Print integers one through n, but print ‘Fizz’ if an integer is divisible by three, ‘Buzz’ if an integer is divisible by five, and ‘FizzBuzz’ if an integer is divisible by both three and five.

Solution

This is the answer I came up with:

// typescript

const f = (num: number): any => {
  return (num % 3 === 0) ? 'Fizz' : num
}

const b = (num: number): any => {
  return (num % 5 === 0) ? 'Buzz' : num
}

const fb = (num: number): any => {
  return ( num % 3 === 0 && num % 5 === 0 ) ? 'FizzBuzz' : num
}

const fizzBuzz = (length: number): any => {
  for(let i = 1; i < length + 1; i++) {
    console.log(f(b(fb(i))))
  }
}

fizzBuzz(100)

Since the functionalities for Fizz, Buzz, and FizzBuzz are separate, it makes sense to create individual functions for each condition. I then created the final fizzBuzz function using composition, which I think was an elegant solution.

Testing

Solution tested in REPL:

https://www.typescriptlang.org/play?#code/MYewdgzgLgBAZjAvDAFGArgWwFww5gIwFMAnASlwEMwBPJAPhgG8BYAKBhhKKnRLFT4YAUhgBmJImQAGMjAD8MAOQAxAJYAvDUpi587AL7t2oSLAJJBWPVmLkqtBs3aduvflcwiYAVkky5RSUAIXQtHRtMQ2M2U2h4C2Q0azxbUgoYajpERlYOLh4+ARRUr1EJKRkYADJq0u8-SphpGEDldS1Q8N1S6LYTcHi4TQ0ujUsUABsiMABzKAALSLsMrKc8zjgQEimeGDVLAEYAbn2YAB4YabnFmABqGBP9u7u5Dc4YOJBpgDpJkFmKDgKAIQNBajIkLILhgRjYcPYw06YQ0KEO0lk7CAA