计算具有特定 XOR 值的子集数

给定一个由 n 个数字组成的数组 arr[] 和一个数字 K，找到 arr[] 的子集的数量，其中元素的异或为 K
例子 ：

Input:   arr[]  = {6, 9, 4,2}, k = 6
Output:  2
The subsets are {4, 2} and {6}

Input:   arr[]  = {1, 2, 3, 4, 5}, k = 4
Output:  4
The subsets are {1, 5}, {4}, {1, 2, 3, 4}
                and {2, 3, 5}

我们强烈建议您在继续解决方案之前单击此处进行练习。

蛮力方法 O(2 ⁿ )：一种简单的方法是生成所有 2 ⁿ个子集并对所有具有 XOR 值 K 的子集进行计数，但这种方法对于 n 的大值效率不高。

动态规划方法 O(n*m)：
我们定义一个数 m 使得 m = pow(2,(log2(max(arr))+1)) – 1。这个数实际上是任何 XOR 子集将获得的最大值。我们通过计算最大数量的位来获得这个数字。我们创建了一个二维数组 dp[n+1][m+1]，这样dp[i][j] 等于来自 arr[0…i-1] 的子集的具有 XOR 值 j 的子集的数量。

我们按如下方式填充 dp 数组：

1. 我们将 dp[i][j] 的所有值初始化为 0。
2. 由于空集的 XOR 为 0，因此设置 dp[0][0] 的值 = 1。
3. 从左到右迭代 arr[i] 的所有值，对于每个 arr[i]，迭代 XOR 的所有可能值，即从 0 到 m(包括两者)并填充 dp 数组如下：
   对于 i = 1 到 n：
   对于 j = 0 到 m：
   dp[i][j] = dp[i-1][j] + dp[i-1][j^arr[i-1]]
   这可以解释为，如果存在一个子集 arr[0…i-2] 的 XOR 值为 j，那么也存在一个子集 arr[0…i-1] 的 XOR 值为 j。此外，如果存在一个子集 arr[0….i-2] 具有 XOR 值 j^arr[i] 那么显然存在一个子集 arr[0…i-1] 具有 XOR 值 j，如 j ^ arr[i- 1] ^ arr[i-1] = j。
4. 计算具有 XOR 值 k 的子集数：由于 dp[i][j] 是来自 arr[0..i-1] 子集的具有 j 作为 XOR 值的子集数，那么来自集合 arr 的子集数[0..n] 具有 XOR 值作为 K 将是 dp[n][K]

C++

// arr dynamic programming solution to finding the number
// of subsets having xor of their elements as k
#include
using namespace std;
 
// Returns count of subsets of arr[] with XOR value equals
// to k.
int subsetXOR(int arr[], int n, int k)
{
    // Find maximum element in arr[]
    int max_ele = arr[0];
    for (int i=1; i max_ele)
           max_ele = arr[i];
 
    // Maximum possible XOR value
    int m = (1 << (int)(log2(max_ele) + 1) ) - 1;
    if( k > m  )
       return 0;
    // The value of dp[i][j] is the number of subsets having
    // XOR of their elements as j from the set arr[0...i-1]
    int dp[n+1][m+1];
    
    // Initializing all the values of dp[i][j] as 0
    for (int i=0; i<=n; i++)
        for (int j=0; j<=m; j++)
            dp[i][j] = 0;
 
    // The xor of empty subset is 0
    dp[0][0] = 1;
 
    // Fill the dp table
    for (int i=1; i<=n; i++)
        for (int j=0; j<=m; j++)
            dp[i][j] = dp[i-1][j] + dp[i-1][j^arr[i-1]];
 
    //  The answer is the number of subset from set
    //  arr[0..n-1] having XOR of elements as k
    return dp[n][k];
}
 
// Driver program to test above function
int main()
{
    int arr[] = {1, 2, 3, 4, 5};
    int k = 4;
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << "Count of subsets is " << subsetXOR(arr, n, k);
    return 0;
}

