# Python code to demonstrate
# method to remove i'th character
# Naive Method

# Initializing String
test_str = "CodeCatch"

# Printing original string
print ("The original string is : " + test_str)

# Removing char at pos 3
# using loop
new_str = ""

for i in range(len(test_str)):
	if i != 2:
		new_str = new_str + test_str[i]

# Printing string after removal
print ("The string after removal of i'th character : " + new_str)

filename = "data.txt"
data = "Hello, World!"

with open(filename, "a") as file:
    file.write(data)

https://codecatch.net/post/06c9f5b7-1e00-40dc-b436-b8cccc4b69be

from typing import Optional

from datetime import datetime

def convert_timestamp_string_to_epoch(timestamp: str) -> Optional[int]:
    epoch_time = None
    time_obj = datetime.strptime(timestamp, "%Y-%m-%d %H:%M:%S.%f")
    epoch_time = int((time_obj - datetime(1970, 1, 1)).total_seconds() * 1000)
    return epoch_time


print(int(convert_timestamp_string_to_epoch("2025-08-01 13:11:47.171")))
#above outputs 1754053907171.0

#how to I remove the .0 ?

#You are given a two-digit integer n. Return the sum of its digits.
#Example
#For n = 29 the output should be solution (n) = 11
def solution(n):
    return (n//10 + n%10)

def clamp_number(num, a, b):
  return max(min(num, max(a, b)), min(a, b))
  
clamp_number(2, 3, 5) # 3
clamp_number(1, -1, -5) # -1