Add 1 to a given number

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 

Method: Adding 1 to a given number by importing add function and without using +,- etc. 

Python3

# Python code
# add 1 to a given number without using +

# importing add function
import operator
#input
n=6
# adding 1 to a given number and print the output
# one line solution
print(operator.add(n,1))

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
// 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
// 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
// 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
# 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
// 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.
?>

Javascript

<script>
// JavaScript code to add
// one to a given number

function addOne( x) {
    let 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*/
document.write(addOne(13));
</script>

Output

14

Time Complexity: O(log₂n)

Auxiliary Space: O(1)

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 the 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.
?>

Javascript

<script>
// JavaScript program for the above approach
function addOne(x)
{
    return (-(~x));
}

// Driver Code
    document.write(addOne(13));

// This code is contributed by susmitakundugoaldanga.
</script>

Output

14

Time Complexity: O(1)

Auxiliary Space: O(1)

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. The same analogy holds for decrement. Note that this method works only if the numbers are stored in 2’s complement form.