AlgoDaily - Fizz Buzz - One Pager Cheat Sheet

One Pager Cheat Sheet

We are given an integer n, and asked to write a function fizzBuzz() that returns the string representation of all numbers from 1 to n using the rules "fizz", "buzz" and "fizzbuzz" for multiples of 3, 5 and both respectively, with a maximum value of 1000, O(n) time complexity and O(1) space complexity.
The key to answering this technical interview question was to identify how to map a problem to its technical implementation, such as using if-else or switch control flows.
We can use the modulo operator (%) to easily determine if one number is a multiple of another - the remainder returned will be 0 if it is.
We can use a conditional flow order of fizzbuzz > fizz > buzz > return number to check if a number is divisible by either 3, 5, both, or neither, with a time complexity of O(n).

This is our final solution.

To visualize the solution and step through the below code, click Visualize the Solution on the right-side menu or the VISUALIZE button in Interactive Mode.

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}
 
var assert = require('assert');
​
function fizzBuzz(n) {
  let finalArr = [];
​
  for (let i = 1; i <= n; i++) {
    if (i % 3 === 0 && i % 5 === 0) {
      finalArr.push('fizzbuzz');
    } else if (i % 3 === 0) {
      finalArr.push('fizz');
    } else if (i % 5 === 0) {
      finalArr.push('buzz');
    } else {
      finalArr.push(i.toString());
    }
  }
​
  return finalArr.join('');
}
​
try {
  assert.equal(fizzBuzz(0), '');
​
  console.log('PASSED: ' + "Expect `fizzBuzz(0)` to equal `''`");
} catch (err) {
  console.log(err);
}
​
try {
  assert.equal(fizzBuzz(1), '1');
​
  console.log('PASSED: ' + "Expect `fizzBuzz(1)` to equal `'1'`");

Got more time? Let's keep going.

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