Subtract 1 without arithmetic operators

Write a program to subtract one from a given number. The use of operators like ‘+’, ‘-‘, ‘*’, ‘/’, ‘++’, ‘–‘ …etc are not allowed. 

Examples:

Input:  12
Output: 11

Input:  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++

// 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

// 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

// 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.

Python3

# 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#

// 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
// 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.
?>

Javascript

<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>

Output

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. 

C++

#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)

C

#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)

Java

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

Python3

def subtractOne(x):

    return ((x << 1) + (~x));

# Driver code
print(subtractOne(13));

# This code is contributed by mits

C#

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

<?php
function subtractOne($x)
{
    return (($x << 1) + (~$x));
}

/* Driver code*/

print(subtractOne(13));

// This code has been contributed by mits
?>

Javascript

<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>

Output

12

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