Edit: This is a variant of the original question, in which instead of removing K number of characters we replace them with any character.

Last week I’ve practiced some codility tests, which turned out a bit more difficult than I expected. For those of you who never had a chance to attend such exams, they consist of 3 tasks of varying difficulty, for which you’re given 90 minutes to complete. In the last test I received tasks which all would be marked as ‘hard’ by a leetcode/hackerrank standard. Please have a look at one of them, which I believe was the most tricky.

Strings with long blocks of repeating characters take much less space if kept in a compressed representation. To obtain the compressed representation, we replace each segment of equal characters in the string with the number of characters in the segment followed by the character(e.x we replace segment “CCCC” with 4C). To avoid increasing the size, we leave the one-letter segments unchanged(the compressed representation of “BC” is the same string – “BC”). For example, the compressed representation of the string “ABBBCCDDCCC” is “A3B2C2D3C”, and the compressed representation of the string “AAAAAAAAAAABXXAAAAAAAAAA” is “11AB2X10A”. Observe that, in the second example, if we replaced the “BXX” segment from the middle of the word with “A”, we would obtain a much shorter compressed representation – “24A”. In order to take advantage of this observation, we decided to modify the compression algorithm. Now before compression, we replace exactly K letters from the input string with letters which would be optimized for the compression algorithm. We would like to know the shortest compressed form that we can generate this way.

Write a function:

class Solution { 
    public int solution(String S, int K) 
    { 
    } 
} 

that given a string S of length N and an integer K, returns the shortest possible length of the compressed representation of S after removing exactly K characters from S.

Example:

Given S = “AAAAAAAAAAABXXAAAAAAAAAA” and K=3, the function should return 3, because after removing “BXX” from S, we are left with “AAAAAAAAAAAAAAAAAAAAA”, which compresses to a representation of length 3 – “21A”.

Requirement:

N is an integer within the range [1...100,000];
K is an integer within the range [0...100,000];
K <= N;
string S consists only of uppercase letters (A-Z)

First of all, I would prefer to avoid starting a discussion on whether or not such tests are good measurement for programming skills. Leaving all doubts aside, the truth is if you want to land a job at one of the top companies, you have to practice a lot, even though writing moderately difficult algorithms probably won’t be a part of your daily routine.

So how to approach this task? I usually start by looking at the boundary conditions:

N is an integer within the range [1...100,000];
K is an integer within the range [0...100,000];
K <= N;
string S consists only of uppercase letters (A-Z)

Ok, we don’t have to consider case sensitivity in char comparisons, fair enough. We may also notice that when given string is less than 3 characters, the result is always going to be equal to its length:

original compressed result
“” “” 0
a a 1
ab ab 2
bb 2b 2

Another thing worth noticing is that when K is equal or 1 less than the string length, the result will be equal to the length of the string expressed as a string + 1:

original k compressed result
aaaaa 0 5a 2
abcde 4 5a 2
abcdefghij 10 10a 3
abcdefghijk 10 11a 3

 

So we start with:

public static int findCompressedLength(string S, int K)
{
    if (S.Length < 3)
        return S.Count();
    else if (K >= S.Count()-1)
        return S.Count().ToString().Length + 1;
}

Boundary cases are done, so now we can focus on the main problem. First, we need to find such permutation of letters, where the sequence of the repeating char is the longest. Possibly we could use a basic brute force solution here with the O(N*K) complexity to calculate K long ranges for all N letters in the string, however, this would surely fail the codility performance criteria. There’s another approach to this type of problems called sliding window or moving window with linear complexity.

How does it work? You start by using two pointers (current position of the char in string and the window start). Then with each loop, you store the char along with its current frequency in a dictionary or hashmap.

if (charsWithFrequencies.ContainsKey(S[i]))
  charsWithFrequencies[S[i]]++;
else
  charsWithFrequencies.Add(S[i], 1);

repetitions = charsWithFrequencies[S[i]];

Then you add a condition which checks of the distance from current index to the beginning of the window – number of the current letter repetitions + 1 (because length is denoted in 0 based convention) is bigger than the number of allowed letter replacement (K). Please take your time to wrap your mind around it.

if (i - windowStart - repetitions + 1 > K)
{
  charsWithFrequencies[S[windowStart]]--;
  windowStart++;
}

If so, you move the window forward and decrease the frequency of the letter, which was placed in a string and the window beginning index.

Great, half of the task is done already. Now, since we know that the algorithm is going to return the longest sequence, each time it’s saving this sequence, let’s save few more properties:

if (i - windowStart + 1 > longestSequence)
  {
    longestSequence = i - windowStart + 1;
    longestSequenceInstance.EndingIndex = i;
    longestSequenceInstance.StartingIndex = i - longestSequence + 1;
    longestSequenceInstance.Letter = S[i];
    longestSequenceInstance.Occurences = longestSequence;
  }

I’ve created a simple struct just for legibility, in which we’ll store the letter with the longest sequence, it’s ending and starting indiced and the number of occurences:

public struct longestSequenceInstance
{
  public char Letter { get; set; }
  public int EndingIndex { get; set; }
  public int StartingIndex { get; set; }
  public int Occurences { get; set; }
}

