Asked in: AMAZON_HACKON
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All Testcases Passed ✔
Solution
#include <iostream>
#include <vector>
#include <map>
#include <algorithm>
// ... rest of solution available after purchaseExplanation
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This problem involves two main parts: first, rearranging the given sequence of books according to specified sorting criteria, and second, querying the total count of vowels (A, E, I, O, U) within a certain segment of this rearranged sequence. To approach this problem, let's break down the key steps and reasoning involved, focusing on the logic rather than implementation details.
1. **Understanding the Input and Goal**
You are given a sequence of books represented by uppercase English alphabets (each alphabet denotes a book type). The length of this sequence is N. Then you have two numbers, L and R, representing a range within the rearranged bookshelf. Your goal is to:
- Rearrange the original sequence of books according to the specified ordering rules.
- Find the total count of vowel-type books (A, E, I, O, U) in the rearranged bookshelf between positions L and R (inclusive).
2. **Rearranging the Bookshelf: Sorting by Frequency and Lexicographic Order**
The first main task is to reorder the books based on frequency and then lex order as a tie-breaker:
- Count the frequency of each distinct book type.
- Sort the book types primarily by their frequency in ascending order (lowest frequency first).
- If two or more book types have the same frequency, sort those types lexicographically in ascending order (alphabetical order).
This sorting will give you the order in which book types should appear on the bookshelf.
3. **Constructing the Rearranged Bookshelf**
Once the book types are sorted by frequency and lex order, the next step is to build the new bookshelf string:
- For each book type in the sorted order, append that book type repeated by its frequency count.
- This will create the final reordered bookshelf, which you will analyze for the query.
For example, if book type 'A' has frequency 2 and book type 'C' has frequency 2, and 'B' has frequency 3, and order is A, C, B, then the new string will be "AACCBBB".
4. **Understanding the Query: L and R Indices on the Reordered Bookshelf**
After rearranging the books, you need to count how many books of vowel types (A, E, I, O, U) exist between indices L and R (1-based indexing) in the rearranged string.
- L and R define a segment within the rearranged bookshelf.
- You want to count the total number of vowels within that segment.
- Note that vowels are only A, E, I, O, and U.
5. **Key Observation About the Counting**
The example clarifies a subtle point: you first identify the book types that fall within the L to R segment in the rearranged bookshelf by positions, then for those book types, you find how many copies exist in the entire original rearranged bookshelf string.
- Let's call the rearranged string 'y' (complete bookshelf).
- The substring y[L:R] consists of some portion of certain book types.
- Identify the distinct book types present in that substring.
- For each vowel book type in that set, count its total frequency (copies) in the entire rearranged string.
- Sum those frequencies to get the answer.
6. **Step-by-Step Thought Process to Solve**
- Read the original list of books.
- Count the frequency of each book type.
- Sort book types by frequency ascending; for ties, by lexicographic order ascending.
- Build the rearranged bookshelf string by concatenating book types repeated by their counts in the sorted order.
- Extract the substring of rearranged string from position L to R.
- Identify the unique book types in that substring.
- For each unique book type, check if it is a vowel.
- For each vowel book type in the substring, add its total frequency from the entire rearranged bookshelf.
- Return the sum as the final answer.
7. **Edge Cases and Constraints**
- If L and R are the same, you only look at one position.
- If there are no vowels in the segment, the output is zero.
- Frequencies can be the same for multiple book types; lex order ensures deterministic sorting.
- The size constraints (N <= 100) allow straightforward frequency counting and sorting without performance concerns.
8. **Validating with Examples**
For the sample test case:
- Frequencies: A=2, B=3, C=2.
- Sorted order by frequency: A(2), C(2), B(3).
- Lex order between A and C because of same frequency puts A before C.
- Reordered bookshelf: AACCBBB.
- L=1, R=2 means substring is "AA".
- Distinct book types in substring: {A}.
- 'A' is a vowel, frequency in entire bookshelf = 2.
- Output: 2.
This matches the provided example output.
9. **Summary of Approach**
The problem boils down to:
- Counting frequencies.
- Sorting book types by frequency and lex order.
- Constructing the rearranged string.
- Finding distinct vowel book types in substring [L, R].
- Summing their total frequencies from the entire rearranged bookshelf.
Following this approach ensures correctness and clarity without unnecessary complexity, and given the constraints, it is efficient and straightforward.
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