C / C++ Interview: Code Optimization

Question

I had an interview today. This question was to optimize the code below. if we will see the code below after for loop there are four steps of "if-else" follows. So, interviewer asked me optimize it to 3 if-else line. I have tried a lot. But could not able to find the solution. Even he told me if you know scripting language then, you can use them also. Please help me in optimizing the same.

int main()
{
    int i = 1;
    for(i; i <= 100; i++)
    {
        if((i % 3 == 0 && i % 5 == 0))
        {cout << "PR\n";}
        else if(i % 3 == 0)
        {cout << "P\n";}
        else if(i % 5 == 0)
        {cout << "R\n";}
        else
        {cout << i <<"\n";}
    }
system("pause");
return 0;
}

This is the [FizzBuzz](http://rosettacode.org/wiki/FizzBuzz) problem (@6502's "F" and "B" reference). — Ken White, Sep 04 '13 at 16:31
Though the assignment is pretty trivial, I daresay that rhe really correct answer is "WTF, outputting to cout takes 10,000 longer than an extra if, that is a useless optimization". In other words, this is probably a catch question. — Damon, Sep 04 '13 at 16:34

6502 · Accepted Answer · 2013-09-04T16:36:26.913

4

This is a well known question... the "FizzBuzz".

You can even solve it without any explicit IFs

const char *messages[] = {"%i\n", "P\n", "R\n", "PR\n"};

for (i=1; i<=100; i++) {
    printf(messages[((i % 3)==0) + 2*((i % 5)==0))], i);
}

edited Sep 04 '13 at 16:36

answered Sep 04 '13 at 16:14

6502

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:- Excellent, I have one question from where did you learn this coding? :) – Rasmi Ranjan Nayak Sep 05 '13 at 04:00
1

Although the claim about avoiding "ifs" is somewhat misleading because internally there is a whole "world" of "ifs" in printf. – Mikhail Apr 17 '17 at 08:07

Óscar López · Answer 2 · 2013-09-04T18:29:45.260

3

Here's one way, in Python:

for i in range(1, 101):
    s = ''
    if i % 3 == 0:
        s += 'P'
    if i % 5 == 0:
        s += 'R'
    if i % 3 != 0 and i % 5 != 0:
        s = i
    print(s)

Equivalently: using a flag, as shown in your own answer:

for i in range(1, 101):
    s, flag = '', False
    if i % 3 == 0:
        flag = True
        s += 'P'
    if i % 5 == 0:
        flag = True
        s += 'R'
    if not flag:
        s = i
    print(s)

Just for fun, a Python version of @6502's answer:

messages = ['{}', 'P', 'R', 'PR']
for i in range(1, 101):
    print(messages[(i%3 == 0) + 2*(i%5 == 0)].format(i))

And finally, my personal favorite (because it's the shortest) - using the Greatest Common Divisor function and a lookup table:

from fractions import gcd
messages = {3:'P', 5:'R', 15:'PR'}
for i in range(1, 101):
    print(messages.get(gcd(i, 15), i))

edited Sep 04 '13 at 18:29

answered Sep 04 '13 at 16:25

Óscar López

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I have put my solution after hours of struggle. Please let me know whether it is optimized one or not? – Rasmi Ranjan Nayak Sep 04 '13 at 16:30
@RasmiRanjanNayak yes, your answer is correct. I updated my answer above with a Python version, both approaches are equivalent and give the same result – Óscar López Sep 04 '13 at 16:37
1

Thanks Oscar, your answers are very much helpful. – Rasmi Ranjan Nayak Sep 05 '13 at 03:59

score 2 · Answer 3 · answered Sep 04 '13 at 16:29

2

I found a solution. Please let me know whether it is good or not?

int main()
{
    int i = 1;int stat=0;
    for(i; i <= 100; i++)
    {
        stat=0;
        if(i%3 == 0){stat++; cout << "P";}
        if(i%5 == 0){stat++; cout << "R";}
        if(stat == 0)cout << i;
        cout << "\n";
    }
system("pause");
return 0;
}

answered Sep 04 '13 at 16:29

Rasmi Ranjan Nayak

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This answer is correct, but the solution to this problem can be written in a simpler way - there's no need to use `if-else if-else` instructions! see my answer for a couple of options – Óscar López Sep 04 '13 at 18:21

score 1 · Answer 4 · answered Nov 23 '15 at 05:01

I really like 6502's answer, but here is a simple solution without extra variables:

for(i = 1; i <= 100; i++)
{
    if(i % 3 != 0 && i % 5 != 0)
    {
        printf("%d\n", i);
        continue;
    }

    if(i % 3 == 0)
        printf("P");

    if(i % 5 == 0)
        printf("R");

    printf("\n");
}

Simeng Yue · Answer 5 · 2014-10-08T06:27:59.507

0

this way only use 3 if

#include <iostream>
using namespace std;

int main()
{
 for (int i = 0; i <= 100; ++i)
 {
   bool fizz = (i % 3) == 0;
   bool buzz = (i % 5) == 0;
   if (fizz)
    cout << "Fizz";
   if (buzz)
    cout << "Buzz";
    if (!fizz && !buzz)
     cout << i;
    cout << endl;
  }
  return 0;
}

edited Oct 08 '14 at 06:27

answered Oct 07 '14 at 07:39

Simeng Yue

11
3

1

You should jot down what exactly is optimized – stealthjong Oct 07 '14 at 07:58

C / C++ Interview: Code Optimization

5 Answers5