Compressed String decoding for Kth character

Given a compressed string composed of lowercase characters and numbers, the task is to decode the compressed string and to find out the character located at the K^th position in the uncompressed string, You can transform the string by following the rule that when you encounter a numeric value "n", repeat the preceding substring starting from index 0 'n' times

Examples:

Input: S = "ab2c3" K = 10
Output: "c"
Explanation:

• When traversing the string we got the first numeric value 2,
• here the preceding Substring =" ab "
• So "ab" is repeated twice, Now the string Will be "ababc3".
• The second numeric value we got is 3,
• Now, our preceding String is "ababc"
• So it will be repeated 3 times.
• Now the expanded string will be "ababc" + "ababc" + "ababc" = "ababcababcababc"
• 10th character is "c", so the output is "c".

Input: S = "z3a2s1" K = 12
Output: -1
Explanation: Expanded string will be "zzzazzzas" and have a length of 9, so output -1 (as the Kth index does not exist)

Naive Approach: The basic way to solve the problem is as follows:

The idea is to generate the expanded string and then find the K^th value of that string. To do so we will use a string "decoded" and insert the compressed string when encounter a string and if we encounter a digit then we will multiply the string the digit times and store it in "decoded" string .

Below is the implementation of the above idea:

c

Time Complexity: O(N),
Auxiliary Space: O(N), where N is the length of the expanded string.

Efficient Approach: To solve the problem without expanding the string using Recursion :

• Initialize a counter to keep track of the current position in the uncompressed string.
• Iterate through the compressed string's characters:
  • If a character is a digit, calculate the potential new position in the uncompressed string without expanding the string.
  • If the new position is greater than or equal to K, the K^th character lies within the repeated portion. Recurse on that portion with an adjusted K value.
    • If the character is not a digit, increment the counter.
    • If the counter reaches or exceeds k, return the current character as the k-th character.
• If the loop completes without finding the k-th character, return -1 to indicate an out-of-bounds position.

Below is the implementation of the above idea:

c

Time Complexity: O(N), where N is the length of the string.
Auxiliary Space: O(1).