(Algorithms in Javascript) Practice. A precedence rule is given as "P>E", which means that letter "P" is followed by letter "E". Write a function, given an array of precedence rules, that finds the word represented by the given rules. Note: Each represented word contains a set of unique characters, i.e. the word does not contain duplicate letters.

practice-precendence-rule.js

/*
we create 2 separate arrays of letters and count 
the number of characters resulting from the 
original precedence array.

we look up the index of each letter from first letter
array and follow the index of the next letter.
*/
function findWord(a){
  console.log(a);
  let split = [];
  let len = a.length;
  let original = a;

  
  let firstLetter = [],
      secondLetter =[],
      currentIndex = 0,
      count = {},
      letter;
  
  //count the number of repetitions for each letter
  while(currentIndex<len){
    firstLetter.push(a[currentIndex].charAt(0));
    secondLetter.push(a[currentIndex].charAt(2));
    
    recordLetter(count, a[currentIndex].charAt(0), a[currentIndex].charAt(2));
    //console.log(count);
    currentIndex++;
  }
  
  //console.log(firstLetter, secondLetter, count);
 
  //The first letter should be in firstLetter array 
  //and has count of 1.
  let first;
  for(let c in count){
    if(count[c] === 1){
      if(firstLetter.indexOf(c) >= 0) first = c;
    }
  }
  
  
  let result = first;
  currentIndex = firstLetter.indexOf(first);
  let times = 0;
  while(times < len){
    result += secondLetter[currentIndex];
    currentIndex = firstLetter.indexOf(secondLetter[currentIndex]);
    
    times++;
  }

  console.log(result);
  return result;
}


function recordLetter(count, letter1, letter2){

  count[letter1] = count[letter1] ? count[letter1]+1 : 1;
  count[letter2] = count[letter2] ? count[letter2]+1 : 1;
  
  return count;
}

findWord(["P>E", "E>R","R>U"]) // PERU
findWord(["I>N", "A>I","P>A","S>P"]) // SPAIN
findWord(["U>N", "G>A", "R>Y", "H>U", "N>G", "A>R"]) // HUNGARY
findWord(["I>F", "W>I", "S>W", "F>T"]) // SWIFT
findWord(["R>T", "A>L", "P>O", "O>R", "G>A", "T>U", "U>G"]) // PORTUGAL
findWord(["U>N", "G>A", "R>Y", "H>U", "N>G", "A>R"]) // HUNGARY
findWord(["I>F", "W>I", "S>W", "F>T"]) // SWIFT
findWord(["R>T", "A>L", "P>O", "O>R", "G>A", "T>U", "U>G"]) // PORTUGAL
findWord(["W>I", "R>L", "T>Z", "Z>E", "S>W", "E>R", "L>A", "A>N", "N>D", "I>T"]) // SWITZERLAND 

I've just found your repo in a Google search. I was trying to solve this one. Here is my answer in Ruby.

def find_word(arr)
  # find initial string
  last_letters = arr.map { |str| str[2] }
  first_letters = arr.map { |str| str[0] }
  initial = first_letters - last_letters
  initial_str = arr.filter { |str| str.start_with?(*initial) }.first
  
  # add first string to new_arr and delete it from the original arr
  new_arr = []
  new_arr.push(initial_str[0], initial_str[2])
  arr.delete(initial_str)

  # iterate, get next char and add to new_arr
  while arr.length >= 1
    arr.each do |str|
      if str.start_with?(new_arr.last)
        new_arr << str[2]
        arr.delete(str)
      end
    end
  end

  return new_arr.join
end

I was asked to write a solution for this problem in just 25 minutes. But, I could not get selected by Toptal. It took me one day to solve the problem partially. I will post coding using java soon.