Recursion performance test

As I've read multiple times, recursion while being more elegant is most of the times slower than a regular loop. I tested this with jsperf on a simple factorial function. Here is the JSperf link if you want to test on your own machine.

Recursion:

 
function factorialize(num) {
  if (num <= 0) {
    return 1;
  }
  return num * (factorialize(num-1));
}

factorialize(5);

Result: 30,406,485 Ops/sec
±0.51%
59% slower



Basic for loop:

function factorialize(num) {
 var result = 1;
  for (i = 1; i <= num; i++) {
 result = result * i;
  }
return result;
}

factorialize(5);

Result:
73,615,404 Ops/sec
±0.47%
fastest

Eloquent Javascript: Minimum

The previous chapter introduced the standard function Math.min that returns its smallest argument. We can do that ourselves now. Write a function min that takes two arguments and returns their minimum.

function min(a, b) {
  if (a < b) {
    return a;
  }
  return b;
}

console.log(min(0, 10));
// → 0
console.log(min(0, -10));
// → -10

Eloquent Javascript - Looping a triangle

function oneMore(num, str) {
  var line = "";
  for (i = 0; i < num; i++) {
    console.log(line += str);
  } 
}
oneMore(7, "#");

Eloquent Javascript - FizzBuzz

function fizzbuzz(num, fizz, buzz) {    
 for (i = 1; i <= num; i++) {    
     if (i % (fizz * buzz) === 0) {
         console.log("FizzBuzz");
     } else if (i % buzz === 0) {
         console.log("Buzz");
     } else if (i % fizz === 0) {
         console.log("Fizz");
     }   
 }  
}    

fizzbuzz(100, 3, 5);

Eloquent Javascript - Chess Board

My solution:

function checkerMaker(size, symbols) {
  
  for (i = 0; i < size; i++) {
    var line = "";
    if (i % 2 === 0) {
      for (j = 0; j < (size/2); j++) {
        line += symbols;
      }
        console.log(line);  
    } else {
      var altSymbols = symbols.split("").reverse().join("");
      for (j = 0; j < (size/2); j++) {
        line += altSymbols;
      }
        console.log(line); 
    }
  }

}

checkerMaker(8, " #");

I knew I was repeating myself when I wrote the second inner for loop. Still havn't got the maths instinct.

Marijn Haverbeke's solution:

function checkerMaker(size, symbol1, symbol2) {
var board = "";
for (var y = 0; y < size; y++) {
  for (var x = 0; x < size; x++) {
    if ((x + y) % 2 == 0)
      board += symbol1;
    else
      board += symbol2;
  }
  board += "\n";
}
console.log(board);
}
checkerMaker(8, " ", "#");

Bonfire: Chunky Monkey

function chunk(arr, size) {

  var multi = [];
  i = 0;
  
  while (i < arr.length){
    multi.push(arr.slice(i,i+=size));
  }
  return multi;
}

chunk(["a", "b", "c", "d"], 1);

Here an alternative where I extracted the increment i=i+size, for more readability (maybe?):

function chunk(arr, size) {

  var multi = [];
  i = 0;
  
  while (i < arr.length){
    multi.push(arr.slice(i,i+size));
    i=i+size;
  }
  return multi;
}

chunk(["a", "b", "c", "d"], 1);

Same with a for loop instead of while:

function chunk(arr, size) {

  var multi = [];
  for (i=0;i < arr.length;i+=size){
    multi.push(arr.slice(i,i+size));
  }
  return multi;
}

chunk(["a", "b", "c", "d"], 2);

Bonfire: Truncate a string

My low bro solution, it's ugly, and still took me time...

function truncate(str, num) {
  
  if(num <= 3){
    return str.slice(0,num) + "...";
  }
   if(str.length > num){
    return str.slice(0,num-3) + "...";
  }
   if(num >= str.length){
    return str;
  }
}

truncate("A-tisket a-tasket A green and yellow basket", 2);

It took me time because I literally translated this instructions "If the length of the string is less than or equal to 3 characters" as this:

  if(str.length <= 3){
    return str.slice(0,num) + "...";
  }

instead of course, of this:

  if(num <= 3){
    return str.slice(0,num) + "...";
  }

akshat01 posted this on Gitter in the HelpBonfires room and it's way more concise:

function truncate(str, num) {

  var i= 0;
  if (num >=3){
    i=3;
  }

  if (str.length> num){
    str= str.slice(0,num-i)+"...";}
  return str;
} 

truncate("A-tisket a-tasket A green and yellow basket", 1);

I don't know what it says about me but it took me a good 15 mins to understand the logic of just returning the str as is, in fact the default value since both "if" statements covered all other cases

Bonfire: Repeat a string repeat a string

My solution:

function repeat(str, num) {
  var singles = [];
  if (num <= 0){
   repeatedStr="";
  }
  for(i=0;i<num;i++){
    singles.push(str);
    repeatedStr = singles.join("");
  }
  return repeatedStr;
}

repeat("abc", 3);

Critics:
- I still make stupid mistakes regarding variables scope and had at first put var singles = []; inside the for loop...
- I repeat myself when instead of directly returning singles.join("") I assign it first to repeatedStr. Only advantage I could see is if we further down the road would want to access this variable.

---

Now, FFC on their wiki, have this similar one:

function repeat(str, num) {
  var newstr = [];
  for (var i = 0; i < num; i++) {
    newstr.push(str);
  };
  return newstr.join('');
}

repeat("abc", 3, "");

Ok, it's still a simple one but more elegant, it gets rid of the if condition by inserting the empty string as an argument. Since we havn't seen this in the previous Javascript Waypoints, I assume it means a sort of "false default" value. I've tried to google it but since I don't know how this notation style is named, it's very hard to find something relevant. If anyone got any idea feel free to tell in the comments.

Bonfire: Return Largest Numbers in Arrays

So, for this bonfire, you get the comparison operators as a link/hint. Nice troll.

function largestOfFour(arr) {
//create empty array to store the biggest numbers the function will return
  var biggest = [];  

//loop through each main array indexes
  for(var i = 0; i < arr.length; i++){
    //create variable to sub-arrays biggest number the function (yet not written) will return
    var temp = 0;
    biggest[i] = temp;
  }
  return biggest;
}
largestOfFour([[1000, 1001, 857, 1], [4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]);

Now add this.

function largestOfFour(arr) {
  var biggest = [];
  for(var i = 0; i < arr.length; i++){
    var temp = 0;
    for(var j = 0; j &lt arr[i].length; j++){
      if(arr[i][j] &gt temp){
        temp = arr[i][j];
      }
    }
    biggest[i] = temp;
  }
  return biggest;
}
largestOfFour([[1000, 1001, 857, 1], [4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]);

See the Pen ClnAe by Katja Hollar (@afkatja) on CodePen