madbernard | Entries tagged with coderbyte

Just did a Coderbyte challenge that was, "given an array of numbers that could be positive or negative but which will not include 0, return 'Arithmetic' for arrays where there's the same distance between all the numbers, 'Geometric' for arrays where there's the same multiple between all the numbers, and -1 for everything else." My answer isn't the best it could possibly be (I feel that I'm using Math.abs poorly, and I should have taken a moment more to make the reduce part bring back the biggest number in the difference array... averages can be deceiving), but looking at answers from unnamed other people, several of them don't even work. This one only compares the gap in numbers of the first two and last two in the array:

function ArithGeo(arr) { 
  var isArithmetic=(arr[1]-arr[0]==arr[arr.length-1]-arr[arr.length-2]);
  var isGeometric=(arr[1]/arr[0]==arr[arr.length-1]/arr[arr.length-2]);
  // code goes here  
  if(!isArithmetic&&!isGeometric){//is not special
    return -1;
  }else{
    return (isArithmetic)?'Arithmetic':'Geometric';
  }
}

It can be defeated with [1,2,45,99,100]. I saw this same pattern three times in like six answers I looked over. Interesting to catch stuff that the tests don't. When I started this post I was feeling a cheerful about not being at the bottom of the heap, but now I'm a bit sad. Anyway, here's my answer. Tell me about better ways of checking that every number in an array is the same... I could have done another for loop, "if difference array [i] !== difference array [0] return false"...

function ArithGeo(arr) { 
  var geo = [];
  var arith = [];
  for (var i = 0; i < arr.length - 1; i++) {
    geo.push(Math.abs(arr[i + 1] / arr[i]));
    arith.push(Math.abs(arr[i + 1] - arr[i]));
  }
  var sumGeo = geo.reduce(function(prev, curr){
    return prev + curr;
  });
  var sumArith = arith.reduce(function(prev, curr){
    return prev + curr;
  });
  if (sumGeo / geo.length === geo[0]) {return 'Geometric';}
  if (sumArith / arith.length === arith[0]) {return 'Arithmetic';}
  return -1; 
}

Here's one I thought would be simple, which wasn't. The problem is, "return a count of all the vowels (not y) in a string." I thought I'd use regex.exec to capture an array of vowels and find the length of the array... But it only ever returned length 2. I see in the Mozilla Developer Network info page that RexExp.length is 2, but I don't know why I'm getting that here. Can someone troubleshoot?

function VowelCount(str) { 
  var re = /([aeiou])/ig;
  var arr = re.exec(str);
  console.log(arr);
  return arr.length; 
}

console.log(VowelCount('moooo'));

So I pivoted to for loops and came up with this,

function VowelCount(str) { 
  var re = /[aeiou]/i;
  var count = 0;
  for (var i = 0; i < str.length; i++) {
    if (re.test(str.charAt(i))) {
      count += 1;
      console.log(i);
  }
  return count; 
}

This also took some troubleshooting, since the count was wrong when I was using the 'g' flag in the regex. I'm also not sure why that was. Other people I looked at also had mangled looking answers, though of course Matt Larsh has something good,

function VowelCount(str) { 
  var vowels = str.match(/[aeiou]/g);
  return vowels.length;
}

Looks like I need to bone up on str.match.

Last evening I had a moment and tried the Coderbyte challenge "when given a string of single letters, plusses, and equals signs, return true only if every letter has a plus on both sides". I came up with a four part if statement reliant on regexes.

function SimpleSymbols(str) {
  if (/(\+\w\+)+/g.test(str) && !/^\w/g.test(str) && !/=\w/g.test(str) && !/\w(?!\+)/g.test(str)) 
  {return true;}
  else {return false;}
}

In English, that's if there is a plus on either side of a letter, and if there is no letter at the start of the string, and if there is no letter with an equals sign immediately in front of it, and if there is no letter that is not followed by a plus sign... Then true.

Ben, who does JS for a living, gave me the learning that if there's more than one regex, it's probably not ideal... "There's so many edge cases". Makes sense. So, later, the Coderbyte challenge appears, "when given a string of lowercase letters, if there is an a followed by a b three spots later, return true". And I'm like, "I'll use a for loop, like we were talking about last night!"( Read more... )

Challenge 5 was, given a number between 1 and 1000, add up all the numbers inclusive in that set. Obviously I could repurpose Matt Larsh's recursive function to do this, so I did so, and it worked first try. Woo! Though, no one else seems to be doing +=. I wonder if this would be super wrong if I wasn't recursing and thus seeing it only once each function call? Or do they just like num + SimpleAdding(stuff)?

function SimpleAdding(num) {
  if (num === 1) {
    return 1;  
  }
  else {
    return num += SimpleAdding(num - 1);
  }      
}

Challenge 6 was one I hadn't yet solved in Free Code Camp: Capitalize every word in a string. I abandoned a couple different paths while trying for speed, and eventually came up with this answer that seems cludgy:

function LetterCapitalize(str) { 
  var foo = str.split('');
  var bar = [];
  bar.push(foo[0].toUpperCase());
  for(var i = 1; i < foo.length; i++) {
    if (foo[i - 1] === ' ') {
      bar.push(foo[i].toUpperCase());
    }
    else {
      bar.push(foo[i]);
    }    
  }
  console.log(bar.join(''));
  return bar.join('');
}
LetterCapitalize('to see how to enter arguments in JavaScript');

I tried replacing the space in the if statement with /\s/ and that didn't work, nor did /\s/g; I'm not sure why not.

