Wherefore art thou suggestions/improvements

Wherefore art thou suggestions/improvements
0.0 0


Hi all,
Tell us what’s happening:
here is my working solution to the problem. I wanted to know is there a better solution to this test, also recommend what can I improve in my code or any other suggestions
link to my code: https://repl.it/@ridafatima15h1/Wherefore-art-thou
Thank You all in advance

Your code so far

function whatIsInAName(collection, source) {
  //**Declarations and initializations**

  //array to be filled
  var arr = []; 
  // Only change code below this line
  //while loop counters
  let j = 0; //Outer 
  let k = 0; //inner 
  //filling Arrays for  algorithm
  let  sourceKeys = Object.keys(source);
  let sourceValues = Object.values(source);
  let collectionKeys = [];
  let collectionValues = [];
  //arguments lengths
  let collectionLength = collection.length;
  let sourceLength = sourceKeys.length;
  for(let j = 0; j < collectionLength; j++){
    collectionKeys[j] = Object.keys(collection[j]);
    collectionValues[j] = Object.values(collection[j]);
  //variables for algorithm
  let checkKey = 0;
  let checkValue;

  while(j < collectionLength){
    while(k < sourceLength){
      //console.log("j: ", j); //Debugging
      //console.log("k: ", k); //Debugging
      checkKey = collectionKeys[j].indexOf(sourceKeys[k]);
      //console.log("checkKey: ",checkKey); //Debugging
      if(checkKey === -1){
        k = sourceLength;
      checkValue =collectionValues[j][checkKey];
      //console.log("checkValue: ", checkValue); //Debugging
      //console.log("sourceValues[k]: ", sourceValues[k]); //Debugging
      if(checkValue != sourceValues[k]){
        checkKey = -1;
        k = sourceLength;
    if(checkKey != -1){
      //console.log("updated arr:", arr); //Debugging
    k = 0;

  console.log("=> Given Inputs:")
  console.log(" source: ", source);
  console.log(" collection: ", collection);
  console.log("sourceLength: ", sourceLength);
  console.log("collectionLength: ", collectionLength);
  console.log("sourceKeys: ", sourceKeys);
  console.log("sourceValues: ", sourceValues);
  console.log("collectionKeys: ", collectionKeys);
  console.log("collectionValues: ", collectionValues);
  console.log("=> Result Output: ")
  console.log(" final updated arr:", arr); 
  // Only change code above this line
  return arr;

whatIsInAName([{ first: "Romeo", last: "Montague" }, { first: "Mercutio", last: null }, { first: "Tybalt", last: "Capulet" }], { last: "Capulet" });
whatIsInAName([{ "apple": 1, "bat": 2 }, { "apple": 1 }, { "apple": 1, "bat": 2, "cookie": 2 }], { "apple": 1, "cookie": 2 });
whatIsInAName([{ "apple": 1 }, { "apple": 1 }, { "apple": 1, "bat": 2 }], { "apple": 1 });

Your browser information:

User Agent is: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/67.0.3396.79 Safari/537.36.

Link to the challenge:


Very good, very good.

Your solution works, but as you asked, I came up with old-school retro two nested for as following

function whatIsInAName(collection, source) {
  // What's in a name?
  var arr = [];
let  sourceKeys = Object.keys(source);
for(var a=0;a<collection.length;a++){
  var push_it=false;
  var ca=collection[a];
  for(var b=0;b<sourceKeys.length;b++){
  return arr;

Since each entry is not dependent to another index, more faster way is search in parallel which I don’t think JS comes with threading support(don’t know)

For more faster code, sorting both keys in collection entries and source could help.

keep going on great work, happy programming.


Hi thank you for the feedback!
I think your approach is much better. Using bracket notion to access source properties by name made it much simple. My solution cluttered arrays . Hopefully I will improve.


Yes, many stuffs of JS are so fancy for me, I tend to code the low old form.

I’m not sure, but I think accessing elements by index number is faster(should be) than accessing by key. But as I stated, JS devs not take care about performance and these stuffs so much, as it’s not used for heavy duties(and should not).

keep going on great work, happy programming.