给定由N个非负整数和整数M组成的数组arr [] ，任务是查找满足条件setBits(X⊕Y)+ 2 * setBits的数组元素的无序对{X，Y}的数量(X＆Y)= M ，其中⊕表示按位XOR， ＆表示按位AND。
例子：
Input: arr[] = {30, 0, 4, 5 }, M = 2
Output: 2
Explanation: The pairs satisfying the necessary conditions are {{3, 0}, {0, 5}}.
Input: arr[] = {1, 2, 3, 4}, M = 3
Output: 3
Explanation: The pairs satisfying the necessary conditions are {{1, 3}, {2, 3}, {3, 4}}.
天真的方法：最简单的方法是从给定的数组生成所有可能的对，并针对每个对检查是否满足必要条件。增加满足给定条件的对的计数，最后打印对的计数。
时间复杂度： O(N 2 )
辅助空间： O(1)
高效的方法：可以基于以下观察来优化上述方法：
- From the property of Bitwise XOR:
- setBits( a⊕ b ) = setBits( a|b ) – setBits( a&b )
- setBits( a|b ) = setBits(a) + setBits(b) – setBits( a&b )
- Therefore, the given equation becomes:
- ( setBits( X|Y ) – setBits( X&Y ) )+( 2 × setBits( X&Y ) ) = M
- setBits( X ) + setBits ( Y ) – setBits( X&Y ) – setBits( X&Y ) + ( 2 × setBits ( X&Y ) ) = M
- setBits( X ) + setBits( Y ) = M
- Therefore, the task reduces to counting the pairs of elements whose sum of set bits is equal to M.
请按照以下步骤解决问题：
- 首先，遍历数组arr []。
- 对于每个数组元素arr [i]，使用其中存在的设置位数对arr [i]进行更新。
- 现在从总和等于M的数组中打印对数。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to count number
// of setbits in the number n
int countsetbits(int n)
{
// Stores the count of setbits
int count = 0;
// Iterate while N is
// greater than 0
while (n) {
// Increment count by 1
// if N is odd
count += n & 1;
// Right shift N
n >>= 1;
}
// Return the count of set bits
return (count);
}
// Function to find total count of
// given pairs satisfying the equation
int countPairs(int a[], int N, int M)
{
for (int i = 0; i < N; i++) {
// Update arr[i] with the count
// of set bits of arr[i]
a[i] = countsetbits(a[i]);
}
// Stores the frequency
// of each array element
unordered_map mp;
// Traverse the array
for (int i = 0; i < N; i++) {
// Increment the count
// of arr[i] in mp
mp[a[i]]++;
}
// Stores the total count of pairs
int count = 0;
// Traverse the array arr[]
for (int i = 0; i < N; i++) {
// Increment count by mp[M - a[i]]
count += mp[M - a[i]];
// If a[i] is equal to M-a[i]
if (a[i] == M - a[i]) {
// Decrment count by 1
count--;
}
}
// Return count/2
return (count / 2);
}
// Driver Code
int main()
{
// Input
int arr[] = { 3, 0, 4, 5 };
int N = sizeof(arr) / sizeof(arr[0]);
int M = 2;
cout << countPairs(arr, N, M);
} Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG{
// Function to count number
// of setbits in the number n
static int countsetbits(int n)
{
// Stores the count of setbits
int count = 0;
// Iterate while N is
// greater than 0
while (n != 0)
{
// Increment count by 1
// if N is odd
count += n & 1;
// Right shift N
n >>= 1;
}
// Return the count of set bits
return (count);
}
// Function to find total count of
// given pairs satisfying the equation
static int countPairs(int[] a, int N, int M)
{
for(int i = 0; i < N; i++)
{
// Update arr[i] with the count
// of set bits of arr[i]
a[i] = countsetbits(a[i]);
}
// Stores the frequency
// of each array element
HashMap mp = new HashMap();
// Traverse the array
for (int i = 0; i < N; i++)
{
if (mp.containsKey(a[i]))
{
mp.put(a[i], mp.get(a[i]) + 1);
}
else
{
mp.put(a[i], 1);
}
}
// Stores the total count of pairs
int count = 0;
// Traverse the array arr[]
for(int i = 0; i < N; i++)
{
// Increment count by mp[M - a[i]]
count += mp.get(M - a[i]);
// If a[i] is equal to M-a[i]
if (a[i] == M - a[i])
{
// Decrment count by 1
count--;
}
}
// Return count/2
return (count / 2);
}
// Driver Code
public static void main(String[] args)
{
// Input
int[] arr = { 3, 0, 4, 5 };
int N = arr.length;
int M = 2;
System.out.println(countPairs(arr, N, M));
}
}
// This code is contributed by avijitmondal1998 Python3
# Python3 Program for the above approach
# Function to count number
# of setbits in the number n
def countSetBits(n):
# Stores the count of setbits
count = 0
# Iterate while N is
# greater than 0
while (n):
# Increment count by 1
# if N is odd
count += n & 1
# Right shift N
n >>= 1
# Return the count of set bits
return count
def countPairs(arr, N, M):
for i in range(0, N):
# Update arr[i] with the count
# of set bits of arr[i]
arr[i] = countSetBits(arr[i])
# Store counts of all elements in a dictionary
mp = {}
for i in range(0, N):
if arr[i] in mp:
mp[arr[i]] += 1
else:
mp[arr[i]] = 1
twice_count = 0
# Iterate through each element and increment
# the count (Notice that every pair is
# counted twice)
for i in range(0, N):
if M - arr[i] in mp.keys():
twice_count += mp[M - arr[i]]
# if (arr[i], arr[i]) pair satisfies the
# condition, then we need to ensure that
# the count is decreased by one such
# that the (arr[i], arr[i]) pair is not
# considered
if (M - arr[i] == arr[i]):
twice_count -= 1
# return the half of twice_count
return int(twice_count / 2)
# Driver code
# Input
arr = [ 3, 0, 4, 5]
N = len(arr)
M = 2
print(countPairs(arr, N, M))
# This cose is contributed by santhoshcharan.C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to count number
// of setbits in the number n
static int countsetbits(int n)
{
// Stores the count of setbits
int count = 0;
// Iterate while N is
// greater than 0
while (n != 0)
{
// Increment count by 1
// if N is odd
count += n & 1;
// Right shift N
n >>= 1;
}
// Return the count of set bits
return (count);
}
// Function to find total count of
// given pairs satisfying the equation
static int countPairs(int[] a, int N, int M)
{
for(int i = 0; i < N; i++)
{
// Update arr[i] with the count
// of set bits of arr[i]
a[i] = countsetbits(a[i]);
}
// Stores the frequency
// of each array element
Dictionary mp = new Dictionary();
// Traverse the array
for(int i = 0; i < N; ++i)
{
// Update frequency of
// each array element
if (mp.ContainsKey(a[i]) == true)
mp[a[i]] += 1;
else
mp[a[i]] = 1;
}
// Stores the total count of pairs
int count = 0;
// Traverse the array arr[]
for(int i = 0; i < N; i++)
{
// Increment count by mp[M - a[i]]
count += mp[M - a[i]];
// If a[i] is equal to M-a[i]
if (a[i] == M - a[i])
{
// Decrment count by 1
count--;
}
}
// Return count/2
return (count / 2);
}
// Driver Code
static public void Main()
{
// Input
int[] arr = { 3, 0, 4, 5 };
int N = arr.Length;
int M = 2;
Console.WriteLine(countPairs(arr, N, M));
}
}
// This code is contributed by sanjoy_62 输出：
2时间复杂度： O(N)
辅助空间： O(1)