Alice decides on an encryption scheme involving two large strings where encryption is dependent on the minimum number of character deletions required to make the two strings anagrams. Can you help her find this number?

Given two strings, X and X, that may or may not be of the same length, determine the minimum number of character deletions required to make X and X anagrams. Any characters can be deleted from either of the strings.

It is guaranteed that and consist of lowercase English alphabetic letters (i.e., through ).

Print a single integer denoting the number of characters you must delete to make the two strings anagrams of each other.

cde

abc

4

// We delete the following characters from our two strings to turn them into anagrams of each other:

// Remove d and e from cde to get c.

// Remove a and b from abc to get c.

// We must delete characters to make both strings anagrams, so we print on a new line.

```
process.stdin.resume();
process.stdin.setEncoding('ascii');
var input_stdin = "";
var input_stdin_array = "";
var input_currentline = 0;
process.stdin.on('data', function (data) {
input_stdin += data;
});
process.stdin.on('end', function () {
input_stdin_array = input_stdin.split("\n");
main();
});
function readLine() {
return input_stdin_array[input_currentline++];
}
function main() {
var a = readLine();
var b = readLine();
/////////////// Start Code Here ////////////////////
aOne = a.split("");
aTwo = b.split("");
// aThree = [];
counter = 0;
// If you reverse engineer, you will find that the number of times it takes to remove elements is the sum the lengths of the arrays, minus the amount of correct unique number combinations.
if (aOne.length < aTwo.length){
for (let j = 0; j < aOne.length; j ++) {
for (let i = 0; i < aTwo.length; i ++ ){
if (aOne[j] === aTwo[i]) {
counter ++;
aTwo[i] = 0; //ensures elements are not counted twice
break;
}
}
}
console.log(aTwo.length + aOne.length - (counter * 2));
} else {
for (let i = 0; i < aTwo.length; i ++) {
for (let j = 0; j < aOne.length; j++) {
if (aTwo[i] === aOne[j]) {
counter++;
aOne[j] = 0; //ensures elements are not counted twice
break;
}
}
}
console.log(aOne.length + aTwo.length - (counter * 2));
}
}
```