How to remove consecutive character from a string which exactly matches k as: "aaabaaa","aaab","abcd" or "aaaba" or "aaabaa

Question

Here is my code:

class RemoveCharacters { func removeKOccurrences(_ data: String, _ k: Int) -> String {
let text = Array(data)
// Get the length let n: Int = text.count
if (n == 0 || k <= 0) { return " " }
var result: String = ""
// Use to count character frequency
var count: [Int] = Array(repeating: 0, count: 256)
var i: Int = 0
while (i < n { count[Int(UnicodeScalar(String(text[i]))!.value)] += 1
i += 1
}
i = 0
while (i < n) {
if (count[Int(UnicodeScalar(String(text[i]))!.value)] != k) {

// Include the resultant character result = result + String(text[i])
}
i += 1
}
return result
}
}
let task: RemoveCharacters = RemoveCharacters()
var d = task.removeKOccurrences("aaabaaa", 1) print(d)

//HERE IS THE TEST CASES:
Test Cases:
"aaabbbc", 3 => "c"
"abbbcddfg", 1 => "bbbdd"
"abcd", 2 => "abcd"
"abcd", 1 => " "
"aaabaaa", 3 => "b"
"aaaab", 3 => "aaaab"
"bbbaabb", 3 => aabb
We need to verify all the cases but out of it 5th and 7th i am unable to verify so help me...m fresher in swift.

Answer 1

See, your program just went above my mind 😅
But the logic is simple

1. You have to input the String and value of k. Add a " " to the end of the string
2. Then you have to take another String variable, suppose New as null and an int variable count=1.
3. Then you run a loop from 1 till ( length of String-1).(use operator <=)
3.1. if(S[i-1]==S[i]) then increment count by 1
3.2. else
3.2.1. if(count != k) then add the substring of that String from (i-count) till i to the variable New
3.2.2. set count=1
4.(outside loop) if New != Null then print the value of variable New else print " "

Answer 2

Ok.tnk u i'll try