Print a given matrix in spiral form

Given a 2D array, print it in spiral form. See the following examples.

Examples: 

Input:  1    2   3   4
        5    6   7   8
        9   10  11  12
        13  14  15  16
Output: 1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10 
Explanation: The output is matrix in spiral format. 

Input:  1   2   3   4  5   6
        7   8   9  10  11  12
        13  14  15 16  17  18
Output: 1 2 3 4 5 6 12 18 17 16 15 14 13 7 8 9 10 11
Explanation :The output is matrix in spiral format.

Best Optimized Method 1: (Simulation Approach)

Intuition:

Draw the path that the spiral makes. We know that the path should turn clockwise whenever it would go out of bounds or into a cell that was previously visited.

Algorithm:

Let the array have R rows and C columns. seen[r] denotes that the cell on the r-th row and c-th column was previously visited. Our current position is (r, c), facing direction di, and we want to visit R x C total cells.

As we move through the matrix, our candidate’s next position is (cr, cc). If the candidate is in the bounds of the matrix and unseen, then it becomes our next position; otherwise, our next position is the one after performing a clockwise turn.

1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10 

Complexity Analysis:

• Time Complexity: O(N)O(N), where NN is the total number of elements in the input matrix. We add every element in the matrix to our final answer.
• Space Complexity: O(N)O(N), the information stored in seen and in ans.

Method 2: This is a simple method to solve the following problem. 

Approach: The problem can be solved by dividing the matrix into loops or squares or boundaries. It can be seen that the elements of the outer loop are printed first in a clockwise manner then the elements of the inner loop is printed. So printing the elements of a loop can be solved using four loops which prints all the elements. Every ‘for’ loop defines a single direction movement along with the matrix. The first for loop represents the movement from left to right, whereas the second crawl represents the movement from top to bottom, the third represents the movement from the right to left, and the fourth represents the movement from bottom to up.

• Algorithm: 
  1. Create and initialize variables k – starting row index, m – ending row index, l – starting column index, n – ending column index
  2. Run a loop until all the squares of loops are printed.
  3. In each outer loop traversal print the elements of a square in a clockwise manner.
  4. Print the top row, i.e. Print the elements of the kth row from column index l to n, and increase the count of k.
  5. Print the right column, i.e. Print the last column or n-1th column from row index k to m and decrease the count of n.
  6. Print the bottom row, i.e. if k < m, then print the elements of m-1th row from column n-1 to l and decrease the count of m
  7. Print the left column, i.e. if l < n, then print the elements of lth column from m-1th row to k and increase the count of l.

Below is the implementation of the above algorithm: 

1 2 3 4 5 6 12 18 17 16 15 14 13 7 8 9 10 11 

Complexity Analysis: 

• Time Complexity: O(m*n). 
  To traverse the matrix O(m*n) time is required.
• Space Complexity: O(1). 
  No extra space is required.

Method 3: (Recursive Approach)

Approach: The above problem can be solved by printing the boundary of the Matrix recursively. In each recursive call, we decrease the dimensions of the matrix. The idea of printing the boundary or loops is the same.

• Algorithm: 
  1. create a recursive function that takes a matrix and some variables (k – starting row index, m – ending row index, l – starting column index, n – ending column index) as parameters
  2. Check the base cases (stating index is less than or equal to ending index) and print the boundary elements in clockwise manner
  3. Print the top row, i.e. Print the elements of kth row from column index l to n, and increase the count of k.
  4. Print the right column, i.e. Print the last column or n-1th column from row index k to m and decrease the count of n.
  5. Print the bottom row, i.e. if k > m, then print the elements of m-1th row from column n-1 to l and decrease the count of m
  6. Print the left column, i.e. if l < n, then print the elements of lth column from m-1th row to k and increase the count of l.
  7. Call the function recursively with the values of starting and ending indices of rows and columns.

Below is the implementation of the above algorithm: 

1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10 

Complexity Analysis: 

• Time Complexity: O(m*n). 
  To traverse the matrix O(m*n) time is required.
• Space Complexity: O(1). 
  No extra space is required.

Method 4: (DFS Recursive Approach)

Approach: Another recursive approach is to consider DFS movement within the matrix (right->down->left->up->right->..->end).

We do this by modifying the matrix itself such that when DFS algorithm visits each matrix cell it’s changed to a value which cannot be contained within the matrix. The DFS algorithm is terminated when it visits a cell such that all of its surrounding cells are already visited. The direction of the DFS search is controlled by a variable. 

Algorithm: 

1. create a DFS function which takes matrix, cell indices and direction
2. check are cell indices pointing to a valid cell (that is, not visited and in bounds)? if not, skip this cell
3. print cell value
4. mark matrix cell pointed by indicates as visited by changing it to a value not supported in the matrix
5. check are surrounding cells valid? if not stop algorithm, else continue
6. if direction given is right then check, is the cell to the right valid? if so, DFS to the right cell given the steps above, else, change the direction to down and DFS downwards given the steps above.
7. else, if the direction given is down then check, is the cell to the down valid? if so, DFS to the cell below given the steps above,                else, change the direction to left and DFS leftwards given the steps above.
8. else, if the direction given is left then check, is the cell to the left valid? if so, DFS to the left cell given the steps above,                           else, change the direction to up and DFS upwards given the steps above.
9. else, if the direction given is up then check, is the cell to the up valid? if so, DFS to the upper cell given the steps above,                           else, change the direction to right and DFS rightwards given the steps above.

Below is an implementation of this algorithm: 

1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10 

Complexity Analysis:

Time Complexity: O(m*n). To traverse the matrix O(m*n) time is required.
Space Complexity: O(1). No extra space is required (without consideration of the stack used by the recursion). 

• Diagonal traversal of Matrix
• Print matrix in antispiral form
• Print a given matrix in zigzag form

Please write comments if you find the above code incorrect, or find other ways to solve the same problem. 

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.