int num; int div; int rem; // assume only positive numbers for(i=0; num>=0 ; num-=div) { i++; rem=num; } printf("\ni: %s%d rem:%d", i, rem );

I had to use recursive subtraction or addition. But, I think I took too long a time to figure out the fastest code and I was typing it out in google docs while the interviewer was on the phone with me. I was nervous. It was my first ever interview. But, it was a heck of a experience.

One way is to iteratively count the number of times Xn = Xn-1 - Y >= Y. No recursion needed. Also, if Y is a power of 2, you can use a right-shift to get the answer...even faster. If the number space is sufficiently small, you can use a lookup table.

int positionOfFirstAndOnlyBitSet(int m) { int pos = -1; int x = 0; for (; x > x) & 1) if (pos == -1) pos = x; else return -1; // found more than one bit } return pos; } int divide(int n, int m) { if (m == 1) return n; if (m == n) return 1; if (m > n) return 0; int pos = positionOfFirstAndOnlyBitSet(m); if (pos != -1) return n >> pos; // how manny times one can multiply m before going over n int x = 1; int mm = m; while (mm <= n) { mm += m; x++; } return x - 1; }

assuming you want to be able to handle doubles, I like the idea of x * pow(y,-1.0); ... why make the answer more difficult for yourself than it needs to be?

I got this question in facebook interview as well. You usually are not allowed to use floats, or pow(), or % And also you have to consider both +, - integers, so Steve M answer is not valid.

Binary search will work ? Let's say you need x/y. And only integral solution is needed i.e. 10/3 = 3 , not 3.33. This won't work if we need float as a result Pseudocode If x > 2) if mid*y == x : return mid if mid*y > x : hi = mid - 1 else: lo = mid + 1 return lo --------------------

Actually, it's quite simple. Recursively! function addTwo(x,y){ if (y === 0) { return x; } return addTwo( x ^ y, (x & y ) << 1); } addTwo(2,3) // 5 OR do it iteratively function addTwo(x, y) { // Iterate till there is no carry. while (y !== 0) { // Carry now contains common set bits of x and y. carry = x & y; // Sum of bits of x and y where at least one of the bits is not set. x = x ^ y; // Carry is shifted by one so that adding it to x gives the required sum. y = carry << 1; } return x; }

X * Y^(-1)

Victor, whats the complexity of your solution ?