One of my abandoned paths was to do a str.replace(regex, function) but I wasn't getting the function to work... So I was super curious to see what Matt Larsh did, and sure enough he did something cool:

function LetterCapitalize(str) { 
  return str.replace(/\b[a-z]/g,function(c){return c.toUpperCase()});
}

That "\b" was something I thought should exist, but didn't find: the signal for "match a word boundary". I just tried /\b\w/g and that works, too; I suppose his way is better, though, since it limits the capitalization to only things that we absolutely know need capitalizing. Unless if \w includes accented letters and other slightly less usual things? My RegEx skillz are not yet l33t and m4d.

Remembering these from last week: Challenge 3 was another I'd done before, "find the longest word in the sentence", and I got a few more refinements in my answer: I used a regular expression to create the first array; and (having discussed it with Ben) I used an empty string as the initializing value. The previous stab at this Reduce, where I was trying to initialize with 0, was comparing apples and oranges in the form of numbers and strings.

function LongestWord(sen) { 
  var arr = sen.split(/(\w+)/g);
  var foo = arr.reduce(function(prev, curr){
    if(prev.length < curr.length){
      return curr;
    } else {return prev;}
  }, '');
  return foo; 
}

Looking over other answers, I think reduce is by far the most elegant path. Wooo! Arrays FTW!

Challenge 4 was ...a journey. ( Read more... )

Factorial happened last week, also; the CoderByte challenge was, we'll test on numbers between 1 and 18, give us that number multiplied by every non-0 integer less than it. Ben-the-housemate suggested that this could be done recursively, so I tried that first, but didn't hack it; so, given the ticking clock, I aborted to the exact same decreasing for-loop plan I'd used last week. Looking through other answers was gratifying... I was among the most elegant! I found the recursive method from user mattlarsh, in a tiny tidy package:

function FirstFactorial(num) { 
  return num<=1?1:num*(FirstFactorial(num-1)); 
}

He's using the ternary operator there, the shortest possible way to say if/then. IF: the num is less than or equal to 1, DO THIS: return 1, AND IF IT'S NOT: multiply the num by FirstFactorial num-1. So once he has a stack of not-quite-resolved functions and the num gets to 1, it returns 1 and then the rest of the stack has all the numbers it needs to work with and can resolve the final return.

My first pass on recursion was the below, and I got stack overflow (infinite program), which I can see now is because the num kept getting bigger but the recursion would only end when the num got to 0.

function FirstFactorial(num) { 
if (num > 0) {
num *= (num - 1);
FirstFactorial(num);
}
  return num; 
}

FirstFactorial(5);

I looked at the Hack Reactor suggestions for what one should be able to do before applying, and headed over to http://www.coderbyte.com to try their easy track. They score submissions based not only on correctness, but on time, which is not at all my bag. I think when I saw that a year ago I just turned away, even though a friend suggested doing it on my own time and then pasting the answer in... I might do that in the future.

Their first question was the exact same as I saw last week; reverse a string. I came up with this, which, amusingly, is not what I did last week. It uses the join method of arrays that I was searching for last week, so I'm pleased that knowledge is in me...

function FirstReverse(str) { 
  var hold = [];
  for(var i = str.length - 1; i >= 0 ; i--){
    hold.push(str[i]);
  }
  var foo = hold.join("");
  return foo; 
}

FirstReverse("stuff");

Also, heeey, check out the respect of spacing in that code block. I thought there were HTML tags about code, and found <code> and thought that was it... Ben the housemate says code isn't really a HTML tag and turned me on to <pre>. Thanks!

Anyway, a good thing about CoderByte: they let you see what other people did, after, answering the socrating question of my friend Catherine about whether I looked at other people's solutions. There's a solution like mine above, but relying on methods of strings and arrays that I wasn't aware of; and that solution lets one do everything in a single line!

function FirstReverse(str) { 
  return str.split('').reverse().join('');
}

Way more elegant.

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Finding bugs in other people's code is surprisingly gratifying
More Regex Complications!
RegEx for Some, Tiny American Flags for Others!
Coderbyte easy 5 and 6
Easy Coderbyte challenges 3 & 4
CoderByte Factorial, me vs others & vs recursion
Beginning CoderByte easy tests

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Javascript programming every day of the week

Entries tagged with coderbyte

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Finding bugs in other people's code is surprisingly gratifying

More Regex Complications!

RegEx for Some, Tiny American Flags for Others!

Coderbyte easy 5 and 6

Easy Coderbyte challenges 3 & 4

CoderByte Factorial, me vs others & vs recursion

Beginning CoderByte easy tests