Lexicographically Smallest String After Applying Operations

Medium

Asked by:

10 views

Topics:

Strings

You are given a string s of even length consisting of digits from 0 to 9, and two integers a and b.

You can apply either of the following two operations any number of times and in any order on s:

Add a to all odd indices of s (0-indexed). Digits post 9 are cycled back to 0. For example, if s = "3456" and a = 5, s becomes "3951".
Rotate s to the right by b positions. For example, if s = "3456" and b = 1, s becomes "6345".

Return the lexicographically smallest string you can obtain by applying the above operations any number of times on s.

A string a is lexicographically smaller than a string b (of the same length) if in the first position where a and b differ, string a has a letter that appears earlier in the alphabet than the corresponding letter in b. For example, "0158" is lexicographically smaller than "0190" because the first position they differ is at the third letter, and '5' comes before '9'.

Example 1:

Input: s = "5525", a = 9, b = 2
Output: "2050"
Explanation: We can apply the following operations:
Start:  "5525"
Rotate: "2555"
Add:    "2454"
Add:    "2353"
Rotate: "5323"
Add:    "5222"
Add:    "5121"
Rotate: "2151"
Add:    "2050"
There is no way to obtain a string that is lexicographically smaller than "2050".

Example 2:

Input: s = "74", a = 5, b = 1
Output: "24"
Explanation: We can apply the following operations:
Start:  "74"
Rotate: "47"
Add:    "42"
Rotate: "24"
There is no way to obtain a string that is lexicographically smaller than "24".

Example 3:

Input: s = "0011", a = 4, b = 2
Output: "0011"
Explanation: There are no sequence of operations that will give us a lexicographically smaller string than "0011".

Constraints:

2 <= s.length <= 100
s.length is even.
s consists of digits from 0 to 9 only.
1 <= a <= 9
1 <= b <= s.length - 1

Try coding it in LeetCode

Lexicographically Smallest String After Applying Operations

Medium

Asked by:

10 views

Topics:

Strings

You are given a string s of even length consisting of digits from 0 to 9, and two integers a and b.

You can apply either of the following two operations any number of times and in any order on s:

Add a to all odd indices of s (0-indexed). Digits post 9 are cycled back to 0. For example, if s = "3456" and a = 5, s becomes "3951".
Rotate s to the right by b positions. For example, if s = "3456" and b = 1, s becomes "6345".

Return the lexicographically smallest string you can obtain by applying the above operations any number of times on s.

A string a is lexicographically smaller than a string b (of the same length) if in the first position where a and b differ, string a has a letter that appears earlier in the alphabet than the corresponding letter in b. For example, "0158" is lexicographically smaller than "0190" because the first position they differ is at the third letter, and '5' comes before '9'.

Example 1:

Input: s = "5525", a = 9, b = 2
Output: "2050"
Explanation: We can apply the following operations:
Start:  "5525"
Rotate: "2555"
Add:    "2454"
Add:    "2353"
Rotate: "5323"
Add:    "5222"
Add:    "5121"
Rotate: "2151"
Add:    "2050"
There is no way to obtain a string that is lexicographically smaller than "2050".

Example 2:

Input: s = "74", a = 5, b = 1
Output: "24"
Explanation: We can apply the following operations:
Start:  "74"
Rotate: "47"
Add:    "42"
Rotate: "24"
There is no way to obtain a string that is lexicographically smaller than "24".

Example 3:

Input: s = "0011", a = 4, b = 2
Output: "0011"
Explanation: There are no sequence of operations that will give us a lexicographically smaller string than "0011".

Constraints:

2 <= s.length <= 100
s.length is even.
s consists of digits from 0 to 9 only.
1 <= a <= 9
1 <= b <= s.length - 1

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of possible values for the characters in the string, and what characters are allowed (e.g., lowercase English letters only)?
What are the specific operations that can be applied to the string, and are there any limitations on how many times each operation can be applied?
If multiple lexicographically smallest strings are achievable, should I return any of them, or is there a specific tie-breaking rule?
Is the input string guaranteed to be non-empty, and what should I return if it is empty or null?
Are there any dependencies or relationships between the different operations that might affect the final result?

Brute Force Solution

Approach

The brute force method explores every possible combination of operations to find the smallest string. It's like trying out every conceivable move in a game until you stumble upon the best outcome. This approach guarantees the correct answer but can take a very long time.

Here's how the algorithm would work step-by-step:

Start with the original string.
Consider all possible operations you can apply to the string. These operations might involve swapping characters or reversing sections of the string.
For each of those operations, apply it to the string, creating a new string.
Now, for each of these new strings, again consider all the possible operations and apply them, creating even more strings.
Keep doing this, going through layer after layer of applying operations, until you've exhausted all the possible combinations within a reasonable number of operations.
After generating all these strings, compare them all to each other.
Find the string that comes first alphabetically (lexicographically). This is your smallest string.

