# Spiral matrix : Print matrix in spiral order

Given a N x M matrix, print all elements in it in spiral order.This problem is also called as spiral matrix problem. For example,

1,2,3,4 5,6,7,8 9,10,11,12 Output should be 1,2,3,4,8,12,11,10,9,5,6,7

Below diagram explains spiral traversal.

Spiral order traversal will be 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

This problem is commonly asked in interviews because it can be easily solved in 30 to 40 minutes and tests your ability to work with array indices very well. Whole problem is about adjusting array indices.

** Spiral matrix algorithm**

Can you figure out what we did in example above? We traversed first row, then last column, last row and then first column. Once done we move inside and our size of matrix now becomes N-2 and M-2 as we have already traversed two rows and column in first round. Again traverse, second row from second column, till newly adjusted column length. Traverse last column starting between newly adjusted rows, as first and last are already traversed. Go on till we have traversed all elements.

Implementation wise, have four variables currentRow, currentColumn, row and column, Adjust these four variables to achieve what is described above.

Figure shows how the traversal happens.

### Spiral matrix implementation

#include<stdio.h> #include<stdlib.h> void spiralMatrix(int a[][5], int row, int col){ int currentRow=0, currentColumn=0; int i; while(currentRow < row && currentColumn <col){ /* Print the first row and next row starts with increased column so increment column count */ for(i=currentColumn; i<col; i++) printf("%d ", a[currentRow][i]); currentRow++; /* Print the last column, next time we need to print second to last column, hence decrement col */ for(i=currentRow; i<row; i++) printf("%d ", a[i][col-1]); col--; /* If we have not reached at the end print column in reverse order*/ if(currentRow<row){ for(i=col-1; i>=currentColumn; i--) printf("%d ", a[row-1][i]); row--; } /* If we have not reached at the end, print row in reverse order */ if(currentColumn<col){ for(i=row-1; i>=currentRow; i--) printf("%d ", a[i][currentColumn]); currentColumn++; } } printf("\n"); } /* Driver program */ int main(){ int a[5][5] = {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}; spiralMatrix(a,5,5); return 0; }

Thanks Prince Mishra for python code.

def spiral_traversal(matrix): size = len(matrix) total = len(matrix)*len(matrix) i = 0 start_row = 0 end_row = size-1 start_col = 0 end_col = size-1 dir = 0 while i<total: if dir == 0: # ltr j = start_col while j <= end_col: print matrix[start_row][j] j+=1 i += 1 start_row +=1 dir = 1 if dir == 1: j = start_row while j <= end_row: print matrix[j][end_col] j+=1 i+=1 end_col -=1 dir = 2 if dir == 2: j = end_col while j>=start_col: print matrix[end_row][j] j-=1 i+=1 end_row -=1 dir = 3 if dir == 3: j = end_row while j>=start_row: print matrix[j][start_col] j-=1 i+=1 start_col+=1 dir = 0 arr = [ [1,2,3,4], [5,6,7,8], [9,10,11,12], [13,14,15,16] ] spiral_traversal(arr)

Complexity of **spiral matrix algorithm ** is O(n^{2}).

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