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