Count the Number of Square-Free Subsets
<p>You are given a positive integer <strong>0-indexed</strong>&nbsp;array <code>nums</code>.</p>

<p>A subset of the array <code>nums</code> is <strong>square-free</strong> if the product of its elements is a <strong>square-free integer</strong>.</p>

<p>A <strong>square-free integer</strong> is an integer that is divisible by no square number other than <code>1</code>.</p>

<p>Return <em>the number of square-free non-empty subsets of the array</em> <strong>nums</strong>. Since the answer may be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>A <strong>non-empty</strong>&nbsp;<strong>subset</strong> of <code>nums</code> is an array that can be obtained by deleting some (possibly none but not all) elements from <code>nums</code>. Two subsets are different if and only if the chosen indices to delete are different.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,4,4,5]
<strong>Output:</strong> 3
<strong>Explanation:</strong> There are 3 square-free subsets in this example:
- The subset consisting of the 0<sup>th</sup> element [3]. The product of its elements is 3, which is a square-free integer.
- The subset consisting of the 3<sup>rd</sup> element [5]. The product of its elements is 5, which is a square-free integer.
- The subset consisting of 0<sup>th</sup> and 3<sup>rd</sup> elements [3,5]. The product of its elements is 15, which is a square-free integer.
It can be proven that there are no more than 3 square-free subsets in the given array.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1]
<strong>Output:</strong> 1
<strong>Explanation:</strong> There is 1 square-free subset in this example:
- The subset consisting of the 0<sup>th</sup> element [1]. The product of its elements is 1, which is a square-free integer.
It can be proven that there is no more than 1 square-free subset in the given array.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length&nbsp;&lt;= 1000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 30</code></li>
</ul>
Bit Manipulation; Array; Math; Dynamic Programming; Bitmask
function squareFreeSubsets(nums: number[]): number {
    const primes: number[] = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29];
    const cnt: number[] = Array(31).fill(0);
    for (const x of nums) {
        ++cnt[x];
    }
    const mod: number = Math.pow(10, 9) + 7;
    const n: number = primes.length;
    const f: number[] = Array(1 << n).fill(0);
    f[0] = 1;
    for (let i = 0; i < cnt[1]; ++i) {
        f[0] = (f[0] * 2) % mod;
    }
    for (let x = 2; x < 31; ++x) {
        if (cnt[x] === 0 || x % 4 === 0 || x % 9 === 0 || x % 25 === 0) {
            continue;
        }
        let mask: number = 0;
        for (let i = 0; i < n; ++i) {
            if (x % primes[i] === 0) {
                mask |= 1 << i;
            }
        }
        for (let state = (1 << n) - 1; state > 0; --state) {
            if ((state & mask) === mask) {
                f[state] = (f[state] + cnt[x] * f[state ^ mask]) % mod;
            }
        }
    }
    let ans: number = 0;
    for (let i = 0; i < 1 << n; ++i) {
        ans = (ans + f[i]) % mod;
    }
    ans -= 1;
    return ans >= 0 ? ans : ans + mod;
}
