Python Program to print hollow half diamond hash pattern

Give an integer N and the task is to print hollow half diamond pattern. Examples:

Input : 6
Output :
# 
# # 
#   # 
#     # 
#       # 
#         # 
#       # 
#     # 
#   # 
# # 
# 

Input : 7
Output :
# 
# # 
#   # 
#     # 
#       # 
#         # 
#           # 
#         # 
#       # 
#     # 
#   # 
# # 
#  

Approach: The idea is to break the pattern into two parts:

• Upper part: For the upper half start the for loop with iterator (i) from 1 to n and one more for loop with iterator (j) from 1 to i.

# 
# # 
#   # 
#     # 
#       # 
#         # 
#           #  

• Lower part: For lower half start the for loop with iterator (n-1) to 1 and loop inside this for will remain same as for the upper half.

#         # 
#       # 
#     # 
#   # 
# # 
#

• Now we have to check this condition that if (i==j) or (j==1) then we have to print “#”, otherwise we have to print ” “(space).

Below is the implementation:

Python3

# python program to print
# hollow half diamond star


# function to print hollow
# half diamond star
def hollow_half_diamond(N):

    # this for loop is for
    # printing upper half
    for i in range( 1, N + 1):
        for j in range(1, i + 1):

            # this is the condition to
            # print "#" only on the
            # boundaries
            if i == j or j == 1:
                print("#", end =" ")

            # print " "(space) on the rest
            # of the area
            else:
                print(" ", end =" ")
        print()

    # this for loop is to print lower half
    for i in range(N - 1, 0, -1):

        for j in range(1, i + 1):

            if j == 1 or i == j:
                print("#", end =" ")

            else:
                print(" ", end =" ")

        print()

# Driver Code
if __name__ == "__main__":
    N = 7
    hollow_half_diamond( N )

Output:

# 
# # 
#   # 
#     # 
#       # 
#         # 
#           # 
#         # 
#       # 
#     # 
#   # 
# # 
# 

Time Complexity: O(N^2)

Space Complexity: O(1) as using constant space

METHOD 2:Using string formatting and list comprehension

APPROACH:

This program prints a hollow half diamond hash pattern of a given size n.

ALGORITHM:

1.Read the value of n from the user.
2.For the upper half of the pattern, use a loop that runs from 1 to n.
3.In each iteration, print a string consisting of a hash (#), followed by zero or more spaces, followed by another hash (#).
4.The number of spaces in each iteration is (i-2), where i is the current iteration number. The first iteration has only one hash, so we need to print a single hash instead of the spaces.
5.Use the rstrip() method to remove any trailing whitespace characters.
6.For the lower half of the pattern, use a loop that runs from n-1 to 1 (in reverse).
7.Use the same logic as in step 3 to print the pattern for the lower half of the diamond.

Python3

n = int(("6"))

# Upper half of the pattern
for i in range(1, n+1):
    print(("# " + "  "*(i-2) + ("#" if i>1 else "")).rstrip())

# Lower half of the pattern
for i in range(n-1, 0, -1):
    print(("# " + "  "*(i-2) + ("#" if i>1 else "")).rstrip())

Output

#
# #
#   #
#     #
#       #
#         #
#       #
#     #
#   #
# #
#

Time complexity: O(n^2) (quadratic) – because we have two nested loops that run n times each.

Space complexity: O(1) – because we only need a few variables to store the input and intermediate values, and the amount of memory used by the program does not depend on the input size.