Javascript Solution Find the Difference

Problem Overview

There are two string s and t.

String t is generated by random shuffling string s and then add one more letter at a random position.

We have to return added letter.

For exact problem visit Leetcode Link;

Sample Input Output

Input: s = "abcd", t = "abcde"
Output: "e"
Explanation: 'e' is the letter that was added.

Javascript Solution

Essentially we will be taking advantage of the Bitwise XOR ^ operator. For every NUMBER X, we have

X^X=0 and X^0=X So the plan is:

Iterate over the shortest array s
Keep accumulating (XOR ^ing) the elements of both s and t, after we have converted them to Numbers with CharCodeAt() so the XOR can work. Elements that appear both in t and s will give us zero ( X^X=0), yet that zero will not affect the total sum (X^0=X)
Return the converted back to character final sum, which is essentially the (desired) character not present in s

Original Solution Link GeorgeChryso;

const findTheDifference = function(s, t) {
    let sum=0 // initialize the accumulator

	// t.length > s.length because there will always be 1 extra letter on t (the one we want to return)
    for (let i = 0; i < s.length; i++) {
		sum^=                       // accumulate with XOR
		t[i].charCodeAt()^          // the converted to Number t[i]
        s[i].charCodeAt()           // along with a converted to Number s[i],

    }
    sum^=t[i].charCodeAt()          // add the remaining letter of t
   return String.fromCharCode(sum)  // return the converted to Character total

};

Quick example:
s='abcd' , t='dcbqa'

Same Algorithm Implementation Using ES6

/**
 * @param {number} x
 * @return {number}
 */
var findTheDifference = function (s, t) {
  return String.fromCharCode(
    [...s, ...t].reduce((acc, c) => {
      return acc ^ c.charCodeAt(0)
    }, 0)
  )
}