• Feb 26, 2023 •wabdelh
0 likes • 2 views
#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]
• Sep 14, 2024 •rgannedo-6205
https://codecatch.net/post/06c9f5b7-1e00-40dc-b436-b8cccc4b69be
• Sep 9, 2023 •AustinLeath
0 likes • 25 views
print("test")
• Nov 19, 2022 •CodeCatch
0 likes • 18 views
def sum_of_powers(end, power = 2, start = 1): return sum([(i) ** power for i in range(start, end + 1)]) sum_of_powers(10) # 385 sum_of_powers(10, 3) # 3025 sum_of_powers(10, 3, 5) # 2925
• Dec 24, 2025 •CodeCatch
0 likes • 0 views
string
• Aug 1, 2025 •AustinLeath
0 likes • 1 view
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 ?