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'`");

Great job getting through this. Let's move on.

If you had any problems with this tutorial, check out the main forum thread here.