If it’s been asked as an interview question at Amazon, LinkedIn, Facebook, Microsoft, AND Apple, you know it’s got to be a good one! Have you solved the challenge productExceptSelf in Interview Practice yet? If not, go give it a shot. Once you’re done, head back here. I’ll walk you through a naive solution, a better solution, and even a few ways to optimize.
…Done? Okay, let’s get into it!
The object of this problem is to calculate the value of a somewhat contrived function. The function
productExceptSelf is given two inputs, an array of numbers
nums and a modulus
m. It should return the sum of all N terms
f(nums, i) modulo
f(nums,i) = nums * nums * .... * nums[i-1] * nums[i+1] * ... * nums[N-1]
We can see this most easily with an example. To calculate
productExceptSelf([1,2,3,4],12) we would calculate:
f([1,2,3,4], 0 ) = 2*3*4 = 24
f([1,2,3,4], 1 ) = 1*3*4 = 12
f([1,2,3,4], 2 ) = 1*2*4 = 8
f([1,2,3,4], 3 ) = 1*2*3 = 6
The sum of all these numbers is
50, so we should return
50 % 12 = 2.
A naive solution
The explanation of the code suggests an implementation:
# Don't use this function! def f(nums,i): ans = 1 for index, n in enumerate(nums): if index != i: ans *= n return ans def productExceptSelf(nums, m): # Add up all the results of the f(nums, i), modulo m return sum([f(nums,i) % m for i in range(len(nums))]) % m
This is technically correct, but the execution time is bad. Each call to the function
f(nums, i) has to do a scan (and multiplication) in the array, so we know the function is
O(N). We call
f(nums,i) a total of
N times, so this function is
Sure enough, this function passes all the test cases. But it gives us a time length execution error on test case #16, so we have to find a more efficient solution.
Division is a better solution (but still not good enough)
A different way of approaching this problem is to find the product of all the numbers, and then divide by the one you are leaving out. We would have to scan to see if any of the numbers were zero first, as we can run into trouble dividing by zero. Essentially, we’d have to deal with that case separately, but it turns out that any array
nums with a zero in it is easy to calculate. (This would be a good extension exercise!)
If we look under the constraints of the code, we are told that
1 <= nums[i], so we don’t have to worry about this case. We can simplify our problem to:
# This version still doesn't run fast enough... # but why?? def productExceptSelf(nums, m): productAll = 1 for n in nums: productAll *= n # these are the f(nums,i) we calculated before f_i = [productAll / n for n in nums] return sum(f_i) % m
Again, we get a time execution error! Note that the running time is much better. We make a pass through the array once to get
productAll, then a pass through the array again to get the
f_i, and one more pass through the array to do the sum. That makes this is a O(N) solution!
Why is the interviewer asking this question?
In other words, what is this question testing? As I mentioned in the introduction, the function we’re calculating is a little contrived. Because it doesn’t seem to have any immediate applicability, the companies asking us this question in interviews are probably looking to see if we know a particular technique or trick.
One of the assumptions that I made when calling the algorithms
) was that multiplication was a constant time operation. This is a reasonable assumption for small numbers, but even for a computer there is a significant difference between calculating
456 x 434
324834750321321355120958 x 934274724556120
There are a couple of math properties of residues (the technical name for the “remainders” the moduli give us) that we can use. One is:
(a + b + c ) % m is the same as
(a % m + b % m + c % m) % m
This is nice because
c%m are all small numbers, so adding them is fast.
The other property is:
(a * b) % m is the same as
((a % m) * (b % m)) % m
That is, I can multiply the remainders of
b after division by
m, and the result I get will have the correct remainder.
At first glance, this doesn’t seem to be saving us much time because we’re doing a lot more operations. We are taking the modulus three times per multiplication, instead of just once! But it turns out that the modulus operation is fast. We more than make up for it by only multiplying small numbers.
So we can change our calculation of
# same as before f_i = [(productAll / n) % m for n in nums] return sum(f_i) % m
This still isn’t good enough to pass the test, but we’re getting there. The problems we still have are:
- The number
productAllis still very large
- Integer division is (relatively) slow
Our next approach will eliminate both of these problems.
Prefix products (aka cumulative products)
We can speed up the execution by building by an array,
prefixProduct, so that
prefixProduct[i] contains the product of the first
i-1 numbers in
nums. We will leave
prefixProduct = 1.
