LeetCode in Kotlin

3115. Maximum Prime Difference

Medium

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 * 105
• 1 <= nums[i] <= 100
• The input is generated such that the number of prime numbers in the nums is at least one.

Solution

import kotlin.math.sqrt

class Solution {
    fun maximumPrimeDifference(nums: IntArray): Int {
        val n = nums.size
        var i = 0
        while (i < n && check(nums[i])) {
            i++
        }
        var j = n - 1
        while (j >= 0 && check(nums[j])) {
            j--
        }
        return j - i
    }

    private fun check(n: Int): Boolean {
        if (n < 2) {
            return true
        }
        var i = 2
        while (i <= sqrt(n.toDouble())) {
            if (n % i == 0) {
                return true
            }
            i++
        }
        return false
    }
}

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