Custom Sort String

You are given two strings order and s. All the characters of order are unique and were sorted in some custom order previously.

Permute the characters of s so that they match the order that order was sorted. More specifically, if a character x occurs before a character y in order, then x should occur before y in the permuted string.

Return any permutation of s that satisfies this property.

Example 1:

Input: order = "cba", s = "abcd"

Output: "cbad"

Explanation: "a", "b", "c" appear in order, so the order of "a", "b", "c" should be "c", "b", and "a".

Since "d" does not appear in order, it can be at any position in the returned string. "dcba", "cdba", "cbda" are also valid outputs.

Example 2:

Input: order = "bcafg", s = "abcd"

Output: "bcad"

Explanation: The characters "b", "c", and "a" from order dictate the order for the characters in s. The character "d" in s does not appear in order, so its position is flexible.

Following the order of appearance in order, "b", "c", and "a" from s should be arranged as "b", "c", "a". "d" can be placed at any position since it's not in order. The output "bcad" correctly follows this rule. Other arrangements like "dbca" or "bcda" would also be valid, as long as "b", "c", "a" maintain their order.

Constraints:

• 1 <= order.length <= 26
• 1 <= s.length <= 200
• order and s consist of lowercase English letters.
• All the characters of order are unique.

def customSortString(order: str, s: str) -> str:
    """Sorts the string 's' according to the order defined in 'order'.

    Args:
        order (str): A string representing the custom order of characters.
        s (str): The string to be sorted.

    Returns:
        str: The sorted string.
    """
    # Count the frequency of each character in s
    count = {}
    for char in s:
        count[char] = count.get(char, 0) + 1

    result = ""
    # Append characters from 'order' based on their frequency in 's'
    for char in order:
        if char in count:
            result += char * count[char]
            del count[char]

    # Append remaining characters (not in 'order') to the result
    for char, freq in count.items():
        result += char * freq

    return result


# Example usage:
order = "cba"
s = "abcd"
print(f"Custom sorted string: {customSortString(order, s)}")  # Output: cbad or similar

order = "bcafg"
s = "abcd"
print(f"Custom sorted string: {customSortString(order, s)}")  # Output: bcad or similar

Explanation:

1. Character Frequency Counting:

• We start by creating a dictionary count to store the frequency of each character in the string s. This allows us to efficiently keep track of how many times each character appears.

1. Ordered String Construction:

• We iterate through the order string. For each character in order, we check if it exists in the count dictionary (meaning it's also present in s).
• If the character is present, we append it to the result string as many times as it appears in s (based on the count dictionary).
• After appending, we remove the character from the count dictionary to avoid processing it again.

1. Appending Remaining Characters:

• After processing all characters from order, there might be some characters left in the count dictionary. These are characters that appear in s but not in order.
• We iterate through the remaining characters in the count dictionary and append them to the result string based on their frequency.

1. Returning the Result:

• Finally, we return the result string, which contains the characters of s sorted according to the order specified in order, with any remaining characters appended at the end.

Naive Solution

A naive solution would involve sorting the string s using a custom comparison function that checks the index of each character in the order string. This would involve a sorting algorithm (e.g., quicksort or mergesort) with a potentially costly comparison function.

def customSortStringNaive(order: str, s: str) -> str:
    """Naive solution to sort string 's' according to 'order'."""
    def compare(a, b):
        try:
            index_a = order.index(a)
        except ValueError:
            index_a = float('inf')  # Place unknown chars at the end
        try:
            index_b = order.index(b)
        except ValueError:
            index_b = float('inf')  # Place unknown chars at the end
        return index_a - index_b

    return ''.join(sorted(s, key=cmp_to_key(compare)))

Big(O) Run-time Analysis

The optimal solution has a time complexity of O(N + M), where N is the length of string s and M is the length of string order.

• The first loop iterates through string s to build the count dictionary, which takes O(N) time.
• The second loop iterates through string order, which takes O(M) time.
• The last loop iterates through the remaining characters in the count dictionary, which in the worst case could be O(N) (if none of the characters in s are in order).
• Therefore, the overall time complexity is O(N + M).

The naive solution, which involves sorting, has a time complexity of O(N log N) due to the sorting algorithm used (e.g., quicksort or mergesort).

Big(O) Space Usage Analysis

The optimal solution has a space complexity of O(1), since the count dictionary will store at most 26 characters (all lowercase English letters), which is constant space.

• The count dictionary stores the frequency of each character in s. In the worst case, it can store all unique characters from s, which is limited to 26 (since s consists of lowercase English letters).
• Therefore, the space complexity is O(1).

The naive solution also has a space complexity of O(N) because the sorted function creates a new list with N number of characters from s.

Edge Cases

1. Empty Strings:

• If either order or s is an empty string, the function should handle it gracefully.
• If s is empty, the function should return an empty string.
• If order is empty, the function should return s as is.

1. order Contains Characters Not in s:

• The function should handle cases where order contains characters that are not present in s. In such cases, those characters should be ignored.

1. s Contains Duplicate Characters:

• The function should correctly handle duplicate characters in s by counting their frequencies and appending them to the result accordingly.

1. order Contains Duplicate Characters (Invalid Input):

• The problem statement specifies that all characters in order are unique. However, if the input violates this constraint, the function should still produce a valid output, although the order might not be strictly as intended.

1. Long Strings:

• The function should be efficient enough to handle long strings (up to the constraint limits).