Python Program to print hollow half diamond hash pattern
Last Updated :
22 Apr, 2023
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
def hollow_half_diamond(N):
for i in range( 1, N + 1):
for j in range(1, i + 1):
if i == j or j == 1:
print("#", end =" ")
else:
print(" ", end =" ")
print()
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()
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"))
for i in range(1, n+1):
print(("# " + " "*(i-2) + ("#" if i>1 else "")).rstrip())
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.
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