FizzBuzz
Easy · rating 1000 · implementation, math
Read an integer n. For each i from 1 to n, print Fizz if i is divisible by 3, Buzz if divisible by 5, FizzBuzz if divisible by both, otherwise the number itself — one per line.
Constraints: 1 ≤ n ≤ 10000
Editorial
Approach
Loop from 1 to n. The only subtlety is order: a number divisible by both 3 and 5 (i.e. by 15) must print FizzBuzz. The cleanest way to guarantee that is to build the output by string concatenation — append Fizz when divisible by 3 and Buzz when divisible by 5 — which handles the combined case for free.
for i in range(1, n+1):
out = ''
if i % 3 == 0: out += 'Fizz'
if i % 5 == 0: out += 'Buzz'
print(out or i)Complexity
Time: O(n). Space: O(1).