1994. The Number of Good Subsets

Author: sergeyleschev

1501-2000/1994. The Number of Good Subsets.ts

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+// Solution by Sergey Leschev
+// 1994. The Number of Good Subsets
+
+function numberOfGoodSubsets(nums: number[]): number {
+    const MOD = 1_000_000_007
+
+    // Primes up to 30
+    const primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29]
+
+    // Map number to its prime factor bitmask
+    const primeBitmask = (num: number): number => {
+        let mask = 0
+        for (let i = 0; i < primes.length; i++) {
+            if (num % primes[i] === 0) {
+                let count = 0
+                while (num % primes[i] === 0) {
+                    count++
+                    num /= primes[i]
+                }
+                if (count > 1) return -1 // Number not valid for distinct prime products
+                mask |= 1 << i
+            }
+        }
+        return mask
+    }
+
+    // Frequency count of numbers
+    const freq: number[] = Array(31).fill(0)
+    for (const num of nums) {
+        freq[num]++
+    }
+
+    // Dynamic Programming
+    const dp: number[] = Array(1 << primes.length).fill(0)
+    dp[0] = 1
+
+    for (let num = 2; num <= 30; num++) {
+        const mask = primeBitmask(num)
+        if (mask === -1 || freq[num] === 0) continue
+
+        for (let state = (1 << primes.length) - 1; state >= 0; state--) {
+            if ((state & mask) === 0) {
+                dp[state | mask] = (dp[state | mask] + dp[state] * freq[num]) % MOD
+            }
+        }
+    }
+
+    // Sum all subsets except the empty one
+    let result = dp.reduce((sum, count) => (sum + count) % MOD, 0) - 1
+
+    // Account for `1` if present
+    if (freq[1] > 0) {
+        // Use BigInt for power calculations
+        const powerOfTwo = BigInt(2) ** BigInt(freq[1]) % BigInt(MOD)
+        result = Number((BigInt(result) * powerOfTwo) % BigInt(MOD))
+    }
+
+    return (result + MOD) % MOD // Ensure non-negative result
+}