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.
// 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;
}
Output:
12
Method 2 (If + is allowed)
We know that the negative number is represented in 2’s complement form on most of the architectures. The idea is to use bitwise operators. Like addition, Write a function Add() that returns sum of two integers. The function should not use any of the arithmetic operators (+, ++, –, -, … etc). Sum of two bits can be obtained by performing XOR (^) of the two bits. Carry bit can be obtained by performing AND (&) of two bits. We can extend this logic for integers. If x and y don’t have set bits at same position(s), then bitwise XOR (^) of x and y gives the sum of x and y. To incorporate common set bits also, bitwise AND (&) is used. Bitwise AND of x and y gives all carry bits. We calculate (x & y) << 1 and add it to x ^ y to get the required result. The idea is to use subtractor logic. 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.
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));
`` }
}
Output:
12