Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.

Example:

Given a = 1 and b = 2, return 3.

**Java Solution**

Given two numbers a and b, a&b returns the number formed by ‘1’ bits on a and b. When it is left shifted by 1 bit, it is the carry.

For example, given a=101 and b=111 (in binary), the a&b=101. a&b << 1 = 1010.

a^b is the number formed by different bits of a and b. a&b=10.

public int getSum(int a, int b) { while(b!=0){ int c = a&b; a=a^b; b=c<<1; } return a; } |

Lol yeah hahahahahahaha

Ok, I understood how it works. If to use regular math, you need to break each of 2 numbers into a sum like (…32+16+8+4+2+1). For example, if we want to add 21 and 37, we break 21=(16+4+1); 33=(32+4+1).

Then we split these into 2 groups: group A duplicate components {+4 +1}; group B single components {+16 +32}.

We multiply duplicate components by 2 and keep single components as is, group A 5 * 2 = 10; group B 16 + 32 = 48.

Now instead of adding 21 and 37, we want to add 10 and 48, but now it is easier. We break these numbers again: 10=(8+2); 48=(32+16). No duplicate components, we stop the loop and just add these 2 numbers using ^: 10+48=58.

This works because operator & finds duplicate components (21 & 37 finds +4 and +1 in our case), operator <<1 multiplies them by 2, operator ^ adds 2 numbers if they don't have duplicate components (10 ^ 48 = 58 in our case).

a– is a = a -1

so you are still using + – operators

if we could use operator ++ and — then following solution would work

public static int sum(int a, int b) {

while (b != 0) {

if (b < 0) {

a--;

b++;

} else {

a++;

b--;

}

}

return a;

}