#1994 - The Number of Good Subsets

Problem Description

You are given an integer array nums. We call a subset of nums good if its product can be represented as a product of one or more distinct prime numbers.

For example, if nums = [1, 2, 3, 4]:
- [2, 3], [1, 2, 3], and [1, 3] are good subsets with products 6 = 2*3, 6 = 2*3, and 3 = 3 respectively.
- [1, 4] and [4] are not good subsets with products 4 = 2*2 and 4 = 2*2 respectively.

Return the number of different good subsets in nums modulo 109 + 7.

A subset of nums is any array that can be obtained by deleting some (possibly none or all) elements from nums. Two subsets are different if and only if the chosen indices to delete are different.

Solution

/**
 * @param {number[]} nums
 * @return {number}
 */
var numberOfGoodSubsets = function(nums) {
  const MOD = 1e9 + 7;
  const primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29];
  const freq = new Array(31).fill(0);
  for (const num of nums) {
    freq[num]++;
  }

  const dp = new Array(1 << primes.length).fill(0);
  dp[0] = 1;
  for (let i = 0; i < freq[1]; i++) {
    dp[0] = (dp[0] * 2) % MOD;
  }

  for (let num = 2; num <= 30; num++) {
    if (freq[num] === 0) continue;
    let mask = 0;
    let valid = true;
    for (let i = 0; i < primes.length; i++) {
      let count = 0;
      let temp = num;
      while (temp % primes[i] === 0) {
        count++;
        temp /= primes[i];
      }
      if (count > 1) {
        valid = false;
        break;
      }
      if (count === 1) {
        mask |= 1 << i;
      }
    }

    if (!valid) continue;

    const prev = dp.slice();
    for (let j = 0; j < 1 << primes.length; j++) {
      if ((j & mask) === 0) {
        dp[j | mask] = (dp[j | mask] + prev[j] * freq[num]) % MOD;
      }
    }
  }

  let result = 0;
  for (let i = 1; i < 1 << primes.length; i++) {
    result = (result + dp[i]) % MOD;
  }

  return result;
};