Java

// Java dynamic programming solution
// to finding the number of subsets
// having xor of their elements as k 
class GFG{
       
// Returns count of subsets of arr[] with 
// XOR value equals to k.
static int subsetXOR(int []arr, int n, int k)
{
     
    // Find maximum element in arr[]
    int max_ele = arr[0];
     
    for(int i = 1; i < n; i++)
        if (arr[i] > max_ele)
            max_ele = arr[i];
   
    // Maximum possible XOR value
    int m = (1 << (int)(Math.log(max_ele) /
                        Math.log(2) + 1) ) - 1;
    if (k > m)
    {
       return 0;  
    }
     
    // The value of dp[i][j] is the number
    // of subsets having XOR of their
    // elements as j from the set arr[0...i-1]
    int [][]dp = new int[n + 1][m + 1];
   
    // Initializing all the values of dp[i][j] as 0
    for(int i = 0; i <= n; i++)
        for(int j = 0; j <= m; j++)
            dp[i][j] = 0;
   
    // The xor of empty subset is 0
    dp[0][0] = 1;
   
    // Fill the dp table
    for(int i = 1; i <= n; i++)
        for(int j = 0; j <= m; j++)
            dp[i][j] = dp[i - 1][j] +
                       dp[i - 1][j ^ arr[i - 1]];
   
    // The answer is the number of
    // subset from set arr[0..n-1]
    // having XOR of elements as k
    return dp[n][k];
}
   
// Driver code
public static void main(String arg[])
{
    int []arr = { 1, 2, 3, 4, 5 };
    int k = 4;
    int n = arr.length;
     
    System.out.println("Count of subsets is " +
                        subsetXOR(arr, n, k));
}
}
 
// This code is contributed by rutvik_56

Python3

# Python 3 arr dynamic programming solution
# to finding the number of subsets having
# xor of their elements as k
import math
 
# Returns count of subsets of arr[] with
# XOR value equals to k.
def subsetXOR(arr, n, k):
     
    # Find maximum element in arr[]
    max_ele = arr[0]
    for i in range(1, n):
        if arr[i] > max_ele :
            max_ele = arr[i]
             
    # Maximum possible XOR value
    m = (1 << (int)(math.log2(max_ele) + 1)) - 1
    if( k > m  ):
       return 0
 
 
    # The value of dp[i][j] is the number
    # of subsets having XOR of their elements
    # as j from the set arr[0...i-1]
 
    # Initializing all the values
    # of dp[i][j] as 0
    dp = [[0 for i in range(m + 1)]
             for i in range(n + 1)]
     
    # The xor of empty subset is 0
    dp[0][0] = 1
 
    # Fill the dp table
    for i in range(1, n + 1):
        for j in range(m + 1):
            dp[i][j] = (dp[i - 1][j] +
                        dp[i - 1][j ^ arr[i - 1]])
 
    # The answer is the number of subset
    # from set arr[0..n-1] having XOR of
    # elements as k
    return dp[n][k]
 
# Driver Code
arr = [1, 2, 3, 4, 5]
k = 4
n = len(arr)
print("Count of subsets is",
       subsetXOR(arr, n, k))
 
# This code is contributed
# by sahishelangia

C#

// C# dynamic programming solution to finding the number
// of subsets having xor of their elements as k
using System;
 
class GFG
{
     
// Returns count of subsets of arr[] with
// XOR value equals to k.
static int subsetXOR(int []arr, int n, int k)
{
    // Find maximum element in arr[]
    int max_ele = arr[0];
    for (int i = 1; i < n; i++)
    if (arr[i] > max_ele)
        max_ele = arr[i];
 
    // Maximum possible XOR value
    int m = (1 << (int)(Math.Log(max_ele,2) + 1) ) - 1;
    if( k > m  ){
       return 0; 
    }
    // The value of dp[i][j] is the number of subsets having
    // XOR of their elements as j from the set arr[0...i-1]
    int [,]dp=new int[n+1,m+1];
 
