algoadvance

Given a string s, sort it in the following manner:

1. Pick the smallest character from s and append it to the result.
2. Pick the smallest character from s which is greater than the last appended character to the result and append it.
3. Repeat step 2 until you cannot pick more characters.
4. Pick the largest character from s and append it to the result.
5. Pick the largest character from s which is smaller than the last appended character to the result and append it.
6. Repeat step 5 until you cannot pick more characters.
7. Repeat the steps from 1 to 6 until you pick all characters from s.

You are required to return the sorted string in the required manner.

Clarifying Questions:

1. Character Set: Is the input string guaranteed to consist only of lowercase English letters?
   • Yes, it’s composed solely of lowercase English letters.
2. Preservation of Duplicates: Should characters that appear multiple times be preserved in the order they appear?
   • Yes, all characters, including duplicates, must be included in the final result as per the given sorting rules.
3. Input Constraints: Are there any constraints on the length of the input string?
   • The string length will be at most (10^5).

Strategy:

1. Character Frequency Count: Use a frequency counter to keep track of the occurrences of each character.

2. Sorted Characters List: Maintain a list of unique characters sorted in ascending order for forward traversal and descending order for backward traversal.

3. Constructing Result: Create a result list and perform the operations according to the steps given in the problem. Alternate between appending characters in ascending and descending order, ensuring you decrease the frequency count and remove empty entries.

Code:

from collections import Counter

def sortString(s: str) -> str:
    # Frequency counter for all characters in the string
    freq = Counter(s)
    
    # Unique characters sorted
    unique_chars_asc = sorted(freq.keys())
    unique_chars_desc = sorted(freq.keys(), reverse=True)
    
    result = []
    
    while freq:
        # Smallest to largest
        for char in unique_chars_asc:
            if char in freq:
                result.append(char)
                freq[char] -= 1
                if freq[char] == 0:
                    del freq[char]
        
        # Largest to smallest
        for char in unique_chars_desc:
            if char in freq:
                result.append(char)
                freq[char] -= 1
                if freq[char] == 0:
                    del freq[char]
    
    return ''.join(result)

# Example usage:
s = "aaaabbbbcccc"
print(sortString(s))  # Output: "abccbaabccba"

Time Complexity:

• Frequency Counter: Counting the frequency of characters takes (O(n)) where (n) is the length of the string.
• Sorting Characters: Sorting the unique characters occurs in (O(k \log k)) where (k) is the number of unique characters (at most 26 for lowercase English letters).
• Result Construction: Constructing the result string involves traversing the string multiple times, thus taking (O(n)).

Therefore, the overall time complexity is approximately (O(n + k \log k)), which simplifies to (O(n)) given that (k) is a constant (26).

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