Given a positive integer n, find count of positive integers i such that 0 <= i <= n and n+i = n^i
Examples :
Input : n = 7 Output : 1 Explanation: 7^i = 7+i holds only for only for i = 0 7+0 = 7^0 = 7 Input: n = 12 Output: 4 12^i = 12+i hold only for i = 0, 1, 2, 3 for i=0, 12+0 = 12^0 = 12 for i=1, 12+1 = 12^1 = 13 for i=2, 12+2 = 12^2 = 14 for i=3, 12+3 = 12^3 = 15
Method 1 (Simple) :
One simple solution is to iterate over all values of i 0<= i <= n and count all satisfying values.
/* C++ program to print count of values such that n+i = n^i */
#include <iostream> using namespace std;
// function to count number of values less than // equal to n that satisfy the given condition int countValues ( int n)
{ int countV = 0;
// Traverse all numbers from 0 to n and
// increment result only when given condition
// is satisfied.
for ( int i=0; i<=n; i++ )
if ((n+i) == (n^i) )
countV++;
return countV;
} // Driver program int main()
{ int n = 12;
cout << countValues(n);
return 0;
} |
/* Java program to print count of values such that n+i = n^i */
import java.util.*;
class GFG {
// function to count number of values
// less than equal to n that satisfy
// the given condition
public static int countValues ( int n)
{
int countV = 0 ;
// Traverse all numbers from 0 to n
// and increment result only when
// given condition is satisfied.
for ( int i = 0 ; i <= n; i++ )
if ((n + i) == (n ^ i) )
countV++;
return countV;
}
/* Driver program to test above function */
public static void main(String[] args)
{
int n = 12 ;
System.out.println(countValues(n));
}
} // This code is contributed by Arnav Kr. Mandal. |
# Python3 program to print count # of values such that n+i = n^i # function to count number # of values less than # equal to n that satisfy # the given condition def countValues (n):
countV = 0 ;
# Traverse all numbers
# from 0 to n and
# increment result only
# when given condition
# is satisfied.
for i in range (n + 1 ):
if ((n + i) = = (n ^ i)):
countV + = 1 ;
return countV;
# Driver Code n = 12 ;
print (countValues(n));
# This code is contributed by mits |
/* C# program to print count of values such that n+i = n^i */ using System;
class GFG {
// function to count number of values
// less than equal to n that satisfy
// the given condition
public static int countValues ( int n)
{
int countV = 0;
// Traverse all numbers from 0 to n
// and increment result only when
// given condition is satisfied.
for ( int i = 0; i <= n; i++ )
if ((n + i) == (n ^ i) )
countV++;
return countV;
}
/* Driver program to test above function */
public static void Main()
{
int n = 12;
Console.WriteLine(countValues(n));
}
} // This code is contributed by anuj_67. |
<?php // PHP program to print count // of values such that n+i = n^i // function to count number // of values less than // equal to n that satisfy // the given condition function countValues ( $n )
{ $countV = 0;
// Traverse all numbers
// from 0 to n and
// increment result only
// when given condition
// is satisfied.
for ( $i = 0; $i <= $n ; $i ++ )
if (( $n + $i ) == ( $n ^ $i ) )
$countV ++;
return $countV ;
} // Driver Code
$n = 12;
echo countValues( $n );
// This code is contributed by m_kit ?> |
<script> /* JavaScript program to print count of values such that n+i = n^i */ // function to count number of values less than // equal to n that satisfy the given condition function countValues (n)
{ let countV = 0;
// Traverse all numbers from 0 to n and
// increment result only when given condition
// is satisfied.
for (let i=0; i<=n; i++ )
if ((n+i) == (n^i) )
countV++;
return countV;
} // Driver program let n = 12;
document.write(countValues(n));
// This code is contributed by Surbhi Tyagi. </script> |
Output:
4
Time Complexity: O(n)
Space Complexity: O(1)
Method 2 (Efficient) :
An efficient solution is as follows
we know that (n+i)=(n^i)+2*(n&i)
So n + i = n ^ i implies n & i = 0
Hence our problem reduces to finding values of i such that n & i = 0. How to find count of such pairs? We can use the count of unset-bits in the binary representation of n. For n & i to be zero, i must unset all set-bits of n. If the kth bit is set at a particular in n, kth bit in i must be 0 always, else kth bit of i can be 0 or 1
Hence, total such combinations are 2^(count of unset bits in n)
For example, consider n = 12 (Binary representation : 1 1 0 0).