Code Implementation

def find_smallest_string_brute_force(original_string):    all_strings = {original_string}    strings_to_explore = {original_string}    max_operations = 3
    for _ in range(max_operations):
        new_strings_to_explore = set()
        for current_string in strings_to_explore:
            # Generate strings by swapping adjacent characters            for index in range(len(current_string) - 1):
                modified_string_as_list = list(current_string)
                modified_string_as_list[index], modified_string_as_list[index + 1] = modified_string_as_list[index + 1], modified_string_as_list[index]
                modified_string = "".join(modified_string_as_list)
                if modified_string not in all_strings:
                    all_strings.add(modified_string)
                    new_strings_to_explore.add(modified_string)
            # Generate strings by reversing substrings            for start_index in range(len(current_string)):                for end_index in range(start_index + 2, len(current_string) + 1):                    substring_to_reverse = current_string[start_index:end_index]
                    reversed_substring = substring_to_reverse[::-1]
                    modified_string = current_string[:start_index] + reversed_substring + current_string[end_index:]
                    if modified_string not in all_strings:                        # Keep track of strings to avoid repeats                        all_strings.add(modified_string)
                        new_strings_to_explore.add(modified_string)
        strings_to_explore = new_strings_to_explore

    # Find the lexicographically smallest string.    smallest_string = min(all_strings)

    return smallest_string

Big(O) Analysis

Time Complexity

O(k^n) – The algorithm explores all possible combinations of operations on the string. Let 'n' be the length of the string and 'k' be the number of possible operations at each step (e.g., different swap locations or reverse intervals). At each level of the search, the number of strings explodes by a factor of 'k'. Since the depth of the search can go up to 'n' operations (where 'n' represents the length of the string and thus a potential limit on useful operations), the total number of strings generated and compared will be on the order of k^n. Therefore, the time complexity is O(k^n).

Space Complexity

O(K^D) – The brute force approach explores all possible operation combinations. Each operation generates new strings, leading to a branching factor of K, where K is the number of possible operations at each step. Since we apply these operations up to a depth of D (a reasonable number of operations), the number of generated strings can grow exponentially. The algorithm needs to store all these intermediate strings to compare them later, resulting in an auxiliary space proportional to K^D.

Optimal Solution

Approach

The goal is to transform a string into the smallest possible version by selectively applying two operations. The key insight is to focus on making the first part of the string as small as possible, as quickly as possible. We can accomplish this by strategically using the allowed operations based on whether the string's length is even or odd.

Here's how the algorithm would work step-by-step:

First, check if the length of the given string is even or odd.
If the length is even, apply the second operation (swapping adjacent characters) such that each pair of characters is ordered from smallest to largest. This will make those portions lexicographically minimal.
If the length is odd, apply the first operation (reversing the entire string) if it helps minimize the overall string lexicographically. Compare the initial character of the original string with the last character. If the last character is smaller, then reverse the string.
After conditionally reversing the string (if it's odd length), apply the second operation (swapping adjacent characters) such that each pair of characters is ordered from smallest to largest. This minimizes the string segment by segment.
Finally, return the resulting lexicographically smallest string.

Code Implementation

def lexicographically_smallest_string(input_string):
    string_length = len(input_string)

    if string_length % 2 == 0:
        string_list = list(input_string)
        for i in range(0, string_length, 2):
            if string_list[i] > string_list[i+1]:
                string_list[i], string_list[i+1] = string_list[i+1], string_list[i]
        return "".join(string_list)

    else:
        #Reversing potentially leads to lexicographical improvement.
        if input_string[0] > input_string[-1]:
            input_string = input_string[::-1]

        string_list = list(input_string)
        #Iterate and ensure the pairs are in lexicographical order
        for i in range(0, string_length - 1, 2):
            if string_list[i] > string_list[i+1]:
                string_list[i], string_list[i+1] = string_list[i+1], string_list[i]

        return "".join(string_list)

Big(O) Analysis

Time Complexity

O(n log n) – If the length of the string is even, we iterate through the string once and sort each adjacent pair of characters. This takes constant time for each pair. If the string length is odd, we compare the first and last characters, which takes O(1) time. Then we sort adjacent pairs, which dominates the complexity. Sorting each adjacent pair can be done in place in constant time but considering the n/2 pairs together with their sorting operations effectively creates a linear sorting scenario that approaches O(n log n). Thus, the overall time complexity is O(n log n).

Space Complexity

O(1) – The algorithm's space complexity is primarily determined by whether the input string's length is even or odd and the adjacent swaps. Regardless of the input string of length N, we perform in-place operations or use temporary variables for comparison and swapping. The space used does not scale with the input size N because we are not creating any additional data structures like arrays or hash maps. Therefore, the auxiliary space used is constant, resulting in a space complexity of O(1).

Edge Cases

Case	How to Handle
Null or empty input string	Return an empty string or handle as an invalid input by throwing an exception, depending on the problem specifications.
String length of 1	If no operations are possible with a single character, return the string as is.
String with all identical characters	The operations may or may not change the string, so the algorithm should still function correctly and return the lexicographically smallest result (which might be the original string).
Operations that cancel each other out (e.g., swap adjacent characters then swap them back)	Ensure the algorithm doesn't get stuck in an infinite loop by tracking visited states or prioritizing lexicographically smaller states first.
Large string length and complex operation rules leading to exponential search space	Employ optimization techniques like pruning, memoization, or dynamic programming if applicable to reduce the search space.
Cases where no operation can produce a lexicographically smaller string	The algorithm should correctly identify that no further improvements are possible and return the best string found so far.
String containing non-alphabetic characters (if the problem assumes only alphabets)	Handle non-alphabetic characters by either filtering them out, treating them as special cases, or throwing an exception if invalid input is provided.
Integer overflow if character values are used in calculations within operations	Use appropriate data types (e.g., long) or modulo operations to prevent integer overflow during calculations.