给定两个数组arr1 []和arr2 [] ，任务是从第一个数组(例如a)中取一个元素，从第二个数组(例如b)中取一个元素。如果通过反转的比特形成的数量等于b，则在一对(A，B)是一个有效的对。
位反转示例：
11以二进制形式写为1011 。将其位反转后，将获得0100 (十进制为4 )。因此(11，4)是有效对，但(4，11)不是，因为将4的数字取反即100-> 011为3后无法获得11 。
例子：
Input: arr1[] = {11, 5, 4}, arr2[] = {1, 4, 3, 11}
Output: 2
(11, 4) and (4, 3) are the only valid pairs.
Input: arr1[] = {43, 7, 1, 99}, arr2 = {5, 1, 28, 20}
Output: 2
方法：
- 取两个空集s1和s2 。
- 在s2中插入arr2 []的所有元素。
- 迭代第一个数组。如果在第一组中不存在该元素，而在第二组中存在通过反转其位而形成的数字，则增加计数并将当前元素插入s1中，以便不再进行计数。
- 最后打印计数值。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the number formed
// by inverting bits the bits of num
int invertBits(int num)
{
// Number of bits in num
int x = log2(num) + 1;
// Inverting the bits one by one
for (int i = 0; i < x; i++)
num = (num ^ (1 << i));
return num;
}
// Function to return the total valid pairs
int totalPairs(int arr1[], int arr2[], int n, int m)
{
// Set to store the elements of the arrays
unordered_set s1, s2;
// Insert all the elements of arr2[] in the set
for (int i = 0; i < m; i++)
s2.insert(arr2[i]);
// Initialize count variable to 0
int count = 0;
for (int i = 0; i < n; i++) {
// Check if element of the first array
// is not in the first set
if (s1.find(arr1[i]) == s1.end()) {
// Check if the element formed by inverting bits
// is in the second set
if (s2.find(invertBits(arr1[i])) != s2.end()) {
// Increment the count of valid pairs and insert
// the element in the first set so that
// it doesn't get counted again
count++;
s1.insert(arr1[i]);
}
}
}
// Return the total number of pairs
return count;
}
// Driver code
int main()
{
int arr1[] = { 43, 7, 1, 99 };
int arr2[] = { 5, 1, 28, 20 };
int n = sizeof(arr1) / sizeof(arr1[0]);
int m = sizeof(arr2) / sizeof(arr2[0]);
cout << totalPairs(arr1, arr2, n, m);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
import java.io.*;
import java.lang.*;
class GFG
{
static int log2(int N)
{
// calculate log2 N indirectly
// using log() method
int result = (int)(Math.log(N) / Math.log(2));
return result;
}
// Function to return the number formed
// by inverting bits the bits of num
static int invertBits(int num)
{
// Number of bits in num
int x = log2(num) + 1;
// Inverting the bits one by one
for (int i = 0; i < x; i++)
num = (num ^ (1 << i));
return num;
}
// Function to return the total valid pairs
static int totalPairs(int arr1[], int arr2[], int n, int m)
{
// Set to store the elements of the arrays
HashSet s1 = new HashSet();
HashSet s2 = new HashSet();
// add all the elements of arr2[] in the set
for (int i = 0; i < m; i++)
s2.add(arr2[i]);
// Initialize count variable to 0
int count = 0;
for (int i = 0; i < n; i++)
{
// Check if element of the first array
// is not in the first set
if (!s1.contains(arr1[i]))
{
// Check if the element formed by inverting bits
// is in the second set
if (s2.contains(invertBits(arr1[i])))
{
// Increment the count of valid pairs and add
// the element in the first set so that
// it doesn't get counted again
count++;
s1.add(arr1[i]);
}
}
}
// Return the total number of pairs
return count;
}
// Driver code
public static void main(String[] args)
{
int arr1[] = { 43, 7, 1, 99 };
int arr2[] = { 5, 1, 28, 20 };
int n = arr1.length;
int m = arr2.length;
System.out.println(totalPairs(arr1, arr2, n, m));
}
}
// This code is contributed by SHUBHAMSINGH10 Python3
# Python3 implementation of the approach
from math import log2;
# Function to return the number formed
# by inverting bits the bits of num
def invertBits(num) :
# Number of bits in num
x = log2(num) + 1;
# Inverting the bits one by one
for i in range(int(x)) :
num = (num ^ (1 << i));
return num;
# Function to return the total valid pairs
def totalPairs(arr1, arr2, n, m) :
# Set to store the elements of the arrays
s1, s2 = set(), set();
# Insert all the elements of
# arr2[] in the set
for i in range(m) :
s2.add(arr2[i]);
# Initialize count variable to 0
count = 0;
for i in range(n) :
# Check if element of the first array
# is not in the first set
if arr1[i] not in s1 :
# Check if the element formed by
# inverting bits is in the second set
if invertBits(arr1[i]) in s2 :
# Increment the count of valid pairs
# and insert the element in the first
# set so that it doesn't get counted again
count += 1;
s1.add(arr1[i]);
# Return the total number of pairs
return count;
# Driver code
if __name__ == "__main__" :
arr1 = [ 43, 7, 1, 99 ];
arr2 = [ 5, 1, 28, 20 ];
n = len(arr1);
m = len(arr2);
print(totalPairs(arr1, arr2, n, m));
# This code is contributed by RyugaC#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
static int log2(int N)
{
// calculate log2 N indirectly
// using log() method
int result = (int)(Math.Log(N) / Math.Log(2));
return result;
}
// Function to return the number formed
// by inverting bits the bits of num
static int invertBits(int num)
{
// Number of bits in num
int x = log2(num) + 1;
// Inverting the bits one by one
for (int i = 0; i < x; i++)
num = (num ^ (1 << i));
return num;
}
// Function to return the total valid pairs
static int totalPairs(int []arr1, int []arr2, int n, int m)
{
// Set to store the elements of the arrays
HashSet s1 = new HashSet();
HashSet s2 = new HashSet();
// add all the elements of arr2[] in the set
for (int i = 0; i < m; i++)
s2.Add(arr2[i]);
// Initialize count variable to 0
int count = 0;
for (int i = 0; i < n; i++)
{
// Check if element of the first array
// is not in the first set
if (!s1.Contains(arr1[i]))
{
// Check if the element formed by inverting bits
// is in the second set
if (s2.Contains(invertBits(arr1[i])))
{
// Increment the count of valid pairs and add
// the element in the first set so that
// it doesn't get counted again
count++;
s1.Add(arr1[i]);
}
}
}
// Return the total number of pairs
return count;
}
// Driver code
public static void Main()
{
int []arr1 = { 43, 7, 1, 99 };
int []arr2 = { 5, 1, 28, 20 };
int n = arr1.Length;
int m = arr2.Length;
Console.Write(totalPairs(arr1, arr2, n, m));
}
}
// This code is contribute by chitranayal 输出：
2