I bumped into it looking at the CodeEval's problems, where it is presented in a slightly more generalized way. We want to list all the numbers from 1 to a given maximum, but replacing the ones that could be divided by two taboo factors, fizz and buzz, with placeholders.

A couple of test cases (written in C++ for GoogleTest), should clarify the requisites.

std::string solution(int fizz, int buzz, int n); TEST(FiBu, CaseGiven1) // 1 { std::string result = solution(3, 5, 10); ASSERT_EQ(result, "1 2 F 4 B F 7 8 F B"); } TEST(FiBu, CaseGiven2) // 2 { std::string result = solution(2, 7, 15); ASSERT_EQ(result, "1 F 3 F 5 F B F 9 F 11 F 13 FB 15"); }1. We want to analyze the numbers in the [1..10] interval, any number that has 3 among its factors should be replaced by a F, and 5 by B. Notice that there is no trailing blank after the last generated element.

2. Here the interval is [1..15], the factor 2 should lead to F, and 7 to B. 14 could be divided by both 2 and 7, so it should be replaced by FB.

There is not much to think about before writing the code. The main issue is how we can check if a given number has fizz (or buzz) among its factors. In C, C++ and related languages we normally use the arithmetic operator % (called "modulus" or "modulo") that returns the remainder of the integer division between the two numbers. When it returns zero, we have an integer divisor.

Here is a possible solution:

std::string solution(int fizz, int buzz, int n) { std::ostringstream oss; for(int i = 1; i <= n; ++ i) { bool fb = false; // 1 if(i % fizz == 0) // 2 { oss << 'F'; fb = true; } if(i % buzz == 0) // 3 { oss << 'B'; fb = true; } if(!fb) // 4 oss << i; if(i != n) // 5 oss << ' '; } return oss.str(); }1. Flag for fizz or buzz detection.

2. If the remainder of the division of the current number by fizz is zero, it is one of its factors.

3. Same for buzz.

4. If nor fizz nor buzz has been detected, the current number in used.

5. I don't want a trailing blank at the end of the string, so I ensure I am not in the last iteration.

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