Lexicographically Smallest String After Operations With Constraint

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Greedy AlgorithmsArrays

You are given a string s and an integer k.

Define a function distance(s₁, s₂) between two strings s₁ and s₂ of the same length n as:

The sum of the minimum distance between s₁[i] and s₂[i] when the characters from 'a' to 'z' are placed in a cyclic order, for all i in the range [0, n - 1].

For example, distance("ab", "cd") == 4, and distance("a", "z") == 1.

You can change any letter of s to any other lowercase English letter, any number of times.

Return a string denoting the lexicographically smallest string t you can get after some changes, such that distance(s, t) <= k.

Example 1:

Input: s = "zbbz", k = 3

Output: "aaaz"

Explanation:

Change s to "aaaz". The distance between "zbbz" and "aaaz" is equal to k = 3.

Example 2:

Input: s = "xaxcd", k = 4

Output: "aawcd"

Explanation:

The distance between "xaxcd" and "aawcd" is equal to k = 4.

Example 3:

Input: s = "lol", k = 0

Output: "lol"

Explanation:

It's impossible to change any character as k = 0.

Constraints:

1 <= s.length <= 100
0 <= k <= 2000
s consists only of lowercase English letters.

Try coding it in LeetCode

Lexicographically Smallest String After Operations With Constraint

Medium

Asked by:

2 views

Topics:

Greedy AlgorithmsArrays

You are given a string s and an integer k.

Define a function distance(s₁, s₂) between two strings s₁ and s₂ of the same length n as:

The sum of the minimum distance between s₁[i] and s₂[i] when the characters from 'a' to 'z' are placed in a cyclic order, for all i in the range [0, n - 1].

For example, distance("ab", "cd") == 4, and distance("a", "z") == 1.

You can change any letter of s to any other lowercase English letter, any number of times.

Return a string denoting the lexicographically smallest string t you can get after some changes, such that distance(s, t) <= k.

Example 1:

Input: s = "zbbz", k = 3

Output: "aaaz"

Explanation:

Change s to "aaaz". The distance between "zbbz" and "aaaz" is equal to k = 3.

Example 2:

Input: s = "xaxcd", k = 4

Output: "aawcd"

Explanation:

The distance between "xaxcd" and "aawcd" is equal to k = 4.

Example 3:

Input: s = "lol", k = 0

Output: "lol"

Explanation:

It's impossible to change any character as k = 0.

Constraints:

1 <= s.length <= 100
0 <= k <= 2000
s consists only of lowercase English letters.

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 are the constraints on the length of the string `s` and the value of `k`?
If the string `s` is empty, what should I return?
Is `k` guaranteed to be non-negative? Can `k` ever be greater than the length of `s`?
If there are multiple lexicographically smallest strings that can be obtained, am I allowed to return any one of them?
Are we limited to only the standard ASCII lowercase English alphabet, or could extended ASCII characters be present in the string `s`?

Brute Force Solution

Approach

The brute force approach to this problem is like trying every single possible combination of operations to see which one gives us the smallest string alphabetically. It's about exploring all possibilities without being smart or efficient, simply checking them all one by one.

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

Start with the original string.
Consider all possible operations you can perform on the string according to the rules.
For each of these operations, create a new string by applying that operation to the original.
Now, for each of these new strings, consider all possible operations you can perform on them (again following the rules).
Keep doing this, generating new strings from existing ones by applying all possible operations at each step.
Since we can't do this forever, we need to decide on a maximum number of operations we'll try or some other stopping condition.
Once we've generated a large number of possible strings, compare them all alphabetically.
The string that comes first alphabetically is the lexicographically smallest string we've found.
Return this string as the result.

