# Get K-th letter of the decoded string formed by repeating substrings

Given a string S containing letter and digit and an integer K where, and . The task is to return the K-th letter of the new string S’.

The new string S’ is formed from old string S by following steps:
1. If the character read is a letter, that letter is added at end of S’.
2. If the character read is a digit, then entire string S’ repeatedly written d-1 more times in total.

Note: The new string is guaranteed to have less than 2^63 letters.

Examples:

Input: S = “geeks2for2”, K = 15
Output: “e”
Explanation: The new string S’ = “geeksgeeksforgeeksgeeksfor”. The 15th letter is “e”.

Input: S = “a2345”, K = 100
Output: “a”
Explanation: The new string S’=”a” repeated 120 times. The 100th letter is “a”.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Let us take a new string like S’ = “geeksgeeksgeeksgeeksgeeks” and an index K = 22, then the answer at K = 22 is the same if K = 2.
In general, when a string is equal to some word with size length repeated some number of times (such as geeks with size = 5 repeated 5 times), then the answer will be same for the index K as it is for the index K % size.

Using this insight and working backwards, we keep track of the size of the new string S’. Whenever the string S’ would equal some word repeated d times, we can reduce K to K % (lengthof(word)).

We first find the length of the new string S’. After, this we’ll work backwards, keeping track of size: the length of the new string after parsing symbols S[0], S[1], …, S[i].

If we see a digit S[i], it means the size of the new string after parsing S[0], S[1], …, S[i-1] will be (size / toInteger(S[i])). Otherwise, it will be size – 1.

Below is the implementation of above approach:

 // CPP implementation of above approach  #include  using namespace std;     // Function to return the K-th letter from new String.  string K_thletter(string S, int K)  {         int N = S.size();      long size = 0;         // finding size = length of new string S'      for (int i = 0; i < N; ++i) {          if (isdigit(S[i]))              size = size * (S[i] - '0');          else             size += 1;      }         // get the K-th letter      for (int i = N - 1; i >= 0; --i) {          K %= size;             if (K == 0 && isalpha(S[i]))              return (string) "" + S[i];             if (isdigit(S[i]))              size = size / (S[i] - '0');          else             size -= 1;      }  }     // Driver program  int main()  {      string S = "geeks2for2";      int K = 15;         cout << K_thletter(S, K);         return 0;  }     // This code is written by Sanjit_Prasad

 // Java implementation of above approach   class GFG   {      // Function to return the K-th letter from new String.       static String K_thletter(String S, int K)      {          int N = S.length();          long size = 0;             // finding size = length of new string S'           for (int i = 0; i < N; ++i)          {              if (Character.isDigit(S.charAt(i)))              {                  size = size * (S.charAt(i) - '0');              }               else             {                  size += 1;              }          }             // get the K-th letter           for (int i = N - 1; i >= 0; --i)           {              K %= size;              if (K == 0 && Character.isAlphabetic(S.charAt(i)))              {                  return (String) "" + S.charAt(i);              }                 if (Character.isDigit(S.charAt(i)))               {                  size = size / (S.charAt(i) - '0');              }               else              {                  size -= 1;              }          }          return null;      }         // Driver program       public static void main(String[] args)       {          String S = "geeks2for2";          int K = 15;          System.out.println(K_thletter(S, K));      }  }     // This code is contributed by Rajput-Ji

 # Python3 implementation of above approach      # Function to return the K-th letter   # from new String.   def K_thletter(S, K):          N = len(S)       size = 0        # finding size = length of new string S'       for i in range(0, N):           if S[i].isdigit():               size = size * int(S[i])           else:              size += 1        # get the K-th letter       for i in range(N - 1, -1, -1):           K %= size              if K == 0 and S[i].isalpha():              return S[i]              if S[i].isdigit():               size = size // int(S[i])           else:              size -= 1    # Driver Code  if __name__ == "__main__":          S = "geeks2for2"     K = 15        print(K_thletter(S, K))      # This code is contributed   # by Rituraj Jain

 // C# implementation of the above approach  using System;              class GFG   {      // Function to return the K-th letter from new String.       static String K_thletter(String S, int K)      {          int N = S.Length;          long size = 0;             // finding size = length of new string S'           for (int i = 0; i < N; ++i)          {              if (char.IsDigit(S[i]))              {                  size = size * (S[i] - '0');              }               else             {                  size += 1;              }          }             // get the K-th letter           for (int i = N - 1; i >= 0; --i)           {              K %= (int)size;              if (K == 0 && char.IsLetter(S[i]))              {                  return (String) "" + S[i];              }                 if (char.IsDigit(S[i]))               {                  size = size / (S[i] - '0');              }               else             {                  size -= 1;              }          }          return null;      }         // Driver code       public static void Main(String[] args)       {          String S = "geeks2for2";          int K = 15;          Console.WriteLine(K_thletter(S, K));      }  }     // This code has been contributed by 29AjayKumar

Output:
e

Time Complexity: O(N), where N is the length of S.

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