# Amazon Online Assessment 2021 (OA) - Decode String Frequency

Amazon Web Services (AWS) is working on a new security feature to help encode text. Consider a string that consists of lowercase English alphabetic letters (i.e., [a-z]) only. The following rules are used to encode all of its characters into string s:

• a is encoded as 1, b is encoded as 2, c is encoded as 3, ..., and i is encoded as 9.
• j is encoded as 10#, k is encoded as 11#, l is encoded as 12#, ..., and z is encoded as 26#.
• If there are two or more consecutive occurrences of any character, then the character count is written within parentheses (i.e., (c), where c is an integer denoting the count of consecutive occurrences being encoded) immediately following the encoded character. For example, consider the following string encodings:
• String "abzx" encoded as s = "1226#24#".
• String "aabccc" is encoded as s = "1(2)23(3)".
• String "bajj" is encoded as s = "2110#(2)".
• String "wwxyzwww" is encoded as s = "23#(2)24#25#26#23#(3)".

Given an encoded string s, determine the character counts for each letter of the original, decoded string. Return an integer array of length 26 where index 0 contains the number of 'a' characters, index 1 contains the number of 'b' characters and so on.

Relevant Amazon OA Problems:

### Input

• s: an encoded string

### Output

an array of 26 integers as described

### Examples

#### Example 1:

Input:

1s = 1226#24#

Output: ``

Explanation:

String "abzx" encoded as s = "1226#24#".

### Constraints

• String s consists of decimal integers from 0 to 9, #, and ()s only.
• 1<=length of s<=10^5.
• It is guaranteed that string s is a valid encoded string.
• 2<=c<=10^4, where c is a parenthetical count of consecutive occurrences of an encoded character.

## Title

### Script

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1  >>> a = [1, 2, 3]
2  >>> a[-1]
3  3

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