Simple Fun #215: Properly Closed Bracket Word

Simple Fun #215: Properly Closed Bracket WordCodewars

Approach

Solve the problem

1. return false remainder is 1 of (input)length divide 2

2. slice half(begin to middle) of array for optimizing

3. repeat through 'some()' for optimizing

4. if process-2 array[i] - 97 is 0 input words[words.length - 1 - i] must 26. repeat this process till not match.

   1. if everything is match 'some()' will return false which means everything is counter value('a' <->'z') so return true via '!(not)' operator

   2. if something is not counter value 'some()' will return true and stop repeating immeadiatly, and we return false via '!(not)' operator

Solution

JavaScript

const closedBracketWord = (word) => {
  if (word.length % 2 === 1) {
    return false;
  }
  
  return ![...word.slice(0, word.length / 2)]
    .some((v, i) => {
      return v.charCodeAt(0) - 97 !== 25 - (word[word.length - 1 - i].charCodeAt(0) - 97);
    });
};