String methods vs. Array methods when comparing strings (Big O)

Question

Let's say that we're comparing two strings, that have a special character that causes something unusual to occur — in this case, the # character deletes the previous character.

Here's one solution that builds out strings and then compares them, using the string methods slice and concat:

const checkSame = (a, b) => {
  let one = ''
  for(let i = 0; i < a.length; i++){
    if(a[i] === '#'){
      one.slice(0, -1)
    } else {
      one.concat(a[i]);
    }
  }
  let two = ''
  for(let i = 0; i < b.length; i++){
    if(b[i] === '#'){
      one.slice(0, -1)
    } else {
      one.concat(b[i]);
    }
  }
  return one === two
};

Here's another solution that uses arrays, and their methods push and pop:

export const compareTwoStrings2 = (a, b) => {
  let one = [];
  for (let i = 0; i < a.length; i++) {
    if (a[i] === "#") {
      one.pop();
    } else {
      one.push(a[i]);
    }
  }
  let two = [];
  for (let i = 0; i < b.length; i++) {
    if (b[i] === "#") {
      two.pop();
    } else {
      two.push(b[i]);
    }
  }
  for(let i = 0; i < one.length; i++){
      if(one[i] !== two[i]) return false;
  }
  return true;
};

The second solution uses an additional loop which makes me think that if given a string of n characters it would take O(3n) time (worst case), and the first would only take O(2n) time.

Is this correct? Is concatenating and slicing off a string more efficient than using arrays? Or does the final comparison of the two strings also take n time as the strings grow in length?

hgb123 · Accepted Answer · 2021-02-04 02:36:46Z

1

O(3n) ~ O(2n) ~ O(n) so basically the worst-case complexity of these two are the same

Ref: Big O notation > Multiplication by a constant

answered Feb 4, 2021 at 2:36

hgb123

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