FizzBuzz
Challenge Description
FizzBuzz is a classic programming problem often used in coding interviews. The goal is to print numbers from 1 to n with the following rules:
- If a number is divisible by 3, print "Fizz".
- If a number is divisible by 5, print "Buzz".
- If a number is divisible by both 3 and 5, print "FizzBuzz".
- Otherwise, print the number itself.
For example:
- 1 => 1
- 2 => 2
- 3 => Fizz
- 5 => Buzz
- 15 => FizzBuzz
Tips to Solve
tip
- Print numbers from 1 to n using a loop.
- Check if the number is divisible by 3 and 5 first, and print "FizzBuzz".
- If the number is only divisible by 3, print "Fizz".
- If the number is only divisible by 5, print "Buzz".
- If none of the above conditions are met, print the number itself.
Step-by-Step Solution
Let's break down the solution into easy steps:
-
Print Numbers from 1 to n: Use a loop to iterate through numbers from 1 to n.
for (let i = 1; i <= n; i++) {
// logic will be implemented here
} -
Check Divisibility by 3 and 5: Use the modulus operator (
%
) to check if the number is divisible by both 3 and 5.if (i % 3 === 0 && i % 5 === 0) {
console.log("FizzBuzz");
} -
Check Divisibility by 3: If the number is not divisible by both 3 and 5, check if it's only divisible by 3.
else if (i % 3 === 0) {
console.log("Fizz");
} -
Check Divisibility by 5: If the number is not divisible by 3, check if it's divisible by 5.
else if (i % 5 === 0) {
console.log("Buzz");
} -
Print the Number: If none of the above conditions are met, print the number itself.
else {
console.log(i);
}
Full Solution Code
Here is the complete solution using the steps above:
const fizzBuzz = (n) => {
for (let i = 1; i <= n; i++) {
if (i % 3 === 0 && i % 5 === 0) {
console.log("FizzBuzz");
} else if (i % 3 === 0) {
console.log("Fizz");
} else if (i % 5 === 0) {
console.log("Buzz");
} else {
console.log(i);
}
}
};
fizzBuzz(15);
output
Output: 1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz