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479. Largest Palindrome Product

Description

Given an integer n, return the largest palindromic integer that can be represented as the product of two n-digits integers. Since the answer can be very large, return it modulo 1337.

 

Example 1:

Input: n = 2
Output: 987
Explanation: 99 x 91 = 9009, 9009 % 1337 = 987

Example 2:

Input: n = 1
Output: 9

 

Constraints:

• 1 <= n <= 8

Solutions

• Java
• C++
• Python
• Go

• class Solution {
      public int largestPalindrome(int n) {
          int mx = (int) Math.pow(10, n) - 1;
          for (int a = mx; a > mx / 10; --a) {
              int b = a;
              long x = a;
              while (b != 0) {
                  x = x * 10 + b % 10;
                  b /= 10;
              }
              for (long t = mx; t * t >= x; --t) {
                  if (x % t == 0) {
                      return (int) (x % 1337);
                  }
              }
          }
          return 9;
      }
  }
  

• class Solution {
  public:
      int largestPalindrome(int n) {
          int mx = pow(10, n) - 1;
          for (int a = mx; a > mx / 10; --a) {
              int b = a;
              long x = a;
              while (b) {
                  x = x * 10 + b % 10;
                  b /= 10;
              }
              for (long t = mx; t * t >= x; --t)
                  if (x % t == 0)
                      return x % 1337;
          }
          return 9;
      }
  };
  

• class Solution:
      def largestPalindrome(self, n: int) -> int:
          mx = 10**n - 1
          for a in range(mx, mx // 10, -1):
              b = x = a
              while b:
                  x = x * 10 + b % 10
                  b //= 10
              t = mx
              while t * t >= x:
                  if x % t == 0:
                      return x % 1337
                  t -= 1
          return 9
  
  

• func largestPalindrome(n int) int {
  	mx := int(math.Pow10(n)) - 1
  	for a := mx; a > mx/10; a-- {
  		x := a
  		for b := a; b != 0; b /= 10 {
  			x = x*10 + b%10
  		}
  		for t := mx; t*t >= x; t-- {
  			if x%t == 0 {
  				return x % 1337
  			}
  		}
  	}
  	return 9
  }
  

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