Read N, print numbers 1..N — but multiples of 3 become "Fizz", multiples of 5 become "Buzz", multiples of both become "FizzBuzz".
The screening question that separates candidates who can code from those who cannot. Trivial in principle, but the fact that a nontrivial minority of professional programmers fail it made it famous.
For each integer i from 1 to N: if i is divisible by 15, print "FizzBuzz"; else if divisible by 3, print "Fizz"; else if divisible by 5, print "Buzz"; else print i.
5
1 2 Fizz 4 Buzz
15
1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 FizzBuzz
Runnable starter code. Click "Try in playground" to open pre-loaded with the test fixtures.
n = int(input())
for i in range(1, n + 1):
if i % 15 == 0: print('FizzBuzz')
elif i % 3 == 0: print('Fizz')
elif i % 5 == 0: print('Buzz')
else: print(i)
import java.util.Scanner;
public class Solution {
public static void main(String[] args) {
int n = new Scanner(System.in).nextInt();
StringBuilder sb = new StringBuilder();
for (int i = 1; i <= n; i++) {
if (i % 15 == 0) sb.append("FizzBuzz");
else if (i % 3 == 0) sb.append("Fizz");
else if (i % 5 == 0) sb.append("Buzz");
else sb.append(i);
sb.append('\n');
}
System.out.print(sb);
}
}
#include <bits/stdc++.h>
using namespace std;
int main() {
int n; cin >> n;
for (int i = 1; i <= n; ++i) {
if (i % 15 == 0) cout << "FizzBuzz";
else if (i % 3 == 0) cout << "Fizz";
else if (i % 5 == 0) cout << "Buzz";
else cout << i;
cout << "\n";
}
}