I wrote logic for "Wordle" game but i think it's too heavy and not very readable. Is there any way to improve it? Simple wordle game with conditions: 1) Green colour - "G" if string1 and string2 indices values are equal. 2) Yellow colour - "Y" if string2 indices value appears in string1. 3) else statement Grey colour - "." 4) Every letter from string1 used for colour only once. My code:

let mstr1 = readLine()!
let mstr2 = readLine()!

var string1 = mstr1.compactMap { $0.asciiValue }
var string2 = mstr2.compactMap { $0.asciiValue }

var dict: Dictionary<UInt8, Int> = [:]

for element in Set(string1) {
    dict[element] = string1.filter { $0 == element }.count
}

var answer = Array(repeatElement(".", count: string1.count))

for i in 0 ..< string1.count {
    if string1[i] == string2[i] {
        answer[i] = "G"
        dict[string1[i]]! -= 1
    }
}

for i in 0 ..< string1.count {
    if dict[string2[i]] != nil && answer[i] != "G"{
        if dict[string2[i]]! > 0 {
            dict[string2[i]]! -= 1
            answer[i] = "Y"
        }
    }
    
    if dict.values.filter { $0 > 0 }.count <= 0 {
        break
    }
}

print(answer.joined())

Input: 
ABBEY
ALGAE
Output: 
G...Y

I wrote logic for "Wordle" game but I think it's too heavy and not very readable.
Is there any way to improve it?

Simple wordle game with conditions:
1) Green colour - "G" if string1 and string2 indices values are equal.
2) Yellow colour - "Y" if string2 indices value appears in string1.
3) else statement Grey colour - "."
4) Every letter from string1 used for colour only once.

My code:

let mstr1 = readLine()!
let mstr2 = readLine()!

var string1 = mstr1.compactMap { $0.asciiValue }
var string2 = mstr2.compactMap { $0.asciiValue }

var dict: Dictionary<UInt8, Int> = [:]

for element in Set(string1) {
    dict[element] = string1.filter { $0 == element }.count
}

var answer = Array(repeatElement(".", count: string1.count))

for i in 0 ..< string1.count {
    if string1[i] == string2[i] {
        answer[i] = "G"
        dict[string1[i]]! -= 1
    }
}

for i in 0 ..< string1.count {
    if dict[string2[i]] != nil && answer[i] != "G"{
        if dict[string2[i]]! > 0 {
            dict[string2[i]]! -= 1
            answer[i] = "Y"
        }
    }
    
    if dict.values.filter { $0 > 0 }.count <= 0 {
        break
    }
}

print(answer.joined())

Input: 
ABBEY
ALGAE
Output: 
G...Y