feat: add solutions to lc problem: No.1994

No.1994.The Number of Good Subsets

Author: yanglbme

solution/1900-1999/1994.The Number of Good Subsets/README.md

@@ -65,7 +65,19 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
-状态压缩 DP。
+**方法一：状态压缩动态规划**
+
+注意到题目中 $nums[i]$ 的范围为 $[1, 30]$，因此我们可以预处理出所有小于等于 $30$ 的质数，即 $[2, 3, 5, 7, 11, 13, 17, 19, 23, 29]$。
+
+好子集中，所有元素的乘积可以表示为一个或多个互不相同的质数的乘积，也即是说，每个质因数最多只能出现一次。因此，我们可以使用一个二进制数来表示一个子集中的质因数，其中二进制数的第 $i$ 位表示质数 $primes[i]$ 是否出现在子集中。
+
+我们可以使用状态压缩动态规划的方法来求解本题。设 $f[i]$ 表示二进制数 $i$ 表示的子集中的质因数的乘积为一个或多个互不相同的质数的乘积的方案数。初始时 $f[0]=1$。
+
+我们在 $[2,..30]$ 的范围内枚举一个数 $x$，如果 $x$ 不在 $nums$ 中，或者 $x$ 为 $4, 9, 25$ 的倍数，那么我们可以直接跳过。否则，我们可以将 $x$ 的质因数用一个二进制数 $mask$ 表示，然后我们从大到小枚举当前的状态 $state$，如果 $state$ 与 $mask$ 按位与的结果为 $mask$，那么我们可以从状态 $f[state \oplus mask]$ 转移到状态 $f[state]$，转移方程为 $f[state] = f[state] + cnt[x] \times f[state \oplus mask]$，其中 $cnt[x]$ 表示 $x$ 在 $nums$ 中出现的次数。
+
+注意，我们没有从数字 $1$ 开始枚举，因为我们可以选择任意个数字 $1$，加入到好子集中。那么最终的答案为 $\sum_{i=1}{2^{10}-1} f[i] \times 2^{cnt[1]}$。
+
+时间复杂度 $O(n + C \times M)$，空间复杂度 $O(M)$。其中 $n$ 为 $nums$ 的长度；而 $C$ 和 $M$ 分别为题目中 $nums[i]$ 的范围和状态的个数，本题中 $C=30$, $M=2^{10}$。
 
 <!-- tabs:start -->
 
@@ -76,29 +88,23 @@
 ```python
 class Solution:
     def numberOfGoodSubsets(self, nums: List[int]) -> int:
-        counter = Counter(nums)
+        primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29]
+        cnt = Counter(nums)
         mod = 10**9 + 7
-        prime = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29]
-        n = len(prime)
-        dp = [0] * (1 << n)
-        dp[0] = 1
+        n = len(primes)
+        f = [0] * (1 << n)
+        f[0] = pow(2, cnt[1])
         for x in range(2, 31):
-            if counter[x] == 0 or x % 4 == 0 or x % 9 == 0 or x % 25 == 0:
+            if cnt[x] == 0 or x % 4 == 0 or x % 9 == 0 or x % 25 == 0:
                 continue
             mask = 0
-            for i, p in enumerate(prime):
+            for i, p in enumerate(primes):
                 if x % p == 0:
                     mask |= 1 << i
-            for state in range(1 << n):
-                if mask & state:
-                    continue
-                dp[mask | state] = (dp[mask | state] + counter[x] * dp[state]) % mod
-        ans = 0
-        for i in range(1, 1 << n):
-            ans = (ans + dp[i]) % mod
-        for i in range(counter[1]):
-            ans = (ans << 1) % mod
-        return ans
+            for state in range((1 << n) - 1, 0, -1):
+                if state & mask == mask:
+                    f[state] = (f[state] + cnt[x] * f[state ^ mask]) % mod
+        return sum(f[i] for i in range(1, 1 << n)) % mod
 ```
 
