LeetCode in Kotlin

2572. Count the Number of Square-Free Subsets

Medium

You are given a positive integer 0-indexed array nums.

A subset of the array nums is square-free if the product of its elements is a square-free integer.

A square-free integer is an integer that is divisible by no square number other than 1.

Return the number of square-free non-empty subsets of the array nums. Since the answer may be too large, return it modulo 109 + 7.

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

Example 1:

Input: nums = [3,4,4,5]

Output: 3

Explanation: There are 3 square-free subsets in this example:

• The subset consisting of the 0^th element [3]. The product of its elements is 3, which is a square-free integer.

• The subset consisting of the 3^rd element [5]. The product of its elements is 5, which is a square-free integer.

• The subset consisting of 0^th and 3^rd 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.

Example 2:

Input: nums = [1]

Output: 1

Explanation: There is 1 square-free subset in this example:

• The subset consisting of the 0^th 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.

Constraints:

• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 30

Solution

class Solution {
    private val primes = intArrayOf(2, 3, 5, 7, 11, 13, 17, 19, 23, 29)
    private val badNums = (1..30).filter { primes.any { p -> it % (p * p) == 0 } }.toSet()
    private val nonPrimes = (2..30).filter { it !in primes }.filter { it !in badNums }.toList()

    private fun gcd(a: Int, b: Int): Int {
        return if (b == 0) a else gcd(b, a % b)
    }

    fun squareFreeSubsets(nums: IntArray): Int {
        val mod: Long = 1_000_000_007
        // Get the frequency map
        val freqMap = nums.toTypedArray().groupingBy { it }.eachCount()
        var dp = mutableMapOf<Int, Long>()
        for (v in nonPrimes) {
            if (v !in freqMap) continue
            val howmany = freqMap[v]!!
            val prev = HashMap(dp)
            dp[v] = (dp[v] ?: 0) + howmany
            for ((product, quantity) in prev)
                if (gcd(product, v) == 1) {
                    dp[product * v] = ((dp[product * v] ?: 0L) + quantity * howmany.toLong()) % mod
                }
        }
        for (v in primes) {
            if (v !in freqMap) continue
            val howmany = freqMap[v]!!.toLong()
            val prev = dp
            dp = mutableMapOf<Int, Long>()
            dp[v] = howmany
            for ((product, quantity) in prev)
                dp[product] = if (product % v != 0) quantity * (1 + howmany) else quantity
        }
        // Getting all possible subsets without `1`
        var res = dp.values.sum() % mod
        // Find the permutations of `1`
        val possible = (1..(freqMap[1] ?: 0)).fold(1L) { sum, _ -> (sum shl 1) % mod }
        res = (res * possible + possible + mod - 1) % mod
        return res.toInt()
    }
}

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