# Python Program to calculate the square root

num = float(input('Enter a number: '))

num_sqrt = num ** 0.5
print('The square root of %0.3f is %0.3f'%(num ,num_sqrt))

from statistics import median, mean, mode

def print_stats(array):
  print(array)
  print("median =", median(array))
  print("mean   =", mean(array))
  print("mode   =", mode(array))
  print()

print_stats([1, 2, 3, 3, 4])
print_stats([1, 2, 3, 3])

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

# Python program for implementation of Bubble Sort

def bubbleSort(arr):
	n = len(arr)

	# Traverse through all array elements
	for i in range(n-1):
	# range(n) also work but outer loop will repeat one time more than needed.

		# Last i elements are already in place
		for j in range(0, n-i-1):

			# traverse the array from 0 to n-i-1
			# Swap if the element found is greater
			# than the next element
			if arr[j] > arr[j+1] :
				arr[j], arr[j+1] = arr[j+1], arr[j]

# Driver code to test above
arr = [64, 34, 25, 12, 22, 11, 90]

bubbleSort(arr)

print ("Sorted array is:")
for i in range(len(arr)):
	print ("%d" %arr[i]),

weigh = lambda a,b: sum(b)-sum(a)
FindCoin = lambda A: 0 if (n := len(A)) == 1 else (m := n//3) * (w := 1 + weigh(A[:m], A[2*m:])) + FindCoin(A[m*w:m*(w+1)])
print(FindCoin([1,1,1,1,1,1,1,2,1]))

# List
lst = [1, 2, 3, 'Alice', 'Alice']

# One-Liner
indices = [i for i in range(len(lst)) if lst[i]=='Alice']

# Result
print(indices)
# [3, 4]