node.js – Check if word has an added/changed letter to a previous word

Question

I'm working on a command for my Discord bot that allows for a game to be played, in which the goal is to post words that either change one letter in the previous word, or add a letter. I'm using this function:

function checkDifferentString(str1, str2) {
  let diff = 0;
  if (str1 === str2) return true;
  let lengthDiff = Math.abs(str1.length - str2.length)
  if (lengthDiff > 1) return false;

  for (let i=0; (i<str1.length || i < str2.length);i++) {
    if (diff > 1) return false;
    if (str1.charAt(i) !== str2.charAt(i)) diff++
  }
  if (diff <= 1) return true
  else return false;
}

and it works fine if you change a letter, or add a letter to the end of the word (e.g. mat->math). But if you add a letter in the word (e.g. mat->malt), it says that the word doesn't follow the rules, even though it does. How can I change the function so it also catches for added letters inside the words?

Answer 1

I think this is more easier to understand.

function isChanged(prev, curr) {
  return (prev.length + 1 === curr.length && curr.slice(0, prev.length) === prev) 
    || (prev.length === curr.length && Array.from(prev).filter((ch, idx) => ch != curr[idx]).length === 1)
}

The first condition is true when add one character. The second conditions is true when there is only one different character.

• Updated (I did misread your question)

To check change one letter, use longest common sequence(https://en.wikipedia.org/wiki/Longest_common_subsequence_problem). If two string has same length N and the length of longest common sequence is N-1, it means one letter is changed. If two string has length N and N+1, and the length of longest common sequesnce is N, it means one letter is added.

function lengtOfLCS(s1, s2) {
  // LCS algorithm..
  return result
}
function isChanged(prev, curr) {
  return (prev.length === curr.length && lengtOfLCS(prev, curr) === prev.length - 1)
    || (prev.length+1 === curr.length && LCSlengtOfLCSprev, curr) === prev.length)
}