Sum of two numbers without using arithmetic operators Geeks For Geeks

Problem Link:

https://practice.geeksforgeeks.org/problems/sum-of-two-numbers-without-using-arithmetic-operators/1#

(It is also the problem of the day for 22-04-2022) 

Solution:

class Solution

{

    int sum(int a, int b)

    {

        return ((a|b)+(a&b));

    }

}

Total Test Cases:310

Time Complexity: O(max number of bits(number 1, number 2))[As we will perform operations and for which we require all the bits.] [GFG Time:0.11/1.13]

Space Complexity: O(1)

Auxiliary Space: O(1)

Approach Used:

Here we will do two things.

1. Take or of two numbers

2. Take and of two numbers

eg if the number is 20 and the other is 40 then

20 = 10100 (5 bits) -> 010100 to convert it into 6 bits as another number is also 6 bits, so and operation will be performed easily and correctly.

40= 101000 (6 bits)

Now taking and of two numbers 

and=010100 & 101000 -> 000000 ie 0

Now doing or operation we get

or=010100 | 101000 -> 111100 ie 60

Now and+or = 0+60 =60 =20+40

Hence we will get correct answer.

If we try to do it with a simple plus(+) operator then

class Solution

{

    int sum(int a , int b)

    {

        return a+b;

        //return ((a|b)+(a&b));

    }

}

Time Complexity:O(1)[GFG Time:0.14/1.13]

Its time will be more than the previous approach because (+)works on numbers while the bitwise operators like & or | works on bits and bits are faster to compute in comparison with one single whole number.

It is the reason the time taken in the previous approach is less in comparison to this approach.

 

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