Impletement multiply, subtract and divide using add operation.

/**

2. Mulitply:

For loop.

1. Subtract:

negate second value and add.

0. Negate: for negative numbers, add 1 to the number until we reach 0 and return count.

for positive numbers, we add -1 to te number until we reach 0.

Optimization-instead of adding 1/-1 (delta), increameant by delta and if adding delta changes the sign, reset delta.

This is O((log a)^2). Longest round would be log a and log a iterations

More optimization: instead of resetting, use a stack to store deltas so far and pop out the last delta and check signs. this way, we don't need to go all the way back to 1/-1. O(log a)

3. Divide:

For loop. Use Multiply function.

if x=a/b => a = bx; find x such that diff between a and bx is minimum.

*/

public class Main {

public static void main(String[] args) {

System.out.println(divide(4,0));

}

public static int negate(int a)

{

int neg = 0;

int newSign = a>0?-1:1; //if positive we add -1 to reach 0.

while(a!=0)

{

neg = neg + newSign;

a = a + newSign;

}

return neg;

}

public static int subtract(int a, int b)

{

return a + negate(b);

}

public static int multiply(int a, int b)

{

int sign = 1;

int sum = 0;

if(a<0 && b<0) { a = negate(a); b = negate(b);}

else if(b<0) { sign = -1; b = negate(b);}

else if(a<0) { sign = -1; a = negate(a);}

while(b!=0)

{

sum+=a;

b = subtract(b,1);

}

if(sign == -1)

{

sum = negate(sum);

}

return sum;

}

public static int divide(int a, int b)

{

if(b==0) return Integer.MIN_VALUE;

int sign = 1;

if(a<0) { sign = -1; a = negate(a);}

if(b<0) { sign = -1; b = negate(b);}

if(a<0 && b<0) { sign = -1; a = negate(a); b = negate(b);}

int div = 1;

int diff = Integer.MAX_VALUE;

int curDiff = Integer.MAX_VALUE;

while(curDiff >= diff)

{

curDiff = subtract(a, multiply(b,div));

if(curDiff>=0 && curDiff<diff)

{

diff = curDiff;

div += 1;

}

}

div = div+(-1);

if(sign == -1)

{

div = negate(div);

}

return div;

}

}