Print concentric rectangular patterns

Question

I am solving interview questions from here.

Problem : Print concentric rectangular pattern in a 2d matrix. The outermost rectangle is formed by A, then the next outermost is formed by A-1 and so on. You will be given number as an argument to the function you need to implement, and you need to return a 2D array.

Example 1: Input: A = 4.

 Output: 
 4 4 4 4 4 4 4 
 4 3 3 3 3 3 4 
 4 3 2 2 2 3 4 
 4 3 2 1 2 3 4 
 4 3 2 2 2 3 4 
 4 3 3 3 3 3 4 
 4 4 4 4 4 4 4 

How can I make this code better?

def pretty_print(num):
 m = n = 2*num -1 ## m x n matrix , length of each row and column
 k = 0 # row start counter
 l = 0 # column start counter
 i = 0 # iterator
 matrix = [[0 for _ in range(n)] for _ in range(m)]
 while k < m and l < n : 
 #insert the first row
 for i in range(l, n) : 
 if matrix[k][i] == 0:
 matrix[k][i] = num # row index constt, change values in columns
 k += 1 # first row printed, so increment row start index
 #insert the last column
 for i in range(k, m) : 
 if matrix[i][n-1]==0:
 matrix[i][n-1] = num # column index constt, change values in rows
 n -= 1 # last column printed, so decrement num of columns
 #insert the last row
 if (k<m): # if row index less than number of rows remaining
 for i in range(n-1, l-1, -1):
 if matrix[m-1][i] == 0:
 matrix[m-1][i] = num # row index constt, insert in columns
 m -= 1 # last row printed, so decrement num of rows
 #insert the first column
 if (l<n): # if column index less than number of columns remaining
 for i in range(m-1, k-1, -1):
 if matrix[i][l] == 0:
 matrix[i][l] = num # column index constt, insert in rows
 l += 1 # first column printed, so increment column start index
 num -= 1 # all elements of value A inserted , so decrement
 return matrix
print pretty_print(6)

Answer 1

Simplify

I find the logic a bit complicated. It could be simpler:

• Loop from -A to +A, let's call the loop variable n
• Take the absolute value of n
• Generate the values in the row:
  • Loop from -A to +A, let's call the loop variable m
  • Use the maximum of abs(n) + 1 and abs(m) + 1

Like this:

def pretty_print(num):
 def print_line(n):
 low = abs(n)
 print(' '.join(str(max(low + 1, abs(i) + 1)) for i in range(-num + 1, num)))
 for i in range(-num + 1, num):
 print_line(i)

Testing

Doctests are awesome, I recommend to use them. Here's the complete solution with doctests, you can run this with python -mdoctest pretty_print.py:

#!/usr/bin/env python
def pretty_print(num):
 """
 >>> pretty_print(1)
 1
 >>> pretty_print(2)
 2 2 2
 2 1 2
 2 2 2
 >>> pretty_print(6)
 6 6 6 6 6 6 6 6 6 6 6
 6 5 5 5 5 5 5 5 5 5 6
 6 5 4 4 4 4 4 4 4 5 6
 6 5 4 3 3 3 3 3 4 5 6
 6 5 4 3 2 2 2 3 4 5 6
 6 5 4 3 2 1 2 3 4 5 6
 6 5 4 3 2 2 2 3 4 5 6
 6 5 4 3 3 3 3 3 4 5 6
 6 5 4 4 4 4 4 4 4 5 6
 6 5 5 5 5 5 5 5 5 5 6
 6 6 6 6 6 6 6 6 6 6 6
 """
 def print_line(n):
 low = abs(n)
 print(' '.join(str(max(low + 1, abs(i) + 1)) for i in range(-num + 1, num)))
 for i in range(-num + 1, num):
 print_line(i)
pretty_print(6)

Style

There are some minor style issues with the posted code. I suggest to follow the PEP8 guidelines.

Answer 2

without the use of 2-d matrix

n =int(input())
for i in range(n):
 temp = n
 for j in range(i):
 print(temp,end='')
 temp = temp -1
 for j in range(2*n-2*i - 1):
 print(n-i,end = '')
 for j in range(i):
 temp = temp+1
 print(temp,end= '')
 print()
for i in range(n-1,0,-1):
 temp = n
 for j in range(i-1):
 print(temp,end='')
 temp = temp -1
 for j in range(2*n-2*i+1):
 print(n-i+1,end= '')
 for j in range(i-1):
 temp = temp+1
 print(temp,end='')
 print()

• Ben A: You should include a short description about how this is an improvement to the original code, and why you chose to do it this way.