Given an array arr[] consisting of N non-negative integers and an integer M, the task is to find the count of unordered pairs {X, Y} of array elements that satisfies the condition setBits(X ⊕ Y) + 2 * setBits(X & Y) = M, where ⊕ denotes the Bitwise XOR and & denotes the Bitwise AND.
Examples:
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}}.
Naive Approach: The simplest approach is to generate all possible pairs from the given array and for each pair, check if the necessary condition is satisfied or not. Increment the count for pairs satisfying the given conditions and finally, print the count of pairs.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find total count of // given pairs satisfying the equation int countPairs( int a[], int N, int M)
{ // Initialize the count to 0
int count = 0;
// Loop through the array
for ( int i = 0; i < N; i++) {
// Loop through the remaining
// elements of the array
for ( int j = i + 1; j < N; j++) {
// Calculate the XOR of
// the pair elements
int x = a[i] ^ a[j];
// Calculate the AND of
// the pair elements
int y = a[i] & a[j];
// Check if the condition is satisfied
// __builtin_popcount() returns no. of set bits
if (__builtin_popcount(x)
+ 2 * __builtin_popcount(y)
== M) {
// increment the count
count++;
}
}
}
// Return the count of pairs
// satisfying the condition
return count;
} // Driver code int main()
{ int arr[] = { 3, 0, 4, 5 };
int N = sizeof (arr) / sizeof (arr[0]);
int M = 2;
// Function Call
cout << countPairs(arr, N, M) << endl;
return 0;
} |
// Java code import java.io.*;
import java.util.*;
public class CountPairs {
// Function to find the total count of
// given pairs satisfying the equation
static int countPairs( int [] a, int N, int M) {
// Initialize the count to 0
int count = 0 ;
// Loop through the array
for ( int i = 0 ; i < N; i++) {
// Loop through the remaining
// elements of the array
for ( int j = i + 1 ; j < N; j++) {
// Calculate the XOR of the pair elements
int x = a[i] ^ a[j];
// Calculate the AND of the pair elements
int y = a[i] & a[j];
// Check if the condition is satisfied
// Integer.bitCount() returns the number of set bits
if (Integer.bitCount(x) + 2 * Integer.bitCount(y) == M) {
// Increment the count
count++;
}
}
}
// Return the count of pairs satisfying the condition
return count;
}
// Driver code
public static void main(String[] args) {
int arr[] = { 3 , 0 , 4 , 5 };
int N = arr.length;
int M = 2 ;
// Function Call
System.out.println(countPairs(arr, N, M));
}
} // This code is contributed by guptapratik |
# 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
# Function to find total count of # given pairs satisfying the equation def countPairs(a, N, M):
#Initialize the count to 0
count = 0
#Loop through the array
for i in range ( 0 ,N):
#Loop through the remaining
#elements of the array
for j in range (i + 1 ,N):
# Calculate the XOR of the pair elements
x = a[i] ^ a[j]
# Calculate the AND of the pair elements
y = a[i] & a[j]
if (countSetBits(x) + 2 * countSetBits(y)) = = M:
#increment the count
count + = 1
# Return the count of pairs satisfying the condition
return count
arr = [ 3 , 0 , 4 , 5 ]
N = len (arr)
M = 2
# Function call pairs = countPairs(arr, N, M)
print (pairs)
|
using System;
namespace CountPairs {
class Program {
// Function to find total count of
// given pairs satisfying the equation
static int CountPairs( int [] a, int N, int M)
{
// Initialize the count to 0
int count = 0;
// Loop through the array
for ( int i = 0; i < N; i++) {
// Loop through the remaining
// elements of the array
for ( int j = i + 1; j < N; j++) {
// Calculate the XOR of
// the pair elements
int x = a[i] ^ a[j];
// Calculate the AND of
// the pair elements
int y = a[i] & a[j];
// Check if the condition is satisfied
// CountSetBits() returns no. of set bits
if (CountSetBits(x) + 2 * CountSetBits(y)
== M) {
// increment the count
count++;
}
}
}
// Return the count of pairs
// satisfying the condition
return count;
}
// Function to count the number of set bits
static int CountSetBits( int num)
{
int count = 0;
while (num > 0) {
count += num & 1;
num >>= 1;
}
return count;
}
// Driver code
static void Main( string [] args)
{
int [] arr = { 3, 0, 4, 5 };
int N = arr.Length;
int M = 2;
// Function Call
Console.WriteLine(CountPairs(arr, N, M));
}
} } |
// Function to find total count of // given pairs satisfying the equation function countPairs(arr, M) {
// Initialize the count to 0
let count = 0;
// Loop through the array
for (let i = 0; i < arr.length; i++) {
// Loop through the remaining
// elements of the array
for (let j = i + 1; j < arr.length; j++) {
// Calculate the XOR of the pair elements
let x = arr[i] ^ arr[j];
// Calculate the AND of the pair elements
let y = arr[i] & arr[j];
// Check if the condition is satisfied
if (popcount(x) + 2 * popcount(y) === M) {
// increment the count
count++;
}
}
}
// Return the count of pairs satisfying the condition
return count;
} // Helper function to count the number of set bits in a number function popcount(num) {
let count = 0;
while (num > 0) {
count += num & 1;
num >>= 1;
}
return count;
} // Driver code let arr = [3, 0, 4, 5]; let M = 2; // Function Call console.log(countPairs(arr, M)); |
Output:
2
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized based on the following observations:
- 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.
Follow the steps below to solve the problem:
- First, traverse the array arr[].
- For every array element arr[i], update arr[i] with the count of set bits present in it.
- Now print the count of pairs from the array whose sum is equal to M.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> 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< int , int > 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]) {
// Decrement 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 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<Integer,
Integer> mp = new HashMap<Integer,
Integer>();
// 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])
{
// Decrement 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 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 code is contributed by santhoshcharan. |
// 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< int ,
int > mp = new Dictionary< int ,
int >();
// 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])
{
// Decrement 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 |
<script> // Javascript program for the above approach // Function to count number // of setbits in the number n function countsetbits(n)
{ // Stores the count of setbits
var 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 function countPairs(a, N, M)
{ for ( var 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
var mp = new Map();
// Traverse the array
for ( var i = 0; i < N; i++) {
// Increment the count
// of arr[i] in mp
if (mp.has(a[i]))
mp.set(a[i], mp.get(a[i])+1)
else
mp.set(a[i], 1)
}
// Stores the total count of pairs
var count = 0;
// Traverse the array arr[]
for ( var 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]) {
// Decrement count by 1
count--;
}
}
// Return count/2
return (count / 2);
} // Driver Code // Input var arr = [3, 0, 4, 5];
var N = arr.length;
var M = 2;
document.write( countPairs(arr, N, M)); </script> |
2
Time Complexity: O(NlogN)
Auxiliary Space: O(N) as using auxiliary space for unordered_map