LeetCode – Divide Two Integers (Java)

Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT.

Analysis

This problem can be solved based on the fact that any number can be converted to the format of the following:

num=a_0*2^0+a_1*2^1+a_2*2^2+...+a_n*2^n

The time complexity is O(logn).

Java Solution

public int divide(int dividend, int divisor) {
    //handle special cases
    if(divisor==0) return Integer.MAX_VALUE;
    if(divisor==-1 && dividend == Integer.MIN_VALUE)
        return Integer.MAX_VALUE;
 
    //get positive values
    long pDividend = Math.abs((long)dividend);
    long pDivisor = Math.abs((long)divisor);
 
    int result = 0;
    while(pDividend>=pDivisor){
        //calculate number of left shifts
        int numShift = 0;    
        while(pDividend>=(pDivisor<<numShift)){
            numShift++;
        }
 
        //dividend minus the largest shifted divisor
        result += 1<<(numShift-1);
        pDividend -= (pDivisor<<(numShift-1));
    }
 
    if((dividend>0 && divisor>0) || (dividend<0 && divisor<0)){
        return result;
    }else{
        return -result;
    }
}

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Sai Nikhil

I think he meant TLE (Time Limit Exceeded)
Tanmay Misra

You are using multiplication in the solution.
Ray

Why the complexity would be O(logn), can you explain more about it? Thank you.
mdlamar

Only problem with this approach is when you get something like 100000000/1 it is really slow cause it has to loop that many times.
Saurabh Rane

This code takes care of everything-

public static int divide(int dividend, int divisor) {
if(divisor == 0) {
return Integer.MAX_VALUE;
} else if(dividend == 0) {
return 0;
}
int sign = 1;
if(divisor < 0) {
sign *= -1;
divisor *= -1;
}
if(divisor == 1) {
return sign*dividend;
}
if(dividend < 0) {
sign *= -1;
dividend *= -1;
}

int sum = 0, count = -1;
while(sum <= dividend) {
sum += divisor;
count++;
}
return count*sign;
}
dev

if dividend is 0 then ? i think a cross check on dividend will be better
Juan Carlos Alvarez

What is LTE?. I like Saurabh solution better, it’s more elegant.
Candis

I tried the same code using ‘int’ for absolue values of dividend and divisor. How come that is giving me ‘Timeout; error??
nelson

The solution would lead to LTE; nevertheless, the code is really concise
Saurabh Rane

Thank you for this code, but I think this can be solved using sum as follows –

public int divide(int dividend, int divisor) {

int sum = 0, count = -1;

while(sum <= dividend) {

sum += divisor;

count++;

}

return count;

}

LeetCode – Divide Two Integers (Java)

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