Code Implementation

def find_lexicographically_smallest_string_brute_force(original_string,
                                                          max_operations):
    all_possible_strings = {original_string}
    queue = [original_string]
    operation_count = 0

    # Limit the search depth to avoid infinite loops.
    while operation_count < max_operations and queue:
        operation_count += 1
        new_strings = set()
        for current_string in queue:
            for index in range(len(current_string) - 1):
                new_string = list(current_string)
                new_string[index], new_string[index + 1] = \
                    new_string[index + 1], new_string[index]
                new_string = "".join(new_string)

                # Avoid revisiting already explored strings.
                if new_string not in all_possible_strings:
                    new_strings.add(new_string)
                    all_possible_strings.add(new_string)
        queue = list(new_strings)

    # Find the lexicographically smallest string among all generated strings.
    lexicographically_smallest_string = min(all_possible_strings)

    return lexicographically_smallest_string

Big(O) Analysis

Time Complexity

O(K * O^K) – Let n be the length of the string and O be the number of possible operations at each step. Since we explore all possible operations up to a maximum depth K, we can think of it as a tree where each node represents a string, and each branch represents applying an operation. At each level, we generate O new strings from each existing string. Thus, at depth K, we can have up to O^K strings, each of which requires O(K) time to generate through string manipulations. Therefore, the total time complexity is O(K * O^K).

Space Complexity

O(N!) or O(M^K) – The brute force approach explores all possible combinations of operations. Each operation creates a new string, and these strings need to be stored for comparison. In the worst-case scenario, the number of possible strings can grow factorially with N, the length of the string (O(N!)), or exponentially such as O(M^K) where M is the number of possible operations that can be performed and K is the number of operations. Therefore, the space complexity is determined by the storage of the generated strings, leading to O(N!) or potentially O(M^K). The auxiliary space grows as more operations are performed and more strings are generated.

Optimal Solution

Approach

The goal is to find the 'smallest' string possible after applying certain allowed changes. We'll analyze the string from left to right, making the best local changes possible to gradually minimize it. This avoids inefficiently exploring every possible combination of changes.

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

Start by examining the initial part of the string.
For each position in the string, check if we can improve the character at that position using the allowed operations, considering the constraint.
If we can make a character smaller without violating the constraint, immediately do it.
Continue moving through the string, always prioritizing smaller characters when making modifications.
By making these small, strategic modifications as we go, we ensure the resulting string is as small as possible in lexicographical order.

Code Implementation

def find_lexicographically_smallest_string(input_string, allowed_changes):
    string_as_list = list(input_string)
    string_length = len(input_string)

    for index in range(string_length):
        for possible_replacement in range(ord('a'), ord(string_as_list[index])):
            replacement_character = chr(possible_replacement)

            # Check if making this replacement is allowed.
            if allowed_changes > 0:
                string_as_list[index] = replacement_character
                allowed_changes -= 1

                # We made the best possible change so stop here
                break

        # Ensures that if we make a change, further attempts are stopped
        else:
            continue
        break

    # Convert modified list of characters back to a string
    smallest_string = "".join(string_as_list)
    return smallest_string

Big(O) Analysis

Time Complexity

O(n^2) – The algorithm iterates through the string of length n from left to right. At each position, it checks whether a 'better' (lexicographically smaller) character can be placed there using the allowed operations and the constraint. This check might involve comparing the current character with other characters within a potentially shrinking window to find a suitable replacement, which in the worst case might be all remaining characters. Thus for each of the n characters we might do up to n operations, leading to approximately n * n/2 comparisons. Therefore the time complexity is O(n^2).

Space Complexity

O(1) – The plain English explanation describes an iterative process that examines the string character by character and modifies it in place. It doesn't mention any auxiliary data structures like arrays, lists, hash maps, or recursion. Therefore, the algorithm's space usage is dominated by a few constant-sized variables for tracking the current position and potentially temporary values during character comparisons or modifications, regardless of the string's length N. Thus, the space complexity is O(1).

Edge Cases

Case	How to Handle
Empty string s	Return an empty string immediately as there's nothing to modify.
k is 0	Return the original string s as no modifications are allowed.
k is equal to or greater than the length of s	Replace all characters with 'a' (or the smallest possible char) since all indices can be modified.
String s contains only the letter 'a'	Return the original string s as no character can be lexicographically smaller.
Long string with k close to the length of s, but not enough to change all letters	The solution efficiently selects the optimal indices to change to obtain the lexicographically smallest result within the k limit.
String s is already lexicographically smallest (e.g., 'aaaab')	Iterate through the string, performing modifications only when a better lexicographical character is possible and within the k limit.
String with repeating characters like 'bbbbba' with a smaller k	Prioritize changing later characters that appear earlier in the string to make the string lexicographically smaller.
Very large string to assess time and space complexity	Ensure the solution uses a time-efficient approach, ideally O(n) or O(n log n), to avoid timeouts and minimizes space complexity.