All possible values of i that can unset all bits of n are 0 0 0/1 0/1 where 0/1 implies either 0 or 1. Number of such values of i are 2^2 = 4.
The following is the program following the above idea.
/* c++ program to print count of values such that n+i = n^i */
#include <bits/stdc++.h> using namespace std;
// function to count number of values less than // equal to n that satisfy the given condition int countValues( int n)
{ // unset_bits keeps track of count of un-set
// bits in binary representation of n
int unset_bits=0;
while (n)
{
if ((n & 1) == 0)
unset_bits++;
n=n>>1;
}
// Return 2 ^ unset_bits
return 1 << unset_bits;
} // Driver code int main()
{ int n = 12;
cout << countValues(n);
return 0;
} |
/* Java program to print count of values such that n+i = n^i */
import java.util.*;
class GFG {
// function to count number of values
// less than equal to n that satisfy
// the given condition
public static int countValues( int n)
{
// unset_bits keeps track of count
// of un-set bits in binary
// representation of n
int unset_bits= 0 ;
while (n > 0 )
{
if ((n & 1 ) == 0 )
unset_bits++;
n=n>> 1 ;
}
// Return 2 ^ unset_bits
return 1 << unset_bits;
}
/* Driver program to test above
function */
public static void main(String[] args)
{
int n = 12 ;
System.out.println(countValues(n));
}
} // This code is contributed by Arnav Kr. Mandal. |
/* C# program to print count of values such that n+i = n^i */
using System;
public class GFG {
// function to count number of values
// less than equal to n that satisfy
// the given condition
public static int countValues( int n)
{
// unset_bits keeps track of count
// of un-set bits in binary
// representation of n
int unset_bits=0;
while (n > 0)
{
if ((n & 1) == 0)
unset_bits++;
n=n>>1;
}
// Return 2 ^ unset_bits
return 1 << unset_bits;
}
/* Driver program to test above
function */
public static void Main(String[] args)
{
int n = 12;
Console.WriteLine(countValues(n));
}
} // This code is contributed by umadevi9616 |
# Python3 program to print count of values such # that n+i = n^i # function to count number of values less than # equal to n that satisfy the given condition def countValues(n):
# unset_bits keeps track of count of un-set
# bits in binary representation of n
unset_bits = 0
while (n):
if n & 1 = = 0 :
unset_bits + = 1
n = n >> 1
# Return 2 ^ unset_bits
return 1 << unset_bits
# Driver code if __name__ = = '__main__' :
n = 12
print (countValues(n))
# This code is contributed by rutvik |
<?php /* PHP program to print count of values such that n+i = n^i */ // function to count number of // values less than equal to n // that satisfy the given // condition function countValues( $n )
{ // unset_bits keeps track
// of count of un-set bits
// in binary representation
// of n
$unset_bits = 0;
while ( $n )
{
if (( $n & 1) == 0)
$unset_bits ++;
$n = $n >> 1;
}
// Return 2 ^ unset_bits
return 1 << $unset_bits ;
} // Driver code $n = 12;
echo countValues( $n );
// This code is contributed // by Anuj_67. ?> |
<script> // Javascript program to print count of values // such that n+i = n^i // Function to count number of values // less than equal to n that satisfy // the given condition function countValues(n)
{ // unset_bits keeps track of count
// of un-set bits in binary
// representation of n
let unset_bits = 0;
while (n > 0)
{
if ((n & 1) == 0)
unset_bits++;
n = n >> 1;
}
// Return 2 ^ unset_bits
return 1 << unset_bits;
} // Driver Code let n = 12; document.write(countValues(n)); // This code is contributed by susmitakundugoaldanga </script> |
Output :
4
Time Complexity: O(log(n))
Space Complexity: O(1)