javascript - Compare two string and count letters that are different - Stack Overflow

I have to write a function that takes in two strings and pares them then returns the number of letters in the strings that are different. For example "ABC" and "DEC" should return 4. My function always turns up one short because of how I am paring them. I've looked but can't seem to find a fix.

I have tried looping through the string without splitting and ended up with the same problem.

function makeAnagram(a, b) {
    let result = 0;

    let a1 = a.split("").sort();
    let b1 = b.split("").sort();
    for(let i = 0; i < b1.length; i++){
        if(a1.indexOf(b1[i] < 0)){
            result += 1;
        }
    }


    return result;
 }

I have to write a function that takes in two strings and pares them then returns the number of letters in the strings that are different. For example "ABC" and "DEC" should return 4. My function always turns up one short because of how I am paring them. I've looked but can't seem to find a fix.

I have tried looping through the string without splitting and ended up with the same problem.

function makeAnagram(a, b) {
    let result = 0;

    let a1 = a.split("").sort();
    let b1 = b.split("").sort();
    for(let i = 0; i < b1.length; i++){
        if(a1.indexOf(b1[i] < 0)){
            result += 1;
        }
    }


    return result;
 }

• javascript
• arrays
• string

Share Improve this question Follow edited Nov 6, 2019 at 9:31 Gowri Pranith Kumar 1,68522 gold badges1212 silver badges2222 bronze badges asked Nov 6, 2019 at 9:21 TEbogoTEbogo 22922 gold badges33 silver badges1212 bronze badges 7

• 1 Your closing bracket is in the wrong place... a1.indexOf(b1[i] < 0) should be a1.indexOf(b1[i]) < 0. Also, there's no need to do the .sort() – freefaller Commented Nov 6, 2019 at 9:25
• 1 how the heck "ABC" and "DEC" should return 4 when there are 3 letters in each word.... – Gibor Commented Nov 6, 2019 at 9:25
• @Gibor - because they're after different letters – freefaller Commented Nov 6, 2019 at 9:26
• @gibor they have C in mon, so the difference is ABDE, which is 4 long. – Matt Ellen Commented Nov 6, 2019 at 9:26
• 1 May I know what would be your result if ABCC and DEC ? Same 4 or 5? – Narandhran Thangavel Commented Nov 6, 2019 at 9:34

 |  Show 2 more ments

4 Answers 4

Sorted by: Reset to default 5

In one line:

("ABC"+"DEC").split('').sort().join('').replace(/(.)\1+/g, "").length

Returns

4

Steps of the program:

1. ("ABC"+"DEC") makes a string with the 2 merged words : ABCDEC

2. ("ABC"+"DEC").split('').sort().join('') makes the characters sorted in the string: ABCCDE. This will enable us to find duplicates easily with regex

3. replace(/(.)\1+/g, "") removes all sequences of 2+ characters, then we get ABDE

4. .length counts the remaining characters, which are the ones with single occurence.

You can do:

Edited as suggestion by @freefaller

const makeAnagram = (a, b) => {
  const arr1 = a.split('')
  const arr2 = b.split('')
  const diff1 = arr1.filter(letter => !arr2.includes(letter))
  const diff2 = arr2.filter(letter => !arr1.includes(letter))
  
  return diff1.length + diff2.length
}

console.log(makeAnagram('ABC', 'DEC'))

An ES6 way to do the same

const makeAnagram = (a, b) => new Set(a + b).size - new Set([...a].filter(x => b.includes(x))).size;
console.log(makeAnagram('ABC', 'DEC')); // prints 4

This is what I would do

Implementation

let makeAnagram = (a,b) => {
  let clubStr = ('' + a).concat(b);
  let sortedStr = clubStr.trim().split('').sort().join('');
  let unmonStr = sortedStr.replace(/(\w)\1+/gi, '');
  return unmonStr.length;
};

You can do same in one liner.

Caller : makeAnagram('ABC', 'DCE')

Ouput : 4