Given

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 ×ばつ 99.

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

Solution

import time
def reverse(num):
 dreverse = 0
 while (num > 0):
 dreverse = dreverse * 10 + num % 10
 num /= 10
 return dreverse
def is_palindrome(num):
 if (num < 0):
 raise ValueError('Non-negative integers only')
 return num == reverse(num)
def main():
 largest_palindrome = 0
 lower = 999
 while lower >= 100:
 upper = 999
 while upper >= lower:
 if upper * lower <= largest_palindrome:
 break
 if is_palindrome(upper * lower):
 largest_palindrome = upper * lower
 upper -= 1
 lower -= 1
 return largest_palindrome
start_time = time.time()
print main()
print("--- %s seconds ---" % (time.time() - start_time))
o/p: 906609
--- 0.0150721073151 seconds --- 

I am including timing metric from now on, this solution seems \$O(N^2)\$ can I improve it further?

Given

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 ×ばつ 99.

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

Solution

import time
def reverse(num):
 dreverse = 0
 while (num > 0):
 dreverse = dreverse * 10 + num % 10
 num /= 10
 return dreverse
def is_palindrome(num):
 if (num < 0):
 raise ValueError('Non-negative integers only')
 return num == reverse(num)
def main():
 largest_palindrome = 0
 lower = 999
 while lower >= 100:
 upper = 999
 while upper >= lower:
 if upper * lower <= largest_palindrome:
 break
 if is_palindrome(upper * lower):
 largest_palindrome = upper * lower
 upper -= 1
 lower -= 1
 return largest_palindrome
start_time = time.time()
print main()
print("--- %s seconds ---" % (time.time() - start_time))
o/p: 906609
--- 0.0150721073151 seconds --- 

I am including timing metric from now on. This solution seems \$O(N^2)\$. Can I improve it further?