without using any extra space.
For example:
For given 2D array :
[ [ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ] ]
After 90 degree rotation in anti clockwise direction, it will become:
[ [ 3, 6, 9 ],
[ 2, 5, 8 ],
[ 1, 4, 7 ] ]
The first line of input contains an integer 'T' representing the number of the test case. Then the test case follows.
The first line of each test case contains an integer 'N' representing the size of the square matrix ARR.
Each of the next 'N' lines contains 'N' space-separated integers representing the elements of the matrix 'ARR'.
For each test case, return the rotated matrix.
You do not need to print anything; it has already been taken care of. Just implement the given function.
1 ≤ T ≤ 50
1 ≤ N ≤ 100
1 ≤ ARR[i][j] ≤ 10^9
Time Limit: 1 sec
2
3
1 2 3
4 5 6
7 8 9
4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
3 6 9
2 5 8
1 4 7
4 8 12 16
3 7 11 15
2 6 10 14
1 5 9 13
(i) The array has been rotated by 90 degrees in an anticlockwise direction as the first row is now the first column inverted and so on for second and third rows.
(ii) The array has been rotated by 90 degrees in an anticlockwise direction as the first row is now first column inverted and so on for second, third and fourth rows.
2
3
7 4 1
8 5 2
9 6 3
4
13 9 5 1
14 10 6 2
15 11 7 3
16 12 8 4
1 2 3
4 5 6
7 8 9
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Try to think of how you can rotate elements of 2D matrix one by one.
O(N^2), where ‘N’ is the side length of the given square matrix.
As we are traversing the matrix single time.
O(1).
As a constant space is needed because we are doing an in-place rotation.
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Interview problems
CPP solution
#include <bits/stdc++.h>
void inplaceRotate(vector<vector<int>> &inputArray)
{
// Write your code here.
for(int i=0;i<inputArray.size();i++)
{
reverse(inputArray[i].begin(),inputArray[i].end());
}
for(int i=0;i<inputArray.size();i++)
{
for(int j=i+1;j<inputArray[0].size();j++)
{
swap(inputArray[i][j],inputArray[j][i]);
}
}
}Interview problems
Inplace rotate matrix 90 degree in java
import java.util.*;
import java.io.*;
public class Solution {
public static void inplaceRotate(int[][] arr, int n) {
// Write your code here.
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr.length; j++) {
//for upper triangular matrix
if (i >= j) {
int temp = arr[i][j];
arr[i][j] = arr[j][i];
arr[j][i] = temp;
}
}
}
int k = 0;
for (int i = arr.length - 1; i >= 0; i--) {
for (int j = 0; j < arr.length; j++) {
//for half row only
if (k <= n/2-1) {
int temp = arr[i][j];
arr[i][j] = arr[k][j];
arr[k][j] = temp;
}
}
k++;
}
}
}
Interview problems
Inplace rotate matrix 90 degree optimize code in C++
#include <bits/stdc++.h>
using namespace std;
void inplaceRotate(vector<vector<int>> &inputArray)
{
int n = inputArray.size();
// Reverse each row
for(int i = 0; i < n; i++){
reverse(inputArray[i].begin(), inputArray[i].end());
}
// Transpose the matrix
for(int i = 0; i < n; i++){
for(int j = i + 1; j < n; j++){
swap(inputArray[i][j], inputArray[j][i]);
}
}
}
Interview problems
Easy python solution
def inplaceRotate(arr, n):
mat = [[0 for j in range(n)] for i in range(n)]
for i in range(n):
arr[i] = arr[i][::-1]
for i in range(n):
for j in range(n):
mat[j][i] = arr[i][j]
return mat
Interview problems
Simple solution
first I transposed the matrix then reverse the rows sequence and you have to check properly how I swaped the rows
Interview problems
Inplace Roatate array by 90 degree anti-clockwise
#include <bits/stdc++.h>
void inplaceRotate(vector<vector<int>> &inputArray)
{
// Write your code here.
int m = inputArray.size() ;
vector<vector<int>> ans(m,vector<int>(m));
for(int i=0;i<m;i++)
{
for(int j=0;j<m;j++)
{
ans[m-j-1][i] = inputArray[i][j];
}
}
for(int i=0;i<m;i++)
{
for(int j=0;j<m;j++)
{
inputArray[i][j] = ans[i][j];
}
}
}
Interview problems
Java Solution
import java.util.* ;
import java.io.*;
public class Solution {
public static void transpose(int arr[][]){
int n = arr.length;
for(int i = 0; i < n ; i++){
for(int j = 0; j < n ; j++){
if(i != j && i < j){
int temp = arr[i][j];
arr[i][j] = arr[j][i];
arr[j][i] = temp;
}
}
}
}
public static void inplaceRotate(int[][] arr, int n) {
// Write your code here.
transpose(arr);
for(int i = 0; i < n ; i++){
int j = 0;
int k = n-1;
while(j < k){
int temp = arr[j][i];
arr[j][i] = arr[k][i];
arr[k][i] =temp;
j++;
k--;
}
}
}
}Interview problems
EASY JAVA SOLUTION
import java.util.* ;
import java.io.*;
public class Solution {
public static void inplaceRotate(int[][] arr, int n) {
for (int i = 0; i < n / 2; i++) {
for (int j = i; j < n - i - 1; j++) {
int temp = arr[i][j];
arr[i][j] = arr[j][n - 1 - i];
arr[j][n - 1 - i] = arr[n - 1 - i][n - 1 - j];
arr[n - 1 - i][n - 1 - j] = arr[n - 1 - j][i];
arr[n - 1 - j][i] = temp;
}
}
}
}
Interview problems
@C++ easy solution beginner friendly :)
void inplaceRotate(vector<vector<int>> &arr)
{
// Write your code here.
int n=arr.size();
int m=arr[0].size();
//Take transpose
for(int i=0;i<n;i++)
{
for(int j=i+1;j<m;j++){
swap(arr[i][j],arr[j][i]);
}
}
//Reverse each column :)
for(int i=0;i<n;i++){
int l=0,h=m-1;
while(l<h){
swap(arr[l][i],arr[h][i]);
l++;
h--;
}
}
}
Interview problems
Easy Solution in C++; please upvote
#include <bits/stdc++.h>
void inplaceRotate(vector<vector<int>> &inputArray)
{
// Write your code here.
//make transpose of the matrix
int n = inputArray.size();
for(int i=0; i<n; i++) {
for(int j=i+1; j<n; j++) {
swap(inputArray[i][j], inputArray[j][i]);
}
}
for(int i=0; i<n/2; i++) {
for(int j=0; j<n; j++) {
swap(inputArray[i][j], inputArray[n-i-1][j]);
}
}
}
