Spiral Matrix II in Python

Suppose we have a positive integer n, we have to generate a square matrix with n2 elements in spiral order. So if n = 5, then the matrix will be −

Let us see the steps −

• set (row1, col1) := (0, 0) and (row2, col2) := (n, n), and create one matrix called res, then fill it with 0s, and set num := 1
• while num <= n2,
  • for i in range col1 to col2,
    • res[row1, i] = num, incase num by 1
    • if num > n2, then break
  • for i in range row1 + 1 to row2,
    • res[i, col2-1] = num, incase num by 1
    • if num > n2, then break
  • for i in range col2 – 2 down to col1 – 1,
    • res[row2 – 1, i] = num, incase num by 1
    • if num > n2, then break
  • for i in range row2 – 2 down to row1,
    • res[i, col1] = num, incase num by 1
  • increase row1 by 1, decrease row2 by 1, increase col1 by 1 and decrease col2 by 1
• return res

Let us see the following implementation to get better understanding −

Example

 Live Demo

class Solution(object):
   def generateMatrix(self, n):
      row1 = 0
      col1 = 0
      row2 = n
      col2 = n
      result = [ [0 for i in range(n)] for j in range(n)]
      num = 1
      while num<=n**2:
         for i in range(col1,col2):
            result[row1][i] = num
            num+=1
         if num > n**2:
            break
         for i in range(row1+1,row2):
            result[i][col2-1] = num
            num+=1
         if num > n**2:
            break
         for i in range(col2-2,col1-1,-1):
            result[row2-1][i] = num
            num+=1
         if num > n**2:
            break
         for i in range(row2-2,row1,-1):
            result[i][col1] = num
            num+=1
            row1+=1
            row2-=1
            col1+=1
            col2-=1
            #print(result)
      return result
ob1 = Solution()
print(ob1.generateMatrix(4))

Input

4

Output

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

Arnab Chakraborty

Updated on: 04-May-2020

2K+ Views

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