F5 Networks Interview Question for Software Engineer / Developers


Country: United States
Interview Type: Phone Interview




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public int numberOfFactors(int n) {
		int totalFactors = 0;
		int i = 1;
		while (i <= n) {
			if (n % i == 0) {
				totalFactors++;
			}
			i++;
		}
		return totalFactors;
	}

Is the total number of factors of N to be returned, or return the actual factors as well?

- shah10 December 13, 2014 | Flag Reply
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-- Filter all factors of n
factorsOfN n = length $ filter ((==0). mod n) [1..n]

-- filter all factors of n from 1..sqrt of n - 1 for squares 
factorsOfN' n = (cSqrt - trSq) - 1 + 2*length (filter ((==0) . mod n) [2.. trSq])
    where 
      trSq = truncate $ sqrt $ fI n
      cSqrt = ceiling $ sqrt $ fI n

- Anonymous December 13, 2014 | Flag Reply
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O(sqrt(n)) algorithm:

factors = 0
for i in range(int(ceil(sqrt(n)))):
  if n % i == 0 and sqrt(n) != i:
    factors += 2
  elif n % i == 0:
    factors += 1

- Anonymous December 13, 2014 | Flag Reply
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Based on Anonymous's O(sqrt(n)) algorithm. Not entirely sure about the ceil(sqrt(n) + 0.0001) part though.

from math import *

def num_factors(n):
    factors = 2
    for i in range(2, int(ceil(sqrt(n) + 0.0001))):
        if n % i == 0:
            factors += 1 if i*i == n else 2
    return factors

print num_factors(12)

- Sunny December 13, 2014 | Flag Reply
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0.0001 is added to ensure sqrt(n) is included in the iteration. The for..range in Python is exclusive at the closing bound.

This is the most efficient solution btw.

- harshjain30 January 30, 2016 | Flag
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public int numberOfFactors(int n) {
int totalNumOfFacts = 0;
for(int i=1;i<=n/2;i++) {
if(n%i == 0)
totalNumOfFacts++;
}
return ++totalNumOfFacts;
}

- Anonymous December 15, 2014 | Flag Reply
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This solution has the best case scenario complexity of O(log(sqrt(n)). The worst case scenario takes O(sqrt(n)) time. The idea is to find prime factors and their numbers. Then we just need to calculate the numbers each prime factor occurs in the input number.

private int FactorsNumber(int n)
        {
            var m = n;
            var res = 1;
            for (int d = 2; d <= Math.Sqrt(m); d++)
            {
                var k = 1;
                while (m % d == 0)
                {
                    m /= d;
                    k++;
                }
                if (k > 1) res *= k;
            }
            if (m > 1) res *= 2;
            return res - 1;
        }

- vladimir.grebennikov December 15, 2014 | Flag Reply


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