The rest is quite simple, we replace part of the original string with the longest subsequence:

var workingString = new StringBuilder(S);
    var replaced = workingString.Remove(longestSequenceInstance.StartingIndex, longestSequenceInstance.Occurences - 1)
                                .Insert(longestSequenceInstance.StartingIndex, String.Concat(Enumerable.Repeat(longestSequenceInstance.Letter, longestSequenceInstance.Occurences-1)))
                                .ToString();
int occurences = 1;
  char currentChar;
  var strBuilder = new StringBuilder("");

  for (var i = 1; i < replaced.Length; i++)
  {
    currentChar = replaced[i];

    if (replaced[i - 1] == replaced[i])
        occurences++;
    else
    {
      if (occurences > 1)
        strBuilder.Append(occurences);

        strBuilder.Append(replaced[i - 1]);

        occurences = 1;
    }

    if (i == replaced.Length - 1)
    {
      if (occurences > 1)
        strBuilder.Append(occurences);

        strBuilder.Append(currentChar);
    }
}

var stringFinal = strBuilder.ToString();
return stringFinal.Count();

Calculation complexity equals to O(N) + O(1) + O(N). Space complexity equals to 2N + 2logN.

Here’s the full listing (C#). Please let me know in comments if you found a more efficient way of solving this task.

public static int findCompressedLength(string S, int K)
{
  if (S.Length < 3)
    return S.Count();
  else if (K >= S.Count()-1)
    return S.Count().ToString().Length + 1;

  Dictionary<char, int> charsWithFrequencies = new Dictionary<char, int>();
  var windowStart = 0;
  int longestSequence = 0;
  var repetitions = 0;
  var longestSequenceInstance = new longestSequenceInstance();

  for (var i = 0; i < S.Length; i++)
  {
    if (charsWithFrequencies.ContainsKey(S[i]))
      charsWithFrequencies[S[i]]++;
    else
      charsWithFrequencies.Add(S[i], 1);

    repetitions = charsWithFrequencies[S[i]];

    if (i - windowStart - repetitions + 1 > K)
      {
        charsWithFrequencies[S[windowStart]]--;
        windowStart++;
      }

    if (i - windowStart + 1 > longestSequence)
    {
       longestSequence = i - windowStart + 1;
       longestSequenceInstance.EndingIndex = i;
       longestSequenceInstance.StartingIndex = i - longestSequence + 1;
       longestSequenceInstance.Letter = S[i];
       longestSequenceInstance.Occurences = longestSequence;
    }
}

   var workingString = new StringBuilder(S);
   var replaced = workingString.Remove(longestSequenceInstance.StartingIndex, longestSequenceInstance.Occurences - 1)
                                        .Insert(longestSequenceInstance.StartingIndex, String.Concat(Enumerable.Repeat(longestSequenceInstance.Letter, longestSequenceInstance.Occurences-1)))
                                        .ToString();
            
      int occurences = 1;
      char currentChar;
      var strBuilder = new StringBuilder("");

      for (var i = 1; i < replaced.Length; i++)
      {
        currentChar = replaced[i];

      if (replaced[i - 1] == replaced[i])
         occurences++;
      else
      {
         if (occurences > 1)
           strBuilder.Append(occurences);

         strBuilder.Append(replaced[i - 1]);

         occurences = 1;
      }

      if (i == replaced.Length - 1)
      {
        if (occurences > 1)
          strBuilder.Append(occurences);
 
        strBuilder.Append(currentChar);
      }
    }

    var stringFinal = strBuilder.ToString();
    return stringFinal.Count();
  }
}

public struct longestSequenceInstance
{
  public char Letter { get; set; }
  public int EndingIndex { get; set; }
  public int StartingIndex { get; set; }
  public int Occurences { get; set; }
}

  • Thomas Chadenga

    How does solution(‘abcabcabca’, 9) -> 1, should return 3 return 3? If your repeating string is 9 characters long then the shortest possible string should be A for the string you passed to the function, shouldn’t it?

    • pawel.p

      Hi Thomas, thanks for the question. According to the task rules if we have a string that is 10 characters long and are able to replace at least 9 characters, we can create a string of any character belonging to that string ex. “aaaaaaaaaa”. Indeed it’s only “a” letter, but we need to state the number of characters therefore ”10a” – string which is 3 characters long.

      • Thomas Chadenga

        I think you completely missunderstood the question. The compression of “aaaaaaaaaa” with a K=9 gives you “a” which is 1, not “10a” because we remove K consecutive letters according to the question.

        Likewise for the string “abcabcabca” with K=9 we remove K consecutive letters then we get “a” or “a” which is 1. Or if I can put it in a more obvious way “xbcabcabca” with K=9 then we get “x” or “a” which is 1.

        • pawel.p

          Hi I think you’re right, the original question says to remove the letters. I solved a variant of this question where you replace k consecutive letters with any letter instead of removing them. Thanks, I’ll update the question to avoid confusion.

  • Norris

    I ran the code for “AABCAPAQAA” with K = 4.
    The result was “10A”.
    However the question specifies that we can only replace exactly K consecutive letters.
    Here the replacements are not consecutive.

    • pawel.p

      Hi, thanks for the input, that’s a good catch. I’ve edited the article and will post an update regarding my misinterpretation. 🙂

This site uses Akismet to reduce spam. Learn how your comment data is processed.