Inplace rotate matrix 90 degree

Easy

0/40

Average time to solve is 12m

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54 upvotes

Asked in companies

Problem statement

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 ]   ]

Detailed explanation ( Input/output format, Notes, Images )

Input Format :

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'.

Output Format:

For each test case, return the rotated matrix.

Note:

You do not need to print anything; it has already been taken care of. Just implement the given function.

Constraints:

1 ≤ T ≤ 50
1 ≤ N ≤ 100
1 ≤ ARR[i][j] ≤ 10^9

Time Limit: 1 sec

Sample Input 1:

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

Sample Output 1:

Explanation of Input 1:

(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.

Sample Input 2:

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

Sample Output 2:

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

Hint 1

Hint 2

Try to think of how you can rotate elements of 2D matrix one by one.

Approaches (2)

Approach 1

Approach 2

Cycle rotation

There are N/2 cycles in a matrix of size ‘N’.
We traverse in the matrix from the outermost cycle, i.e. (0,0) to innermost cycle i.e. ((N/2)-1, (N/2)-1).
For each cycle, we’ll swap the elements of the matrix in a group of four elements i.e. for each ‘i’ <= ‘j’ < ‘N-i-1’ for each 0 <= ‘i’ <= ‘(N/2)-1’ we swap:
- ‘ARR[i] [j]’ with ‘ARR[j, N-1-i]’
- ‘ARR[j, N-1-i]’ with ‘ARR[N-1-i, N-1-j]’
- ‘ARR[N-1-i, N-1-j]’ with ‘ARR[N-1-j,i]’
- ‘ARR[N-1-j,i]’ with ‘ARR[i][j]’
At the end of these loops, we’ll get rotated matrix.
Print the matrix.

Time Complexity

O(N^2), where ‘N’ is the side length of the given square matrix.

As we are traversing the matrix single time.

Space Complexity

O(1).

As a constant space is needed because we are doing an in-place rotation.

(100% EXP penalty)

Inplace rotate matrix 90 degree

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