Given an integer N. The task is to count the numbers of pairs of integers A and B such that A + B + N = A ^ B ^ N and A and B are less than N.
Examples:
Input: N = 2
Output: 3
Explanation:-
For N = 2
2 XOR 0 XOR 0 = 2+0+0
2 XOR 0 XOR 1 = 2+0+1
2 XOR 0 XOR 2 != 2+0+2
2 XOR 1 XOR 0 = 2+1+0
2 XOR 1 XOR 1 != 2+1+1
2 XOR 1 XOR 2 != 2+1+2
2 XOR 2 XOR 0 != 2+2+0
2 XOR 2 XOR 1 != 2+2+1
2 XOR 2 XOR 2 != 2+2+2
So (0, 0), (0, 1) and (1, 0) are the required pairs. So the output is 3.
Input: N = 4
Output: 9
Approach :
To make the sum of three numbers equal to the xor of three number with one of the number given we can do following:-
- Represent the fixed number in binary form.
- Traverse the binary expansion of the fixed number.
- If you find a 1 there is only one condition i.e. you take the other two number’s binary bits as 0 and 0.
- If you find a 0 there will be three conditions i.e. either you can have binary bits as (0, 0), (1, 0)
or (0, 1).
- The following triplets of bits will never go for a carry so they are valid.
- So the answer will be 3^(number of zeros in binary representation).
Below is the implementation of the above approach:
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std;
// Defining ull to unsigned long long int typedef unsigned long long int ull;
// Function to calculate power of 3 ull calculate( int bit_cnt)
{ ull res = 1;
while (bit_cnt--) {
res = res * 3;
}
return res;
} // Function to return the count of the // unset bit ( zeros ) int unset_bit_count(ull n)
{ int count = 0;
while (n) {
// Check the bit is 0 or not
if ((n & 1) == 0)
count++;
// Right shifting ( dividing by 2 )
n = n >> 1;
}
return count;
} // Driver Code int main()
{ ull n;
n = 2;
int count = unset_bit_count(n);
ull ans = calculate(count);
cout << ans << endl;
return 0;
} |
// Java implementation of the approach import java.util.*;
class GFG
{ // Function to calculate power of 3 static long calculate( int bit_cnt)
{ long res = 1 ;
while (bit_cnt-- > 0 )
{
res = res * 3 ;
}
return res;
} // Function to return the count of the // unset bit ( zeros ) static int unset_bit_count( long n)
{ int count = 0 ;
while (n > 0 )
{
// Check the bit is 0 or not
if ((n & 1 ) == 0 )
count++;
// Right shifting ( dividing by 2 )
n = n >> 1 ;
}
return count;
} // Driver Code public static void main(String[] args)
{ long n;
n = 2 ;
int count = unset_bit_count(n);
long ans = calculate(count);
System.out.println(ans);
} } // This code is contributed by PrinciRaj1992 |
# Python3 implementation of the approach # Function to calculate power of 3 def calculate(bit_cnt):
res = 1 ;
while (bit_cnt > 0 ):
bit_cnt - = 1 ;
res = res * 3 ;
return res;
# Function to return the count of the # unset bit ( zeros ) def unset_bit_count(n):
count = 0 ;
while (n > 0 ):
# Check the bit is 0 or not
if ((n & 1 ) = = 0 ):
count + = 1 ;
# Right shifting ( dividing by 2 )
n = n >> 1 ;
return count;
# Driver Code if __name__ = = '__main__' :
n = 2 ;
count = unset_bit_count(n);
ans = calculate(count);
print (ans);
# This code contributed by Rajput-Ji |
// C# implementation of the approach using System;
class GFG
{ // Function to calculate power of 3 static long calculate( int bit_cnt)
{ long res = 1;
while (bit_cnt-- > 0)
{
res = res * 3;
}
return res;
} // Function to return the count of the // unset bit (zeros) static int unset_bit_count( long n)
{ int count = 0;
while (n > 0)
{
// Check the bit is 0 or not
if ((n & 1) == 0)
count++;
// Right shifting ( dividing by 2 )
n = n >> 1;
}
return count;
} // Driver Code public static void Main(String[] args)
{ long n;
n = 2;
int count = unset_bit_count(n);
long ans = calculate(count);
Console.WriteLine(ans);
} } // This code is contributed by 29AjayKumar |
<script> // Javascript implementation of // the above approach // Function to calculate power of 3 function calculate(bit_cnt)
{ let res = 1;
while (bit_cnt--) {
res = res * 3;
}
return res;
} // Function to return the count of the // unset bit ( zeros ) function unset_bit_count(n)
{ let count = 0;
while (n) {
// Check the bit is 0 or not
if ((n & 1) == 0)
count++;
// Right shifting ( dividing by 2 )
n = n >> 1;
}
return count;
} // Driver Code let n;
n = 2;
let count = unset_bit_count(n);
let ans = calculate(count);
document.write(ans);
</script> |
3
Time Complexity: O(Number of unset_bits).
Auxiliary Space: O(1).