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. 

#include <iostream>
#include <functional>
using namespace std;

int main() {
    // input
    int n = 6;

    // adding 1 to a given number and print the output
    // one line solution
    std::function<int(int)> addOne = [](int x) -> int { return x + 1; };
    cout << addOne(n) << endl;
    return 0;
}

import java.util.function.IntUnaryOperator;

public class AddOne {
  public static void main(String[] args) {
    // input
    int n = 6;

    // adding 1 to a given number and print the output
    // one line solution
    IntUnaryOperator addOne = x -> x + 1;
    System.out.println(addOne.applyAsInt(n));
  }
}

# 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))

using System;
// C# equivalent
public class AddOne
{
    public static void Main(string[] args)
    {
        // input
        int n = 6;

        // adding 1 to a given number and print the output
        // one line solution
        System.Func<int, int> addOne = x => x + 1;
        System.Console.WriteLine(addOne(n));
    }
}

// input
const n = 6;

// adding 1 to a given number and print the output
// one line solution
const addOne = (x) => x + 1;
console.log(addOne(n));

7

Time Complexity: O(1)
Auxiliary Space: O(1)

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++ 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++ 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 code to add 
// one to a given number 
import java.io.*;
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 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# 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 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.
?>

<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>

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)). 

#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

#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 code to Add 1 to a given number
import java.io.*;
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 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# 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 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.
?>

<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>

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.