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

1808. Maximize Number of Nice Divisors

Level

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;
        }
    }
    
  • // 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;
        }
    };
    
  • print("Todo!")
    

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