Expand compressed string range

Question

Say you have two strings: exp->aaccbb comp-> aa2bb

This example will return true as there are 2 chars between aa and bb. I have written this code snippet and am wondering if there is a better way do solve this problem and if I am missing any edge cases.

Also, what would be the fastest way to solve this problem with respect to time/place complexity? The solution is a 2 pointer approach with linear complexity. Follow up -> either side can have compressions or both at the same time.

const isNumber = (ch)=>{
  return Number(ch) == ch
}

const isExp = (exp, comp)=>{
  let i=j=0;


  while(i<exp.length && j<comp.length){
    if(comp[j] !== exp[i] && !isNumber(comp[j])){
      return false
    }
    else if(isNumber(comp[j])){

     let currentCompIdx = j;

     /*Find consexutive numbers*/
      while(isNumber(comp[currentCompIdx+1]))currentCompIdx++;

      /*Get Number from the compressed string*/
      const moveIndexBy = Number(comp.slice(j,currentCompIdx+1));
            /*set j pointer to index after the number*/
            j=currentCompIdx+1;

      /*Check if the letters after number is same*/
      if(comp[j] !== exp[i+moveIndexBy]){
        return false
      }
     i++;      
     j++;

    }
    else{
      i++;
      j++;
    }
  }
  return true
}

/*Returns true as it has 2 chars between aa and cc*/
alert(isExp("aaccbb", "aa2bb"))

/*Returns true as it has 2 chars between aa and cc and 4 chars between bb and cc*/
alert(isExp("aaccbbqqqqcc", "aa2bb4cc"))

/*Returns true as it has 10 chars between aa and cc */
alert(isExp("aaccbbqqqqsacc", "aa10cc"))

/*Returns false as the exp string as only 1 char between aa and bb */
alert(isExp("aacbb", "aa2bb"))
```

Answer 1

Notice how close the syntax of your compressed strings resemble that of a regular expression:

   aa2bb => aa.{2}bb
aa2bb4cc => aa.{2}bb.{4}cc
  aa10cc => aa.{10}cc

Therefore you can easily create a regexp from your compressed string and test the expanded string against it:

function isExp(exp, comp) {
  const regexp = RegExp(`^${comp.replace(/\d+/g, '.{$&}')}$`)
  return regexp.test(exp);
}