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Formatted question description: https://leetcode.ca/all/1808.html

1808. Maximize Number of Nice Divisors

Hard

Description

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 10^9 + 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 <= 10^9

Solution

If primeFactors <= 3, simply return primeFactors.

For greater values, consider the prime factors of n. The smallest nice divisor of n is the product of all distinct prime factors of n. For example, the prime factors of 12 are 2,2,3, so the smmalest nice divisor of 12 is 2*3=6. If n = p1^a1 * p2^a2 * ... * pk*ak, where a1 + a2 + ... + ak <= primeFactors, then n has a1 * a2 * ... * ak nice divisors. The problem is to find the maximum value of a1 * a2 * ... * ak.

To get the maximum value, there should be as many numbers 3 as possible, and the remaining numbers should be 2 (at most two numbers 2). Calculate the product using pow function and deal with large values accordingly.

• class Solution {
static final int MODULO = 1000000007;

public int maxNiceDivisors(int primeFactors) {
if (primeFactors <= 3)
return primeFactors;
int quotient = primeFactors / 3;
int remainder = primeFactors % 3;
if (remainder == 0)
return (int) (pow(3, quotient) % MODULO);
else if (remainder == 1)
return (int) (pow(3, quotient - 1) * 4 % MODULO);
else
return (int) (pow(3, quotient) * 2 % MODULO);
}

public long pow(long x, int n) {
long power = 1;
while (n > 0) {
if (n % 2 == 1)
power = power * x % MODULO;
x = x * x % MODULO;
n /= 2;
}
return power;
}
}

############

class Solution {
public int maxNiceDivisors(int primeFactors) {
if (primeFactors < 4) {
return primeFactors;
}
final int mod = (int) 1e9 + 7;
if (primeFactors % 3 == 0) {
return (int) qmi(3, primeFactors / 3, mod);
}
if (primeFactors % 3 == 1) {
return (int) (4 * qmi(3, primeFactors / 3 - 1, mod) % mod);
}
return (int) (2 * qmi(3, primeFactors / 3, mod) % mod);
}

private long qmi(long a, long k, long p) {
long res = 1;
while (k != 0) {
if ((k & 1) == 1) {
res = res * a % p;
}
k >>= 1;
a = a * a % p;
}
return res;
}
}

• // OJ: https://leetcode.com/problems/maximize-number-of-nice-divisors/
// Time: O(logN)
// Space: O(1)
class Solution {
long mod = 1e9+7;
public:
long modpow(long base, long exp) {
long ans = 1;
while (exp > 0) {
if (exp & 1) ans = (ans * base) % mod;
base = (base * base) % mod;
exp >>= 1;
}
return ans;
}
int maxNiceDivisors(int N) {
if (N <= 3) return N;
long ans;
switch (N % 3) {
case 0: ans = modpow(3, N / 3); break;
case 1: ans = 2 * 2 * modpow(3, N / 3 - 1); break;
case 2: ans = 2 * modpow(3, N / 3); break;
}
return ans % mod;
}
};

• class Solution:
def maxNiceDivisors(self, primeFactors: int) -> int:
mod = 10**9 + 7
if primeFactors < 4:
return primeFactors
if primeFactors % 3 == 0:
return pow(3, primeFactors // 3, mod) % mod
if primeFactors % 3 == 1:
return 4 * pow(3, primeFactors // 3 - 1, mod) % mod
return 2 * pow(3, primeFactors // 3, mod) % mod


• func maxNiceDivisors(primeFactors int) int {
if primeFactors < 4 {
return primeFactors
}
const mod int = 1e9 + 7
if primeFactors%3 == 0 {
return qmi(3, primeFactors/3, mod)
}
if primeFactors%3 == 1 {
return 4 * qmi(3, primeFactors/3-1, mod) % mod
}
return 2 * qmi(3, primeFactors/3, mod) % mod
}

func qmi(a, k, p int) int {
res := 1
for k != 0 {
if k&1 == 1 {
res = res * a % p
}
k >>= 1
a = a * a % p
}
return res
}

• /**
* @param {number} primeFactors
* @return {number}
*/
var maxNiceDivisors = function (primeFactors) {
if (primeFactors < 4) {
return primeFactors;
}
const mod = 1e9 + 7;
const qpow = (a, n) => {
let ans = 1;
for (; n; n >>= 1) {
if (n & 1) {
ans = Number((BigInt(ans) * BigInt(a)) % BigInt(mod));
}
a = Number((BigInt(a) * BigInt(a)) % BigInt(mod));
}
return ans;
};
const k = Math.floor(primeFactors / 3);
if (primeFactors % 3 === 0) {
return qpow(3, k);
}
if (primeFactors % 3 === 1) {
return (4 * qpow(3, k - 1)) % mod;
}
return (2 * qpow(3, k)) % mod;
};