############################################################################################################################

#

# Problem 4

#

# A palindromic number reads the same both ways. The largest palindrome made from the product

# of two 2-digit numbers is 9009 = 91 × 99.

#

# Find the largest palindrome made from the product of two 3-digit numbers.

# Ans: 906609

############################################################################################################################

import numpy as np

def is_palindrome_ver1(n: int) -> bool:

'''

Args:

n (int >= 0): natural number palindrome candidate

Returns:

(bool): True if palindrome, False otherwise

'''

if n < 0:

raise ValueError("Must input a natural number")

split = [i for i in str(n)] # Convert each digit to an element in a list

length = len(split)

a = split[:length//2] # Slice the first half

b = []

if length % 2 == 0: # Is the length even or odd?

b += split[length//2:]

else:

b += split[length//2 + 1:]

b.reverse() # Reverse the second half slice

return a == b # Do the two sides match exactly?

def is_palindrome_ver2(x: int) -> bool:

'''

Args:

n (int >= 0): natural number palindrome candidate

Returns:

(bool): True if palindrome, False otherwise

'''

if x < 0:

raise ValueError("Must input a natural number")

string = str(x)

length = len(string)

beginning, reversed_end = string[:length//2], string[::-1][length//2:]

return string in beginning + reversed_end

def findLargestPalindrome(d=3):

'''

Args:

d (int > 0): number of digits

Returns:

largest (int): the largest palindrome made from the product of two d-digit numbers

Notes: This is a weird way to solve this problem.

'''

# Highest and lowest d-digit numbers e.g. If d=3 then h=999 and l=100

h, l = int('9'*d), 10**(d-1)

x = np.array([[*range(l,h+1)]]) # All of the d-digit numbers row vector

y = np.reshape(x, (len(x[0]),1)) # Column vector version

z = y.dot(x) # The multiplication table of d-digit numbers

calc = np.vectorize(is_palindrome_ver2)(z) # Apply the isPal function to multiplication table

return max(z[calc]) # Filter to get palindromes pick get the biggest

if __name__ == "__main__":

print(findLargestPalindrome(3))