LeetCode in Kotlin

1808. Maximize Number of Nice Divisors

Hard

You are given a positive integer primeFactors. You are asked to construct a positive integer n that satisfies the following conditions:

• The number of prime factors of n (not necessarily distinct) is at most primeFactors.
• The number of nice divisors of n is maximized. Note that a divisor of n is nice if it is divisible by every prime factor of n. For example, if n = 12, then its prime factors are [2,2,3], then 6 and 12 are nice divisors, while 3 and 4 are not.

Return the number of nice divisors of n. Since that number can be too large, return it modulo 109 + 7.

Note that a prime number is a natural number greater than 1 that is not a product of two smaller natural numbers. The prime factors of a number n is a list of prime numbers such that their product equals n.

Example 1:

Input: primeFactors = 5

Output: 6

Explanation: 200 is a valid value of n. It has 5 prime factors: [2,2,2,5,5], and it has 6 nice divisors: [10,20,40,50,100,200]. There is not other value of n that has at most 5 prime factors and more nice divisors.

Example 2:

Input: primeFactors = 8

Output: 18

Constraints:

• 1 <= primeFactors <= 109

Solution

@Suppress("NAME_SHADOWING")
class Solution {
    private fun modPow(b: Long, e: Int, m: Int): Long {
        var b = b
        var e = e
        if (m == 1) {
            return 0
        }
        if (e == 0 || b == 1L) {
            return 1
        }
        b %= m.toLong()
        var r: Long = 1
        while (e > 0) {
            if (e and 1 == 1) {
                r = r * b % m
            }
            e = e shr 1
            b = b * b % m
        }
        return r
    }

    fun maxNiceDivisors(pf: Int): Int {
        val mod = 1000000007
        val st = intArrayOf(0, 1, 2, 3, 4, 6)
        return if (pf < 5) pf else (modPow(3, pf / 3 - 1, mod) * st[3 + pf % 3] % mod).toInt()
    }
}

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