Write a program to add one to a given number. The use of operators like ‘+’, ‘-‘, ‘*’, ‘/’, ‘++’, ‘–‘ …etc are not allowed.
Examples:
Input: 12
Output: 13
Input: 6
Output: 7
This question can be approached by using some bit magic. Following are different methods to achieve the same using bitwise operators.
Method 1
To add 1 to a number x (say 0011000111), flip all the bits after the rightmost 0 bit (we get 0011000000). Finally, flip the rightmost 0 bit also (we get 0011001000) to get the answer.
C++
#include <bits/stdc++.h>
using namespace std;
int addOne(int x)
{
int m = 1;
while( x & m )
{
x = x ^ m;
m <<= 1;
}
x = x ^ m;
return x;
}
int main()
{
cout<<addOne(13);
return 0;
}
|
C
#include <stdio.h>
int addOne(int x)
{
int m = 1;
while( x & m )
{
x = x ^ m;
m <<= 1;
}
x = x ^ m;
return x;
}
int main()
{
printf("%d", addOne(13));
getchar();
return 0;
}
|
Java
class GFG {
static int addOne(int x)
{
int m = 1;
while( (int)(x & m) >= 1)
{
x = x ^ m;
m <<= 1;
}
x = x ^ m;
return x;
}
public static void main(String[] args)
{
System.out.println(addOne(13));
}
}
|
Python3
def addOne(x) :
m = 1;
while(x & m):
x = x ^ m
m <<= 1
x = x ^ m
return x
n = 13
print addOne(n)
|
C#
using System;
class GFG {
static int addOne(int x)
{
int m = 1;
while( (int)(x & m) == 1)
{
x = x ^ m;
m <<= 1;
}
x = x ^ m;
return x;
}
public static void Main()
{
Console.WriteLine(addOne(13));
}
}
|
PHP
<?php
function addOne($x)
{
$m = 1;
while( $x & $m )
{
$x = $x ^ $m;
$m <<= 1;
}
$x = $x ^ $m;
return $x;
}
echo addOne(13);
?>
|
Output:
14
Method 2
We know that the negative number is represented in 2’s complement form on most of the architectures. We have the following lemma hold for 2’s complement representation of signed numbers.
Say, x is numerical value of a number, then
~x = -(x+1) [ ~ is for bitwise complement ]
(x + 1) is due to addition of 1 in 2’s complement conversion
To get (x + 1) apply negation once again. So, the final expression becomes (-(~x)).
C++
#include <bits/stdc++.h>
using namespace std;
int addOne(int x)
{
return (-(~x));
}
int main()
{
cout<<addOne(13);
return 0;
}
|
C
#include<stdio.h>
int addOne(int x)
{
return (-(~x));
}
int main()
{
printf("%d", addOne(13));
getchar();
return 0;
}
|
Java
class GFG
{
static int addOne(int x)
{
return (-(~x));
}
public static void main(String[] args)
{
System.out.printf("%d", addOne(13));
}
}
|
Python3
def addOne(x):
return (-(~x));
print(addOne(13))
|
C#
using System;
class GFG
{
static int addOne(int x)
{
return (-(~x));
}
public static void Main()
{
Console.WriteLine(addOne(13));
}
}
|
PHP
<?php
function addOne($x)
{
return (-(~$x));
}
echo addOne(13);
?>
|
Javascript
<script>
function addOne(x)
{
return (-(~x));
}
document.write(addOne(13));
</script>
|
Output:
14
Example :
Assume the machine word length is one *nibble* for simplicity.
And x = 2 (0010),
~x = ~2 = 1101 (13 numerical)
-~x = -1101
Interpreting bits 1101 in 2’s complement form yields numerical value as -(2^4 – 13) = -3. Applying ‘-‘ on the result leaves 3. Same analogy holds for decrement. Note that this method works only if the numbers are stored in 2’s complement form.
https://www.youtube.com/watch?v=wP5P2
-5n8iU
Thanks to Venki for suggesting this method.
Please write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem
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