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

#84 48 13 20 61 20 33 97 34 45 6 63 71 66 24 57 92 74 6 25 51 86 48 15 64 55 77 30 56 53 37 99 9 59 57 61 30 97 50 63 59 62 39 32 34 4 96 51 8 86 10 62 16 55 81 88 71 25 27 78 79 88 92 50 16 8 67 82 67 37 84 3 33 4 78 98 39 64 98 94 24 82 45 3 53 74 96 9 10 94 13 79 15 27 56 66 32 81 77
# xor a list of integers to find the lonely integer
res = a[0]
    for i in range(1,len(a)):
        res = res ^ a[i]

def print_x_pattern(size):
    i,j = 0,size - 1

    while j >= 0 and i < size:

        initial_spaces = ' '*min(i,j)
        middle_spaces = ' '*(abs(i - j) - 1)
        final_spaces = ' '*(size - 1 - max(i,j))

        if j == i:
            print(initial_spaces + '*' + final_spaces)
        else:
            print(initial_spaces + '*' + middle_spaces + '*' + final_spaces)

        i += 1
        j -= 1

print_x_pattern(7)

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 ?

def key_of_min(d):
  return min(d, key = d.get)

key_of_min({'a':4, 'b':0, 'c':13}) # b

{"output":{"name":"format_final_json_response","arguments":{"output":{}}}}