delete character of 2 string to make them anagram in typescript-CodePudding

I tried to solve one of the Hackerrank challenges String: Making Anagrams

I have 2 strings like this:

let a: string = "fcrxzwscanmligyxyvym"
let b: string = "jxwtrhvujlmrpdoqbisbwhmgpmeoke"

and I have a function that only passed 3 tests!!! :

function makeAnagram(a: string, b: string): number {
    type Map = {
        [key :string] : number
    }
   let string1 = a.split('')
   let string2 = b.split('')
   let map : Map = {}
   let deletedCount = 0
   let firstCount = 0
   let isMoreThanTwo = false
   
   for(let i of string1) {
       map[i] = (map[i] | 0)   1
   }
   
   for(let i of string2){
       map[i] = (map[i] | 0)   1
   }
   
   for(let i in map) {
       if(map[i] == 1) {
           deletedCount  
           firstCount  
       } else if(map[i] > 2) {
           isMoreThanTwo = true
           deletedCount  = (map[i] - 1)
       }
       
   }
   
   return isMoreThanTwo ? deletedCount   firstCount : deletedCount

is there any other solution to count deleted characters? and can you guys give me some advice, thank you

CodePudding user response：

I've just solved this problem but before passing all test cases I came to a solution where I used

for (let [k, v] of map) {
  remain  = v;
}

which failed 13/16 test cases, then I debugged the program and came to know that sometimes when we subtract 1 from the previous of this step, It goes less than 0 so Yo also have to handle this case, I handles as

for (let [k, v] of map) {
  remain  = v < 0 ? v * -1 : v;
}

Now, all test cases are passed

function makeAnagram(a, b) {
  const [small, big] = a.length < b.length ? [a, b] : [b, a];
  const map = new Map();
  for (let c of small) {
    map.set(c, (map.get(c) ?? 0)   1);
  }

  let remain = 0;
  for (let c of big) {
    !map.has(c) ? remain   : map.set(c, map.get(c) - 1);
  }

  for (let [k, v] of map) {
    remain  = v < 0 ? v * -1 : v;
  }

  return remain;
}

console.log(makeAnagram("fast", "sofasty"));