Add 1 to a given number

Difficulty Level : Medium
Last Updated : 19 Aug, 2019

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++

// C++ code to add add  
// one to a given number  
#include <bits/stdc++.h> 
using namespace std; 
  
int addOne(int x)  
{  
    int m = 1;  
      
    // Flip all the set bits  
    // until we find a 0  
    while( x & m )  
    {  
        x = x ^ m;  
        m <<= 1;  
    }  
      
    // flip the rightmost 0 bit  
    x = x ^ m;  
    return x;  
}  
  
/* Driver program to test above functions*/
int main()  
{  
    cout<<addOne(13);  
    return 0;  
}  
  
// This code is contributed by rathbhupendra 

C

// C++ code to add add 
// one to a given number  
#include <stdio.h> 
  
int addOne(int x) 
{ 
    int m = 1; 
      
    // Flip all the set bits  
    // until we find a 0  
    while( x & m ) 
    { 
        x = x ^ m; 
        m <<= 1; 
    } 
      
    // flip the rightmost 0 bit  
    x = x ^ m; 
    return x; 
} 
  
/* Driver program to test above functions*/
int main() 
{ 
    printf("%d", addOne(13)); 
    getchar(); 
    return 0; 
} 

Java

// Java code to add add 
// one to a given number  
class GFG { 
  
    static int addOne(int x) 
    { 
        int m = 1; 
          
        // Flip all the set bits  
        // until we find a 0  
        while( (int)(x & m) >= 1) 
        { 
            x = x ^ m; 
            m <<= 1; 
        } 
      
        // flip the rightmost 0 bit  
        x = x ^ m; 
        return x; 
    } 
      
    /* Driver program to test above functions*/
    public static void main(String[] args) 
    { 
        System.out.println(addOne(13)); 
    } 
} 
  
// This code is contributed by prerna saini. 

Python3

# Python3 code to add 1 
# one to a given number  
def addOne(x) : 
      
    m = 1; 
    # Flip all the set bits 
    # until we find a 0  
    while(x & m): 
        x = x ^ m 
        m <<= 1
      
    # flip the rightmost  
    # 0 bit  
    x = x ^ m 
    return x 
  
# Driver program 
n = 13
print addOne(n) 
  
# This code is contributed by Prerna Saini. 

C#

// C# code to add one 
// to a given number  
using System; 
  
class GFG { 
  
    static int addOne(int x) 
    { 
        int m = 1; 
          
        // Flip all the set bits  
        // until we find a 0  
        while( (int)(x & m) == 1) 
        { 
            x = x ^ m; 
            m <<= 1; 
        } 
      
        // flip the rightmost 0 bit  
        x = x ^ m; 
        return x; 
    } 
      
    // Driver code 
    public static void Main() 
    { 
        Console.WriteLine(addOne(13)); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP code to add add 
// one to a given number  
  
  
function addOne($x) 
{ 
    $m = 1; 
      
    // Flip all the set bits  
    // until we find a 0  
    while( $x & $m ) 
    { 
        $x = $x ^ $m; 
        $m <<= 1; 
    } 
      
    // flip the rightmost 0 bit  
    $x = $x ^ $m; 
    return $x; 
} 
  
// Driver Code 
echo addOne(13); 
  
// This code is contributed by vt_m. 
?> 

Output:

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));  
}  
  
/* Driver code*/
int main()  
{  
    cout<<addOne(13);  
    return 0;  
}  
  
  
// This code is contributed by rathbhupendra 

C

#include<stdio.h> 
  
int addOne(int x) 
{ 
   return (-(~x)); 
} 
  
/* Driver program to test above functions*/
int main() 
{ 
  printf("%d", addOne(13)); 
  getchar(); 
  return 0; 
} 

Java

// Java code to Add 1 to a given number 
class GFG 
{ 
    static int addOne(int x) 
    { 
         return (-(~x)); 
    } 
      
    // Driver program 
    public static void main(String[] args) 
    { 
        System.out.printf("%d", addOne(13)); 
    } 
} 
  
// This code is contributed  
// by Smitha Dinesh Semwal 

Python3

# Python3 code to add 1 to a given number 
  
def addOne(x): 
    return (-(~x)); 
  
  
# Driver program  
print(addOne(13)) 
  
# This code is contributed by Smitha Dinesh Semwal 

C#

// C# code to Add 1  
// to a given number 
using System; 
  
class GFG 
{ 
    static int addOne(int x) 
    { 
        return (-(~x)); 
    } 
      
    // Driver program 
    public static void Main() 
    { 
        Console.WriteLine(addOne(13)); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP Code to Add 1  
// to a given number 
  
function addOne($x) 
{ 
return (-(~$x)); 
} 
  
// Driver Code 
echo addOne(13); 
  
// This code is contributed by vt_m. 
?> 

Output:

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.

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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