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LESSON 177 OF 410 · LOOPS - CHALLENGES

FizzBuzz in Java: Solution with a for Loop and the % Operator

Practice O(n) Read the full explanation

Given a positive integer n print each number from 1 to n, each on a new line.
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers which are multiples of both 3 and 5, print "FizzBuzz" instead of the number.

Example:

Input: n = 15
Output:
1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz

Note:

Your algorithm should run in O(n) time and use O(1) space.


Hints:

In order to print the items, we should iterate through the numbers 1 to n and check for each of them what condition is satisfied to print accordingly. How can we do this?

Let's use a for loop and a variable i to iterate through the numbers 1 to n: 
for each i from 1 to n:
	print either i or "Fizz" or "Buzz" or "FizzBuzz"
Inside the loop, we need to decide what should be printed between Fizz, Buzz, FizzBuzz and i.
How can we do this?

Using sequential if - else statements along with the modulo (%) operator, which gives the remainder of a division (i % 15 == 0 means that i is divisible by 15):
for each i from 1 to n:
	if i is divisible by 15, print "FizzBuzz"
	otherwise, if i is divisible by 3, print "Fizz"
	otherwise, if i is divisible by 5, print "Buzz"
	otherwise, print i

We have a for loop that performs exactly n iterations, thus the Time Complexity is O(n).
The only extra space we use is the iterator variable i inside the loop, thus the Extra Space used is O(1).

Video Solution:

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    JAVA · LOOPS - CHALLENGES

    Fizz Buzz in Java, explained

    What the FizzBuzz problem asks

    FizzBuzz is one of the most common first coding exercises. Given a positive integer n, the method fizzBuzz(int n) prints the numbers from 1 to n, one per line, with three exceptions:

    • a multiple of 3 prints Fizz instead of the number;
    • a multiple of 5 prints Buzz instead of the number;
    • a multiple of both 3 and 5 prints FizzBuzz.

    The method returns nothing (void). Its result is the text it prints. For n = 15 the output is 1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, each item on its own line. The required cost is O(n) time and O(1) extra space.

    Approach: one loop, one decision per number

    The numbers are independent of each other, so there is nothing to remember between them. A for loop visits each number from 1 to n, and inside the loop an if / else if chain picks what to print.

    The test for "multiple of 3" is the remainder operator: i % 3 == 0 is true when dividing i by 3 leaves nothing over. A number that is a multiple of both 3 and 5 is a multiple of 15, because 3 and 5 share no factor. So i % 15 == 0 identifies the FizzBuzz case with a single test.

    For example, 9 % 3 is 0 because 3 goes into 9 exactly three times, while 10 % 3 is 1 because 10 is three threes plus 1 left over. Only a remainder of 0 means "multiple". Both operands here are int values, so the result is an int as well, and the comparison with 0 gives a boolean the if can use directly.

    The order of the checks matters. Every multiple of 15 is also a multiple of 3 and of 5, and an else if chain stops at the first condition that is true. If the check for 3 came first, 15 would print Fizz and never reach the FizzBuzz branch. The most specific case, 15, goes first.

    Step by step

    1. Loop from 1 to n inclusive. The loop variable starts at 1, not 0, and the condition is i <= n so that n itself is included.
    2. Test for 15 first. If i % 15 == 0, print FizzBuzz.
    3. Then test for 3. If i % 3 == 0, print Fizz.
    4. Then test for 5. If i % 5 == 0, print Buzz.
    5. Otherwise print the number itself. System.out.println(i) prints the int and ends the line.

    The solution in Java

    class Solution {
    public void fizzBuzz(int n) {
        for (int i = 1; i <= n; i++) {
            if (i % 15 == 0) {
                System.out.println("FizzBuzz");
            } else if (i % 3 == 0) {
                System.out.println("Fizz");
            } else if (i % 5 == 0) {
                System.out.println("Buzz");
            } else {
                System.out.println(i);
            }
        }
    }
}

    A few Java details are worth knowing:

    • println versus print. System.out.println adds a line break after its argument, which is what "each on a new line" requires. System.out.print does not.
    • One method, two kinds of argument. println is overloaded, so the same call prints a String such as "Fizz" and an int such as i. No conversion is needed.
    • Exact text. The words are case-sensitive: "Fizzbuzz" or "Fizz Buzz" would not match the expected output.

    An alternative: build the word from parts

    Some programmers avoid the test for 15 by building the output string. For each number, start with an empty string, append Fizz if the number is a multiple of 3 and Buzz if it is a multiple of 5, and print the number only when the string is still empty:

    for (int i = 1; i <= n; i++) {
    String out = "";
    if (i % 3 == 0) out += "Fizz";
    if (i % 5 == 0) out += "Buzz";
    System.out.println(out.isEmpty() ? String.valueOf(i) : out);
}

    It prints the same lines. For 15 both if statements run, so the string becomes FizzBuzz. This version scales better if a rule such as "multiples of 7 print Bazz" is added later, because each rule is one more independent if rather than a new combination to place in the chain.

    Complexity

    The loop runs exactly n times and each iteration does a fixed amount of work (a few remainder operations and one print), so the time is O(n). The only variable the method creates is the loop counter i, so the extra space is O(1). The method prints as it goes and never stores the sequence.

    Common mistakes

    • Checking 3 or 5 before 15. With this order the output for 15 is wrong:
    if (i % 3 == 0) System.out.println("Fizz");
else if (i % 5 == 0) System.out.println("Buzz");
else if (i % 15 == 0) System.out.println("FizzBuzz"); // never reached
else System.out.println(i);

    For i = 15 this prints Fizz, because the first condition is already true. The FizzBuzz branch can never run.

    • Starting at 0. 0 % 15 == 0 is true, so a loop that starts at 0 prints a stray FizzBuzz first.
    • Using < instead of <=. The last number, n, is silently missing.
    • Using = instead of ==. i % 3 = 0 does not compile, because = assigns and == compares.
    • Returning a list instead of printing. The method is void and is checked by what it prints, so collecting the strings in a list and returning them does not do the job.

    Testing

    Try n = 1, which prints only 1; n = 3, which prints 1, 2, Fizz; n = 5, which ends with Buzz; and n = 15, which ends with FizzBuzz. If those four are right, the three special cases and the plain case are all covered.

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