Miller-Rabin Primality Test and Caching II
A bit overkill to use MRP here. Also, there is probably a way to do it a little faster by scanning from left, then right, and stopping at the first prime number from each side. Code is down below, cheers, ACC. 3115. Maximum Prime Difference Medium 76 6 Add to List Share You are given an integer array nums . Return an integer that is the maximum distance between the indices of two (not necessarily different) prime numbers in nums . Example 1: Input: nums = [4,2,9,5,3] Output: 3 Explanation: nums[1] , nums[3] , and nums[4] are prime. So the answer is |4 - 1| = 3 . Example 2: Input: nums = [4,8,2,8] Output: 0 Explanation: nums[2] is prime. Because there is just one prime number, the answer is |2 - 2| = 0 . Constraints: 1 <= nums.length <= 3 * 10 5 1 <= nums[i] <= 100 The input is generated such that the number of prime numbers in the nums is at least one. public class Solution { public static Hashtable primes100 = null; public int MaximumP