Javascript Interview Question: Stairs

es6-steps.js

/*
 A simple iterative solution to building steps.
 Uses String.prototype.repeat which is an ES2015 feature.
*/
function steps(n) {
  for (let i = 1; i <= n; i++) {
    let step = '#'.repeat(i) + ' '.repeat(n - i);
    console.log(step);
  }
}

//Try it
steps(5);

steps.js

/*
Create a stair case.

Pseudo Code:
To create a staircase, 

Draw a grid to visualize the question.

'#   '  
'##  '
'### '
'####'

For rows 0 to n
  - Create a string variable to hold the characters that make up the step.
  - For cols 0 to n
    - Check if the col is less than or equal to row. 
      - If so print a #.
      - Otherwise, print a space.

*/

function steps(n) {
  for (let row = 0; row < n; row++) {
    let step = '';
    for (let col = 0; col < n; col++) {
      step += (col <= row) ? '#' : ' ';
    }
    console.log(step);
  }
}

//Try it
steps(5);

/*
Create a stair case using recursion.

First, determine the base case. Base case can be found in the first for loop. Once the 
row === n, the base case condition has been met. Exit function.

Second, determine when we are at the end of a row. If that condition is met, print the step to 
the console, call the steps function with arguments: (a) n, (b) row + 1. Call return here.
 
Third, append the # or space character. Call the steps function with three arguments: (a) n, (b) row, and (c) step.

Note:
We can replace the col variable from the first example with step.length. They mirror each other 
which means we do not need to track the col varaible.
 
For any arguments we provide other than n (n is required), we can assume a reasonable default 
and initialize the variable in the function signature.
*/

function steps(n, row = 0, step = '') {

  //Base case.
  if (n === row) {
    return;
  }
  
  //Is it time for a new row?
  if (n === step.length) {
    console.log(step);
    steps(n, row + 1);
    return;
  }
  
  //Append a character.
  step += (step.length <= row) ? '#' : ' ';
  steps(n, row, step);
}

steps(5);