 ### **Java**
@@ -107,40 +113,38 @@ class Solution:
 
 ```java
 class Solution {
-    private static final int MOD = (int) 1e9 + 7;
-
     public int numberOfGoodSubsets(int[] nums) {
-        int[] counter = new int[31];
+        int[] primes = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29};
+        int[] cnt = new int[31];
         for (int x : nums) {
-            ++counter[x];
+            ++cnt[x];
+        }
+        final int mod = (int) 1e9 + 7;
+        int n = primes.length;
+        long[] f = new long[1 << n];
+        f[0] = 1;
+        for (int i = 0; i < cnt[1]; ++i) {
+            f[0] = (f[0] * 2) % mod;
         }
-        int[] prime = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29};
-        int n = prime.length;
-        long[] dp = new long[1 << n];
-        dp[0] = 1;
-        for (int x = 2; x <= 30; ++x) {
-            if (counter[x] == 0 || x % 4 == 0 || x % 9 == 0 || x % 25 == 0) {
+        for (int x = 2; x < 31; ++x) {
+            if (cnt[x] == 0 || x % 4 == 0 || x % 9 == 0 || x % 25 == 0) {
                 continue;
             }
             int mask = 0;
             for (int i = 0; i < n; ++i) {
-                if (x % prime[i] == 0) {
-                    mask |= (1 << i);
+                if (x % primes[i] == 0) {
+                    mask |= 1 << i;
                 }
             }
-            for (int state = 0; state < 1 << n; ++state) {
-                if ((mask & state) > 0) {
-                    continue;
+            for (int state = (1 << n) - 1; state > 0; --state) {
+                if ((state & mask) == mask) {
+                    f[state] = (f[state] + cnt[x] * f[state ^ mask]) % mod;
                 }
-                dp[mask | state] = (dp[mask | state] + counter[x] * dp[state]) % MOD;
             }
         }
         long ans = 0;
         for (int i = 1; i < 1 << n; ++i) {
-            ans = (ans + dp[i]) % MOD;
-        }
-        while (counter[1]-- > 0) {
-            ans = (ans << 1) % MOD;
+            ans = (ans + f[i]) % mod;
         }
         return (int) ans;
     }
@@ -153,71 +157,79 @@ class Solution {
 class Solution {
 public:
     int numberOfGoodSubsets(vector<int>& nums) {
-        vector<int> counter(31);
-        for (int& x : nums) ++counter[x];
-        vector<int> prime = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29};
-        const int MOD = 1e9 + 7;
-        int n = prime.size();
-        vector<long long> dp(1 << n);
-        dp[0] = 1;
-        for (int x = 2; x <= 30; ++x) {
-            if (counter[x] == 0 || x % 4 == 0 || x % 9 == 0 || x % 25 == 0) continue;
+        int primes[10] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29};
+        int cnt[31]{};
+        for (int& x : nums) {
+            ++cnt[x];
+        }
+        int n = 10;
+        const int mod = 1e9 + 7;
+        vector<long long> f(1 << n);
+        f[0] = 1;
+        for (int i = 0; i < cnt[1]; ++i) {
+            f[0] = f[0] * 2 % mod;
+        }
+        for (int x = 2; x < 31; ++x) {
+            if (cnt[x] == 0 || x % 4 == 0 || x % 9 == 0 || x % 25 == 0) {
+                continue;
+            }
             int mask = 0;
-            for (int i = 0; i < n; ++i)
-                if (x % prime[i] == 0)
-                    mask |= (1 << i);
-            for (int state = 0; state < 1 << n; ++state) {
-                if ((mask & state) > 0) continue;
-                dp[mask | state] = (dp[mask | state] + counter[x] * dp[state]) % MOD;
+            for (int i = 0; i < n; ++i) {
+                if (x % primes[i] == 0) {
+                    mask |= 1 << i;
+                }
+            }
+            for (int state = (1 << n) - 1; state; --state) {
+                if ((state & mask) == mask) {
+                    f[state] = (f[state] + 1LL * cnt[x] * f[state ^ mask]) % mod;
+                }
             }
         }
         long long ans = 0;
-        for (int i = 1; i < 1 << n; ++i) ans = (ans + dp[i]) % MOD;
-        while (counter[1]--) ans = (ans << 1) % MOD;
-        return (int)ans;
+        for (int i = 1; i < 1 << n; ++i) {
+            ans = (ans + f[i]) % mod;
+        }
+        return ans;
     }
 };
 ```
 
 ### **Go**
 
 ```go
-func numberOfGoodSubsets(nums []int) int {
-	counter := make([]int, 31)
+func numberOfGoodSubsets(nums []int) (ans int) {
+	primes := []int{2, 3, 5, 7, 11, 13, 17, 19, 23, 29}
+	cnt := [31]int{}
 	for _, x := range nums {
-		counter[x]++
+		cnt[x]++
 	}
 	const mod int = 1e9 + 7
-	prime := []int{2, 3, 5, 7, 11, 13, 17, 19, 23, 29}
-	n := len(prime)
-	dp := make([]int, 1<<n)
-	dp[0] = 1
-	for x := 2; x <= 30; x++ {
-		if counter[x] == 0 || x%4 == 0 || x%9 == 0 || x%25 == 0 {
+	n := 10
+	f := make([]int, 1<<n)
+	f[0] = 1
+	for i := 0; i < cnt[1]; i++ {
+		f[0] = f[0] * 2 % mod
+	}
+	for x := 2; x < 31; x++ {
+		if cnt[x] == 0 || x%4 == 0 || x%9 == 0 || x%25 == 0 {
 			continue
 		}
 		mask := 0
-		for i, p := range prime {
+		for i, p := range primes {
 			if x%p == 0 {
-				mask |= (1 << i)
+				mask |= 1 << i
 			}
 		}
-		for state := 0; state < 1<<n; state++ {
-			if (mask & state) > 0 {
-				continue
+		for state := 1<<n - 1; state > 0; state-- {
+			if state&mask == mask {
+				f[state] = (f[state] + f[state^mask]*cnt[x]) % mod
 			}
-			dp[mask|state] = (dp[mask|state] + counter[x]*dp[state]) % mod
 		}
 	}
-	ans := 0
 	for i := 1; i < 1<<n; i++ {
-		ans = (ans + dp[i]) % mod
-	}
-	for counter[1] > 0 {
-		ans = (ans << 1) % mod
-		counter[1]--
+		ans = (ans + f[i]) % mod
 	}
-	return ans
+	return
 }
 ```