• Aug 1, 2025 •AustinLeath
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import re _proposal_regex = r'(?:(?:(IKE|ESP):)?[\w/]+(?:/NO_EXT_SEQ)?(?:, ?(IKE|ESP):[\w/]+(?:/NO_EXT_SEQ)?)*)?' _proposals_re = rf'(?P<proposals>{_proposal_regex}|)' pattern = rf'received proposals: {_proposals_re}' match = re.match(pattern, 'received proposals: ') print(match.group('proposals') if match else "No match") # Prints "No match"
• Jun 26, 2025 •AustinLeath
def format_timestamp(timestamp_epoch): """ Convert epoch timestamp to formatted datetime string without using datetime package. Args: timestamp_epoch (int/float): Unix epoch timestamp (seconds since 1970-01-01 00:00:00 UTC) Returns: str: Formatted datetime string in 'YYYY-MM-DD HH:MM:SS' format """ # Constants for time calculations SECONDS_PER_DAY = 86400 SECONDS_PER_HOUR = 3600 SECONDS_PER_MINUTE = 60 # Handle negative timestamps and convert to integer timestamp = int(timestamp_epoch) # Calculate days since epoch and remaining seconds days_since_epoch = timestamp // SECONDS_PER_DAY remaining_seconds = timestamp % SECONDS_PER_DAY # Calculate hours, minutes, seconds hours = remaining_seconds // SECONDS_PER_HOUR remaining_seconds %= SECONDS_PER_HOUR minutes = remaining_seconds // SECONDS_PER_MINUTE seconds = remaining_seconds % SECONDS_PER_MINUTE # Calculate date (simplified, ignoring leap seconds) year = 1970 days = days_since_epoch while days >= 365: is_leap = (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0) days_in_year = 366 if is_leap else 365 if days >= days_in_year: days -= days_in_year year += 1 # Month lengths (non-leap year for simplicity, adjusted later for leap years) month_lengths = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31] if (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0): month_lengths[1] = 29 month = 0 while days >= month_lengths[month]: days -= month_lengths[month] month += 1 # Convert to 1-based indexing for month and day month += 1 day = days + 1 # Format the output string return f"{year:04d}-{month:02d}-{day:02d} {hours:02d}:{minutes:02d}:{seconds:02d}" # Example timestamp (Unix epoch seconds) timestamp = 1697054700 formatted_date = format_timestamp(timestamp) print(formatted_date + " UTC") # Output: 2023-10-11 18:45:00
• Nov 19, 2022 •CodeCatch
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# Python program for implementation of Bogo Sort import random # Sorts array a[0..n-1] using Bogo sort def bogoSort(a): n = len(a) while (is_sorted(a)== False): shuffle(a) # To check if array is sorted or not def is_sorted(a): n = len(a) for i in range(0, n-1): if (a[i] > a[i+1] ): return False return True # To generate permuatation of the array def shuffle(a): n = len(a) for i in range (0,n): r = random.randint(0,n-1) a[i], a[r] = a[r], a[i] # Driver code to test above a = [3, 2, 4, 1, 0, 5] bogoSort(a) print("Sorted array :") for i in range(len(a)): print ("%d" %a[i]),
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def print_pyramid_pattern(n): # outer loop to handle number of rows # n in this case for i in range(0, n): # inner loop to handle number of columns # values changing acc. to outer loop for j in range(0, i+1): # printing stars print("* ",end="") # ending line after each row print("\r") print_pyramid_pattern(10)
# Python program for implementation of Radix Sort # A function to do counting sort of arr[] according to # the digit represented by exp. def countingSort(arr, exp1): n = len(arr) # The output array elements that will have sorted arr output = [0] * (n) # initialize count array as 0 count = [0] * (10) # Store count of occurrences in count[] for i in range(0, n): index = (arr[i]/exp1) count[int((index)%10)] += 1 # Change count[i] so that count[i] now contains actual # position of this digit in output array for i in range(1,10): count[i] += count[i-1] # Build the output array i = n-1 while i>=0: index = (arr[i]/exp1) output[ count[ int((index)%10) ] - 1] = arr[i] count[int((index)%10)] -= 1 i -= 1 # Copying the output array to arr[], # so that arr now contains sorted numbers i = 0 for i in range(0,len(arr)): arr[i] = output[i] # Method to do Radix Sort def radixSort(arr): # Find the maximum number to know number of digits max1 = max(arr) # Do counting sort for every digit. Note that instead # of passing digit number, exp is passed. exp is 10^i # where i is current digit number exp = 1 while max1/exp > 0: countingSort(arr,exp) exp *= 10 # Driver code to test above arr = [ 170, 45, 75, 90, 802, 24, 2, 66] radixSort(arr) for i in range(len(arr)): print(arr[i]),
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# function which return reverse of a string def isPalindrome(s): return s == s[::-1] # Driver code s = "malayalam" ans = isPalindrome(s) if ans: print("Yes") else: print("No")