按位与等于零计算三元组

给定一个由N 个整数组成的整数数组A[] ，找到索引 (i, j, k) 的三元组数，使得A[i] & A[j] & A[k]为0 (<0 ≤ i , j, k ≤ N 和&表示按位与运算符。

例子：

Input: A[]={2, 1, 3}
Output: 12
Explanation: The following i, j, k triples can be chosen whose bitwise AND is zero:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2

Input: A[]={21, 15, 6}
Output: 0
Explanation: No such triplets exist.

方法：解决这个问题的思路是使用一个Map来处理数组求解元素。请按照以下步骤解决问题：

• 初始化 Map 以存储A[i] 和 A[j]的每个可能值的频率。此外，用0初始化一个变量answer ，以存储所需的计数。
• 遍历数组，对于每个数组元素，遍历映射并检查每个映射是否为键，如果它与当前数组元素的按位 AND是否为 0。对于发现为真的每个数组元素，按关键字的频率增加答案。
• 遍历完数组后，打印answer 。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
#include 
using namespace std;
 
// Function to find the number of
// triplets whose Bitwise AND is 0.
int countTriplets(vector& A)
{
    // Stores the count of triplets
    // having bitwise AND equal to 0
    int cnt = 0;
 
    // Stores frequencies of all possible A[i] & A[j]
    unordered_map tuples;
 
    // Traverse the array
    for (auto a : A)
 
        // Update frequency of Bitwise AND
        // of all array elements with a
        for (auto b : A)
            ++tuples[a & b];
 
    // Traverse the array
    for (auto a : A)
 
        // Iterate the map
        for (auto t : tuples)
 
            // If bitwise AND of triplet
            // is zero, increment cnt
            if ((t.first & a) == 0)
                cnt += t.second;
 
    // Return the number of triplets
    // whose Bitwise AND is 0.
    return cnt;
}
 
// Driver Code
int main()
{
 
    // Input Array
    vector A = { 2, 1, 3 };
 
    // Function Call
    cout << countTriplets(A);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to find the number of
// triplets whose Bitwise AND is 0.
static int countTriplets(int []A)
{
   
    // Stores the count of triplets
    // having bitwise AND equal to 0
    int cnt = 0;
 
    // Stores frequencies of all possible A[i] & A[j]
    HashMap tuples = new HashMap();
 
    // Traverse the array
    for (int a : A)
 
        // Update frequency of Bitwise AND
        // of all array elements with a
        for (int b : A)
        {
            if(tuples.containsKey(a & b))
                tuples.put(a & b, tuples.get(a & b) + 1);
            else
                tuples.put(a & b, 1);
        }
 
    // Traverse the array
    for (int a : A)
 
        // Iterate the map
        for (Map.Entry t : tuples.entrySet())
 
            // If bitwise AND of triplet
            // is zero, increment cnt
            if ((t.getKey() & a) == 0)
                cnt += t.getValue();
 
    // Return the number of triplets
    // whose Bitwise AND is 0.
    return cnt;
}
 
// Driver Code
public static void main(String[] args)
{
 
    // Input Array
    int []A = { 2, 1, 3 };
 
    // Function Call
    System.out.print(countTriplets(A));
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python3 program for the above approach
 
# Function to find the number of
# triplets whose Bitwise AND is 0.
def countTriplets(A) :
 
    # Stores the count of triplets
    # having bitwise AND equal to 0
    cnt = 0;
 
    # Stores frequencies of all possible A[i] & A[j]
    tuples = {};
 
    # Traverse the array
    for a in A:
 
        # Update frequency of Bitwise AND
        # of all array elements with a
        for b in A:
            if (a & b) in tuples:
                tuples[a & b] += 1;               
            else:
                tuples[a & b] = 1;
 
    # Traverse the array
    for a in A:
         
        # Iterate the map
        for t in tuples:
 
            # If bitwise AND of triplet
            # is zero, increment cnt
            if ((t & a) == 0):
                cnt += tuples[t];
 
    # Return the number of triplets
    # whose Bitwise AND is 0.
    return cnt;
 
# Driver Code
if __name__ ==  "__main__" :
 
    # Input Array
    A = [ 2, 1, 3 ];
 
    # Function Call
    print(countTriplets(A));
 
    # This code is contributed by AnkThon

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
// Function to find the number of
// triplets whose Bitwise AND is 0.
static int countTriplets(int []A)
{
   
    // Stores the count of triplets
    // having bitwise AND equal to 0
    int cnt = 0;
 
    // Stores frequencies of all possible A[i] & A[j]
    Dictionary tuples = new Dictionary();
 
    // Traverse the array
    foreach (int a in A)
 
        // Update frequency of Bitwise AND
        // of all array elements with a
        foreach (int b in A)
        {
            if(tuples.ContainsKey(a & b))
                tuples[a & b] = tuples[a & b] + 1;
            else
                tuples.Add(a & b, 1);
        }
 
    // Traverse the array
    foreach (int a in A)
 
        // Iterate the map
        foreach (KeyValuePair t in tuples)
 
            // If bitwise AND of triplet
            // is zero, increment cnt
            if ((t.Key & a) == 0)
                cnt += t.Value;
 
    // Return the number of triplets
    // whose Bitwise AND is 0.
    return cnt;
}
 
// Driver Code
public static void Main(String[] args)
{
 
    // Input Array
    int []A = { 2, 1, 3 };
 
    // Function Call
    Console.Write(countTriplets(A));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript program for the above approach
 
// Function to find the number of
// triplets whose Bitwise AND is 0.
function countTriplets(A)
{
    // Stores the count of triplets
    // having bitwise AND equal to 0
    var cnt = 0;
 
    // Stores frequencies of all possible A[i] & A[j]
    var tuples = new Map();
 
    // Traverse the array
    A.forEach(a => {
         
        // Update frequency of Bitwise AND
        // of all array elements with a
        A.forEach(b => {
             
            if(tuples.has(a & b))
                tuples.set(a & b, tuples.get(a & b)+1)
            else
                tuples.set(a & b, 1)
        });
    });
 
    // Traverse the array
    A.forEach(a => {
         
        // Update frequency of Bitwise AND
        // of all array elements with a
        tuples.forEach((value, key) => {
             
            // If bitwise AND of triplet
            // is zero, increment cnt
            if ((key & a) == 0)
                cnt += value;
        });
    });
     
 
    // Return the number of triplets
    // whose Bitwise AND is 0.
    return cnt;
}
 
// Driver Code
// Input Array
var A = [2, 1, 3];
// Function Call
document.write( countTriplets(A));
 

输出：

12

时间复杂度： O(max(M, N ² )) 其中M是给定数组中存在的最大元素
辅助空间： O(M)

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