    // Initializing all the values of dp[i][j] as 0
    for (int i = 0; i <= n; i++)
        for (int j = 0; j <= m; j++)
            dp[i, j] = 0;
 
    // The xor of empty subset is 0
    dp[0, 0] = 1;
 
    // Fill the dp table
    for (int i = 1; i <= n; i++)
        for (int j = 0; j <= m; j++)
            dp[i, j] = dp[i-1, j] + dp[i-1, j^arr[i-1]];
 
    // The answer is the number of subset from set
    // arr[0..n-1] having XOR of elements as k
    return dp[n, k];
}
 
    // Driver code
    static public void Main ()
    {
        int []arr = {1, 2, 3, 4, 5};
        int k = 4;
        int n = arr.Length;
        Console.WriteLine ("Count of subsets is " + subsetXOR(arr, n, k));
    }
}
 
// This code is contributed by jit_t.

PHP

 $max_ele)
        $max_ele = $arr[$i];
 
    // Maximum possible XOR value
    $m = (1 << (int)(log($max_ele,
                    2) + 1) ) - 1;
    if( $k > $m  ){
       return 0;
    }
    // The value of dp[i][j] is the
    // number of subsets having
    // XOR of their elements as j
    // from the set arr[0...i-1]
     
    // Initializing all the
    // values of dp[i][j] as 0
    for ($i = 0; $i <= $n; $i++)
        for ($j = 0; $j <= $m; $j++)
            $dp[$i][$j] = 0;
 
    // The xor of empty subset is 0
    $dp[0][0] = 1;
 
    // Fill the dp table
    for ($i = 1; $i <= $n; $i++)
        for ( $j = 0; $j <= $m; $j++)
            $dp[$i][$j] = $dp[$i - 1][$j] +
                          $dp[$i - 1][$j ^
                          $arr[$i - 1]];
 
    // The answer is the number
    // of subset from set arr[0..n-1]
    // having XOR of elements as k
    return $dp[$n][$k];
}
 
// Driver Code
$arr = array (1, 2, 3, 4, 5);
$k = 4;
$n = sizeof($arr);
echo "Count of subsets is " ,
     subsetXOR($arr, $n, $k);
 
// This code is contributed by ajit
?>

Javascript

 
    // Javascript dynamic programming solution
    // to finding the number of subsets
    // having xor of their elements as k
       
    // Returns count of subsets of arr[] with
    // XOR value equals to k.
    function subsetXOR(arr, n, k)
    {
 
        // Find maximum element in arr[]
        let max_ele = arr[0];
 
        for(let i = 1; i < n; i++)
            if (arr[i] > max_ele)
                max_ele = arr[i];
 
        // Maximum possible XOR value
        let m = (1 << parseInt(Math.log(max_ele) /
        Math.log(2) + 1, 10) ) - 1;
        if (k > m)
        {
           return 0; 
        }
 
        // The value of dp[i][j] is the number
        // of subsets having XOR of their
        // elements as j from the set arr[0...i-1]
        let dp = new Array(n + 1);
 
        // Initializing all the values of dp[i][j] as 0
        for(let i = 0; i <= n; i++)
        {
            dp[i] = new Array(m + 1);
            for(let j = 0; j <= m; j++)
            {
                dp[i][j] = 0;
            }
        }
 
        // The xor of empty subset is 0
        dp[0][0] = 1;
 
        // Fill the dp table
        for(let i = 1; i <= n; i++)
            for(let j = 0; j <= m; j++)
                dp[i][j] = dp[i - 1][j] +
                dp[i - 1][j ^ arr[i - 1]];
 
        // The answer is the number of
        // subset from set arr[0..n-1]
        // having XOR of elements as k
        return dp[n][k];
    }
     
    let arr = [ 1, 2, 3, 4, 5 ];
    let k = 4;
    let n = arr.length;
      
    document.write("Count of subsets is " + subsetXOR(arr, n, k));
     

输出 ：

Count of subsets is 4

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