Write a program to subtract one from a given number. The use of operators like ‘+’, ‘-‘, ‘*’, ‘/’, ‘++’, ‘–‘ …etc are not allowed.
Examples:
Input: 12
Output: 11Input: 6
Output: 5
Method 1
To subtract 1 from a number x (say 0011001000), flip all the bits after the rightmost 1 bit (we get 0011001111). Finally, flip the rightmost 1 bit also (we get 0011000111) to get the answer.
Steps to solve this problem:
1. declare a variable m=1.
2. while x AND m is not equal to 1:
*update x as x XOR m.
*m<<=1.
3. update x as x XOR m.
4. return x.
// C++ code to subtract // one from a given number #include <iostream> using namespace std;
int subtractOne( int x)
{ int m = 1;
// Flip all the set bits
// until we find a 1
while (!(x & m))
{
x = x ^ m;
m <<= 1;
}
// Flip the rightmost 1 bit
x = x ^ m;
return x;
} // Driver code int main()
{ cout << subtractOne(13) << endl;
return 0;
} // This code is contributed by noob2000 |
// C code to subtract // one from a given number #include <stdio.h> int subtractOne( int x)
{ int m = 1;
// Flip all the set bits
// until we find a 1
while (!(x & m)) {
x = x ^ m;
m <<= 1;
}
// flip the rightmost 1 bit
x = x ^ m;
return x;
} /* Driver program to test above functions*/ int main()
{ printf ( "%d" , subtractOne(13));
return 0;
} |
// Java code to subtract // one from a given number import java.io.*;
class GFG
{ static int subtractOne( int x)
{ int m = 1 ;
// Flip all the set bits
// until we find a 1
while (!((x & m) > 0 ))
{
x = x ^ m;
m <<= 1 ;
}
// flip the rightmost
// 1 bit
x = x ^ m;
return x;
} // Driver Code public static void main (String[] args)
{ System.out.println(subtractOne( 13 ));
} } // This code is contributed // by anuj_67. |
# Python 3 code to subtract one from # a given number def subtractOne(x):
m = 1
# Flip all the set bits
# until we find a 1
while ((x & m) = = False ):
x = x ^ m
m = m << 1
# flip the rightmost 1 bit
x = x ^ m
return x
# Driver Code if __name__ = = '__main__' :
print (subtractOne( 13 ))
# This code is contributed by # Surendra_Gangwar |
// C# code to subtract // one from a given number using System;
class GFG
{ static int subtractOne( int x)
{ int m = 1;
// Flip all the set bits
// until we find a 1
while (!((x & m) > 0))
{
x = x ^ m;
m <<= 1;
}
// flip the rightmost
// 1 bit
x = x ^ m;
return x;
} // Driver Code public static void Main ()
{ Console.WriteLine(subtractOne(13));
} } // This code is contributed // by anuj_67. |
<?php // PHP code to subtract // one from a given number function subtractOne( $x )
{ $m = 1;
// Flip all the set bits
// until we find a 1
while (!( $x & $m ))
{
$x = $x ^ $m ;
$m <<= 1;
}
// flip the
// rightmost 1 bit
$x = $x ^ $m ;
return $x ;
} // Driver Code echo subtractOne(13);
// This code is contributed // by anuj_67. ?> |
<script> // JavaScript code to subtract // one from a given number function subtractOne(x)
{ let m = 1;
// Flip all the set bits
// until we find a 1
while (!(x & m)) {
x = x ^ m;
m <<= 1;
}
// flip the rightmost 1 bit
x = x ^ m;
return x;
} /* Driver program to test above functions*/ document.write(subtractOne(13));
// This code is contributed by Surbhi Tyagi. </script> |
12
Time Complexity: O(log n), where n is the number of bits in x
Auxiliary Space: O(1)
Method 2 (If + is allowed)
We know that the negative number is represented in 2’s complement form on most of the architectures. We have the following lemma hold for 2’s complement representation of signed numbers.
Say, x is numerical value of a number, then
~x = -(x+1) [ ~ is for bitwise complement ]
Adding 2x on both the sides,
2x + ~x = x – 1
To obtain 2x, left shift x once.
#include <bits/stdc++.h> using namespace std;
int subtractOne( int x) { return ((x << 1) + (~x)); }
/* Driver program to test above functions*/ int main()
{ cout<< subtractOne(13);
return 0;
} // This code is contributed by Aditya Kumar (adityakumar129) |
#include <stdio.h> int subtractOne( int x) { return ((x << 1) + (~x)); }
/* Driver program to test above functions*/ int main()
{ printf ( "%d" , subtractOne(13));
return 0;
} // This code is contributed by Aditya Kumar (adityakumar129) |
import java.io.*;
class GFG
{ static int subtractOne( int x)
{
return ((x << 1 ) + (~x));
}
/* Driver code*/
public static void main(String[] args)
{
System.out.printf( "%d" , subtractOne( 13 ));
}
} // This code has been contributed by 29AjayKumar |
def subtractOne(x):
return ((x << 1 ) + (~x));
# Driver code print (subtractOne( 13 ));
# This code is contributed by mits |
using System;
class GFG
{ static int subtractOne( int x)
{
return ((x << 1) + (~x));
}
/* Driver code*/
public static void Main(String[] args)
{
Console.Write( "{0}" , subtractOne(13));
}
} // This code contributed by Rajput-Ji |
<?php function subtractOne( $x )
{ return (( $x << 1) + (~ $x ));
} /* Driver code*/ print (subtractOne(13));
// This code has been contributed by mits ?> |
<script> function subtractOne(x)
{ return ((x << 1) + (~x));
} /* Driver program to test above functions*/ document.write((subtractOne(13)));
// This is code is contributed by Mayank Tyagi </script> |
12
Time Complexity: O(1)
Auxiliary Space: O(1)