# Python code to demonstrate
# method to remove i'th character
# Naive Method

# Initializing String
test_str = "CodeCatch"

# Printing original string
print ("The original string is : " + test_str)

# Removing char at pos 3
# using loop
new_str = ""

for i in range(len(test_str)):
	if i != 2:
		new_str = new_str + test_str[i]

# Printing string after removal
print ("The string after removal of i'th character : " + new_str)

list_1 = [1,2,3,4,5,6,7,8,9]
cubed = map(lambda x: pow(x,3), list_1)
print(list(cubed))

#Results
#[1, 8, 27, 64, 125, 216, 343, 512, 729]

#Python 3: Fibonacci series up to n
def fib(n):
 a, b = 0, 1
 while a < n:
     print(a, end=' ')
     a, b = b, a+b
 print()
fib(1000)

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

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]))

my_list = [1, 2, 3, 4, 5]
removed_element = my_list.pop(2)  # Remove and return element at index 2
print(removed_element)  # 3
print(my_list)  # [1, 2, 4, 5]

last_element = my_list.pop()  # Remove and return the last element
print(last_element)  # 5
print(my_list)  # [1, 2, 4]