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
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"));