... ... prefixProduct =  * len(nums) for i in range(1,len(nums)): prefixProduct[i] = prefixProduct[i-1]*nums[i-1] ...
The neat thing about this array is that
prefixProduct[i] contains the product of all elements of the array up to
i, not including
i. If we also made a
suffixProduct such that
suffixProduct[i] was equal to all the product of all numbers in
nums past index position
i, then the productExceptSelf for number
i would just be the product of all numbers except the ith one =
prefixProduct[i] * suffixProduct[i]
We have eliminated one of the costly operations: division! We can also avoid seeing large numbers in the multiplication as well, by changing the step inside the loop to contain a modulus.
Our new solution is:
def productExceptSelf(nums, m): prefixProduct = *len(nums) suffixProduct = *len(nums) # setup the cumulative product from left and right for i in range(1,len(nums)): # Need parenthesis, as % has higher precedence than * prefixProduct[i] = (prefixProduct[i-1] * nums[i-1]) % m suffixProduct[-i-1] = (suffixProduct[-i] * nums[-i]) % m total = 0 for i in range(len(nums)): # start at the end, with prefixProduct -1 # and scan right total += (prefixProduct[i]*suffixProduct[i]) % m return total % m
This finally works! We’ve eliminated all multiplication by big numbers (but still have multiplications by small numbers), and no divisions at all. But we can still do better…
For the technical interview, an even better solution
It turns out that we don’t need to have a
suffixProduct. We can build it as we go! This is the accumulator pattern:
def productExceptSelf(nums, m): prefixProduct = *len(nums) suffixProduct = 1 # now this is just a number # setup the cumulative product from left and right for i in range(1,len(nums)): # Need parenthesis, as % has higher precedence than * prefixProduct[i] = (prefixProduct[i-1] * nums[i-1]) % m total = 0 for i in range(len(nums)): # start at the end, with prefixProduct -1 # and scan right total += (prefixProduct[-1 - i]*suffixProduct) % m suffixProduct = (suffixProduct * nums[-1-i]) % m # now multiply suffixProduct by the number that # was excluded return total % m
The main things you’re being asked to think about in this task are:
- Arithmetic operations aren’t always constant time. Multiplying big numbers is much slower than multiplying small numbers.
- Operations are not all the same speed. Integer modulus is very fast, addition and multiplication are fast, while division is (relatively) slow.
- Some number theory: You can multiply the residues of numbers, instead of the numbers themselves. But you cannot divide by residues, or divide the residues unless you have certain guarantees about divisibility.
- The idea of precomputing certain operations, which is where the
Other problems that use the cumulative or prefix techniques are finding the lower and upper quartiles of an array, or finding the equilibrium point of an array. (I cover prefix sums in a lot more detail in this article.)
Footnote: Horner’s Method
One of the solutions presented used a method of calculation known as Horner’s method. Take the cubic
f(x) = 2 x^3 + 3 x^2 + 2 x + 6
f(3) naively would require 8 multiplications (every power
n copies of
x multiplied together, and then they are multiplied by a coefficient), and three additions. There is a lot of wasted calculation here, because when we calculate
x^3 we calculate
x^2 in the process! We could store the powers of
x separately to reduce the number of multiplications.
Horner’s method is a way of doing this without using additional storage. The idea is, for example, that we can use operator precedence to store numbers for us:
3 x^2 + 2 x + 6 = (3 * x + 2) * x + 6
The left side has a (naive) count of 4 multiplications and 2 additions, while the right side has 2 multiplications and 2 additions. Moving to the cubic is even more dramatic:
f(x) = 2 x^3 + 3x^2 + 2 x + 6 = ( (2 * x + 3) * x + 2 ) * x + 6
This takes our 8 multiplications and 3 additions to only 3 multiplications and 3 additions!
The shortest solution so far, submitted by CodeFighter k_lee, uses Horner’s method, along with taking moduli at the different steps. See if you can decipher it.
def productExceptSelf(nums, m): p = 1 g = 0 for x in nums: g = (g * x + p) % m p = (p * x) % m return g
Did your solution for this Interview Practice challenge look different than mine? How did you approach the problem? Let us know over on the CodeSignal forum!