Find the smallest prime palindrome greater than or equal to N.

Recall that a number is prime if it's only divisors are 1 and itself, and it is greater than 1.

For example, 2,3,5,7,11 and 13 are primes.

Recall that a number is a palindrome if it reads the same from left to right as it does from right to left.

For example, 12321 is a palindrome.

Example 1:

Input: 6
Output: 7

Example 2:

Input: 8
Output: 11

Example 3:

Input: 13
Output: 101

Note:

  • 1 <= N <= 10^8
  • The answer is guaranteed to exist and be less than 2 * 10^8.

Approach #1: Math. [Java]

class Solution {
public int primePalindrome(int N) {
if (8 <= N && N <= 11) return 11;
for (int x = 1; x < 100000; ++x) {
String s = Integer.toString(x), r = new StringBuilder(s).reverse().toString().substring(1);
int y = Integer.parseInt(s + r);
if (y >= N && isPrime(y)) return y;
}
return -1;
} boolean isPrime(int x) {
if (x < 2 || x % 2 == 0) return x == 2;
for (int i = 3; i * i <= x; i += 2) {
if (x % i == 0) return false;
}
return true;
}
}

  

Analysis:

All palindrome with even digits is multiple of 11.

We can prove as follow:

11 % 11 == 0

1111 % 11 == 0

111111 % 11 == 0

11111111 % 11 == 0

So:

1001 % 11 = (1111 - 11 * 10) % 11 == 0

100001 % 11 = (111111 - 1111 * 10) % 11 == 0

10000001 % 11 = (11111111 - 111111 * 10) % 11 == 0

For any palindrome with even digits:

abcddcba % 11

= (a * 10000001 + b * 100001 * 10 + c * 1001 * 100 + d * 11 * 1000) % 11

= 0

All palindrome with even digits is multiple of 11.

So among them, 11 is the only one prime

if (8 <= N <= 11) return 11

For other cases, we consider only palindrome with odd digits.

Time Complexity:

O(10000) to check all numbers 1 - 10000.

isPrime function is O(sqrt(x)) in worst case.

But only sqrt(N) worst cases for 1 <= x <= N

In general it's O(logx)

Reference:

https://leetcode.com/problems/prime-palindrome/discuss/146798/Search-Palindrome-with-Odd-Digits

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