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 binary search function
def binary_search(arr, target):
    left = 0
    right = len(arr) - 1
    while left <= right:
        mid = (left + right) // 2
        if arr[mid] == target:
            return mid
        elif arr[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
    return -1

# Usage
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
target = 7
result = binary_search(arr, target)
if result != -1:
    print(f"Element is present at index {result}")
else:
    print("Element is not present in array")

#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]

# Python program for implementation of Selection
# Sort
import sys
A = [64, 25, 12, 22, 11]

# Traverse through all array elements
for i in range(len(A)):
	
	# Find the minimum element in remaining
	# unsorted array
	min_idx = i
	for j in range(i+1, len(A)):
		if A[min_idx] > A[j]:
			min_idx = j
			
	# Swap the found minimum element with
	# the first element		
	A[i], A[min_idx] = A[min_idx], A[i]

# Driver code to test above
print ("Sorted array")
for i in range(len(A)):
	print("%d" %A[i]),

from collections import Counter

def find_parity_outliers(nums):
  return [
    x for x in nums
    if x % 2 != Counter([n % 2 for n in nums]).most_common()[0][0]
  ]

find_parity_outliers([1, 2, 3, 4, 6]) # [1, 3]

def when(predicate, when_true):
  return lambda x: when_true(x) if predicate(x) else x

double_even_numbers = when(lambda x: x % 2 == 0, lambda x : x * 2)
print(double_even_numbers(2)) # 4
print(double_even_numbers(1)) # 1