CF 102697005 - Fizz Buzz

Rating: -
Tags: -
Solve time: 2m
Verified: yes

Solution

Problem Understanding

We are given one integer N, with 1 <= N <= 1000. We must classify that number according to divisibility by 3 and 5.

If N is divisible by both 3 and 5, the required output is FizzBuzz. If it is divisible only by 3, we print Fizz. If it is divisible only by 5, we print Buzz. If neither condition holds, the program prints nothing. The output is thus determined entirely by the two divisibility tests.

The bound of 1000 makes the problem extremely small. Even an algorithm that inspected every integer from 1 through N would perform at most 1000 iterations, which is comfortably inside a one-second limit. There is no realistic complexity pressure here, so the cleanest solution is to test the two divisibility conditions directly in constant time.

The main edge case is a number divisible by both values. For input 15, the correct output is FizzBuzz, not Fizz or Buzz. A careless implementation that checks divisibility by 3 first and immediately prints Fizz would never reach the test for 5.

Another edge case is a number divisible by neither value. For input 1, the correct output is empty. The program should not print a word, a zero, or an extra message. The official sample uses exactly this case.

The boundary values also behave normally. For input 1, nothing is printed, while for input 1000, the number is divisible by 5 but not by 3, so the output is Buzz.

Approaches

A straightforward brute-force approach could generate every integer from 1 through N and test each one for divisibility, stopping when it reaches N. This would be correct because the final iteration examines exactly the number whose classification we need. With the actual constraint N <= 1000, its worst case is only 1000 iterations, so even this approach is easily fast enough.

There is no meaningful input size at which that particular brute-force method becomes too slow under the stated constraints. If the bound were increased dramatically, scanning all previous integers would become unnecessary work because the classification of N depends only on N % 3 and N % 5. We can remove the entire scan and perform those two tests directly.

The key detail is the order of the conditions. Divisibility by both 3 and 5 is more specific than divisibility by either individual number. Since a number divisible by both also satisfies the test for 3 and the test for 5, the combined case must be checked first. After that, the individual cases can be handled independently.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(N)	O(1)	Accepted for `N <= 1000`, but unnecessary
Optimal	O(1)	O(1)	Accepted

Algorithm Walkthrough

Read the integer N. There is only one test case, so no outer test-case loop is needed.
Check whether N is divisible by both 3 and 5. This means N % 3 == 0 and N % 5 == 0. If so, print FizzBuzz and finish. This condition must come first because every number divisible by both also passes each individual test.
If the combined condition was false, check whether N is divisible by 3. If so, print Fizz.
Otherwise, check whether N is divisible by 5. If so, print Buzz.
If none of the conditions matched, print nothing. The program can simply terminate without producing output.

Why it works

The algorithm considers exactly the four possible divisibility states of N: divisible by both 3 and 5, divisible only by 3, divisible only by 5, or divisible by neither. The first condition captures the only overlapping case before either individual condition can claim it. Once that case is excluded, the remaining tests are mutually exclusive, so exactly the required output is produced for every valid input.

Python Solution

import sys
input = sys.stdin.readline

n = int(input())

if n % 3 == 0 and n % 5 == 0:
    print("FizzBuzz")
elif n % 3 == 0:
    print("Fizz")
elif n % 5 == 0:
    print("Buzz")

The first condition checks both remainders at once. Python's % operator gives the remainder after division, so a remainder of zero is precisely the test for divisibility.

The elif structure is significant. Once FizzBuzz has been printed, no later condition is evaluated. This prevents 15, for example, from being classified as merely Fizz.

There is no explicit final else branch. When the number is divisible by neither 3 nor 5, the required output is empty, so doing nothing is exactly the required behavior.

Integer overflow is not a concern because the input is at most 1000, and the implementation performs only two small remainder operations.

Worked Examples

For the first example, N = 1, neither divisibility test succeeds.

N	`N % 3`	`N % 5`	Condition	Output
1	1	1	Neither	empty

This demonstrates the case where the correct output contains no characters. The absence of an else print is intentional.

For the second example, N = 15, both remainders are zero.

N	`N % 3`	`N % 5`	Condition	Output
15	0	0	Both	`FizzBuzz`

This demonstrates why the combined condition has to be checked before the individual conditions. If the program tested divisibility by 3 first, it would incorrectly stop at Fizz.

The other official examples follow the same logic: 3 produces Fizz, and 5 produces Buzz.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Only two divisibility tests are needed
Space	O(1)	Only the input integer is stored

The input is bounded by 1000, but the solution does not depend on that small bound at all. Even if the allowed value of N were made much larger, the number of operations would remain constant. The solution therefore fits comfortably within the one-second and 256 MB limits stated by the problem.

Test Cases

import sys
import io

def solve():
    input = sys.stdin.readline
    n = int(input())

    if n % 3 == 0 and n % 5 == 0:
        print("FizzBuzz")
    elif n % 3 == 0:
        print("Fizz")
    elif n % 5 == 0:
        print("Buzz")

def run(inp: str) -> str:
    old_stdin = sys.stdin
    old_stdout = sys.stdout

    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()

    try:
        solve()
        return sys.stdout.getvalue()
    finally:
        sys.stdin = old_stdin
        sys.stdout = old_stdout

# Provided samples
assert run("1\n") == "", "sample 1"
assert run("3\n") == "Fizz\n", "sample 2"
assert run("5\n") == "Buzz\n", "sample 3"
assert run("15\n") == "FizzBuzz\n", "sample 4"

# Custom cases
assert run("2\n") == "", "minimum non-multiple case"
assert run("1000\n") == "Buzz\n", "maximum input"
assert run("30\n") == "FizzBuzz\n", "multiple of both"
assert run("999\n") == "Fizz\n", "large multiple of 3 only"
assert run("995\n") == "Buzz\n", "large multiple of 5 only"

Test input	Expected output	What it validates
`2`	empty	A value divisible by neither number
`1000`	`Buzz`	Maximum allowed input and divisibility by 5
`30`	`FizzBuzz`	Combined divisibility condition
`999`	`Fizz`	Large value divisible by 3 but not 5
`995`	`Buzz`	Large value divisible by 5 but not 3

Edge Cases

For 1, the algorithm computes 1 % 3 = 1 and 1 % 5 = 1. Neither condition succeeds, so it reaches the end without printing anything. This matches the required empty output.

For 15, the algorithm computes both remainders as zero. The first condition succeeds immediately, producing FizzBuzz. The individual Fizz and Buzz branches are skipped, preventing the overlapping case from being misclassified.

For 1000, the algorithm gets 1000 % 3 = 1 and 1000 % 5 = 0. The combined condition fails because the number is not divisible by 3, while the final divisibility test succeeds, producing Buzz. This also confirms that the upper input boundary needs no special handling.

If you want, I can also turn this into a more compact Codeforces-style editorial while preserving the same correctness reasoning.