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Find K’th Character of Decrypted String
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# Find K’th Character of Decrypted String

Difficulty: MEDIUM
Contributed By
Avg. time to solve
33 min

Problem Statement

#### You have been given an Encrypted String where repetitions of substrings are represented as substring followed by the count of substrings.

``````Example: String "aabbbcdcdcd" will be encrypted as "a2b3cd3".
``````

#### Note :

``````Input string will always be lowercase characters without any spaces.

If the count of a substring is 1 then also it will be followed by Integer '1'.
Example: "aabcdee" will be Encrypted as "a2bcd1e2"
This means it's guaranteed that each substring is followed by some Integer.

Also, the frequency of encrypted substring can be of more than one digit. For example, in "ab12c3", ab is repeated 12 times. No leading 0 is present in the frequency of substring.

The frequency of a repeated substring can also be in parts.
Example: "aaaabbbb" can also have "a2a2b3b1" as Encrypted String.
``````

#### Input format :

``````The first line of each test case contains an Encrypted String 'S'.

The second line contains the integer value 'K'.
``````

#### Output format :

``````For each test case print, the 'K'th character of Decrypted String obtained from Encrypted String 'S'.
``````

#### Note :

``````You are not required to print the output explicitly, it has already been taken care of. Just implement the function.
``````

#### Constraints :

``````2 <= N <= 10^5
1 <= K <= M
1 <= M <= 10^18

Where 'N' is the length of String 'S' and 'M' is the length of the Decrypted String of 'S'.
All the characters of String 'S' are lowercase English letters.

Time Limit: 1sec
``````

#### Sample Input 1 :

``````a2b3cd3
8
``````

#### Sample Output 1 :

``````c
``````

#### Explanation to Sample Input 1 :

``````S = "a2b3cd3"
Decrypted String of S = "aabbbcdcdcd"
According to 1-based indexing for S, the 8th character is 'c'.
``````

#### Sample Input 2 :

``````ab12c3
20
``````

#### Sample Output 2 :

``````b
``````

#### Explanation to Sample Input 2 :

``````S = "ab12c3"
Decrypted String of S = "ababababababababababababccc"
So 20th character is 'b'.
``````
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