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Lexicographically Smallest Equivalent String

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You are given two strings of the same length s1 and s2 and a string baseStr.

We say s1[i] and s2[i] are equivalent characters.

  • For example, if s1 = "abc" and s2 = "cde", then we have 'a' == 'c', 'b' == 'd', and 'c' == 'e'.

Equivalent characters follow the usual rules of any equivalence relation:

  • Reflexivity: 'a' == 'a'.
  • Symmetry: 'a' == 'b' implies 'b' == 'a'.
  • Transitivity: 'a' == 'b' and 'b' == 'c' implies 'a' == 'c'.

For example, given the equivalency information from s1 = "abc" and s2 = "cde", "acd" and "aab" are equivalent strings of baseStr = "eed", and "aab" is the lexicographically smallest equivalent string of baseStr.

Return the lexicographically smallest equivalent string of baseStr by using the equivalency information from s1 and s2.

Example 1:

Input: s1 = "parker", s2 = "morris", baseStr = "parser"
Output: "makkek"

Example 2:

Input: s1 = "hello", s2 = "world", baseStr = "hold"
Output: "hdld"

Example 3:

Input: s1 = "leetcode", s2 = "programs", baseStr = "sourcecode"
Output: "aauaaaaada"

Constraints:

  • 1 <= s1.length, s2.length, baseStr <= 1000
  • s1.length == s2.length
  • s1, s2, and baseStr consist of lowercase English letters.

How would you implement a function to find the lexicographically smallest equivalent string of baseStr given s1 and s2?

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:

  1. What are the possible characters in the strings `s1`, `s2`, and `baseStr`? Are they all lowercase English letters?
  2. What are the maximum lengths of the strings `s1`, `s2`, and `baseStr`? Is there a limit to the combined length of `s1` and `s2`?
  3. If `s1` and `s2` have conflicting equivalencies (e.g., 'a' is equivalent to 'b' and 'a' is equivalent to 'c', but 'b' and 'c' aren't explicitly linked), how should I resolve these?
  4. Is it possible for `s1` and `s2` to be empty strings?
  5. If a character in `baseStr` is not present in `s1` or `s2`, should it remain unchanged in the output string?

Brute Force Solution

Approach

The brute force approach to finding the lexicographically smallest equivalent string involves systematically trying every possible mapping between characters. We check each mapping to see if it produces an equivalent string and then determine the lexicographically smallest one from those equivalent strings.

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

  1. Consider every possible way to group letters as being equivalent. For example, 'a' and 'b' are the same, 'c' and 'd' are the same, and so on.
  2. For each of these groupings, create a new string based on these equivalencies.
  3. Compare all of these new strings.
  4. Pick the string that comes first alphabetically. This is the lexicographically smallest equivalent string.

Code Implementation

def lexicographically_smallest_equivalent_string_brute_force(source_string_one, source_string_two, base_string):    all_equivalent_strings = []
    # Explore all possible character mappings, a combinatorial problem
    import itertools
    alphabet = 'abcdefghijklmnopqrstuvwxyz'
    for character_mapping in itertools.permutations(alphabet):
        character_map = dict(zip(alphabet, character_mapping))
        # Create a string based on this character mapping
        equivalent_string_one = ''.join(character_map[character] for character in source_string_one)
        equivalent_string_two = ''.join(character_map[character] for character in source_string_two)
        # Check if the strings are equivalent based on current mapping
        if sorted(equivalent_string_one) == sorted(equivalent_string_two):
            mapped_base_string = ''.join(character_map[character] for character in base_string)
            all_equivalent_strings.append(mapped_base_string)
    # If no equivalent strings are found, return original base string
    if not all_equivalent_strings:
        return base_string
    # Find the lexicographically smallest equivalent string
    smallest_equivalent_string = min(all_equivalent_strings)
    return smallest_equivalent_string

Big(O) Analysis

Time Complexity
O((26! * n) * 26 + n)The brute force approach iterates through all possible character mappings. There are 26! (26 factorial) ways to group the 26 letters of the alphabet. For each of these mappings, we create a new equivalent string of length n, costing O(n). After creating the string, we also map the original letters to the lowest lexicographical letter which could be another pass of 26 operations. Finally, we need to compare the n-length string of the given input to the string we derive from the grouping which takes O(n) time, for each grouping. This gives us approximately (26! * n) * 26 + n, which simplifies to O((26! * n) * 26 + n) since 26! dominates.
Space Complexity
O(1)The brute force approach, as described, considers 'every possible way to group letters'. While conceptually this sounds expansive, the provided description doesn't explicitly create and store all possible groupings or equivalent strings in memory simultaneously. There are no explicit auxiliary data structures like arrays, hash maps, or trees mentioned that would scale with the input size (N, where N could be the length of the input string or the number of unique characters). Therefore, the space complexity is considered constant as the algorithm primarily manipulates the input string (or its copies) and doesn't allocate significant additional memory scaling with input size, thereby leading to O(1) space complexity.

Optimal Solution

Approach

The goal is to find the smallest possible string by making characters equivalent. We treat the characters as belonging to groups and merge groups to find the smallest representation for each character. This involves creating connections and finding the ultimate representative for each group, ensuring the smallest character represents the group.

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

  1. Think of each character as belonging to its own group initially.
  2. When two characters are declared equivalent, merge their groups. This means connecting them in a way that shows they belong together.
  3. To find the smallest representation for a character, trace its group connections back to the ultimate parent or representative of its group.
  4. The character that is earliest in the alphabet within that group becomes the new representative of the whole group.
  5. For each character in the input string, replace it with the representative character of its group to form the final, lexicographically smallest string.

Code Implementation

def lexicographically_smallest_equivalent_string(string_one, string_two, base_string):

    parent = {char: char for char in 'abcdefghijklmnopqrstuvwxyz'}

    def find(character):
        # Find the ultimate parent representative of the given character.
        if parent[character] != character:
            parent[character] = find(parent[character])
        return parent[character]

    def union(character_one, character_two):
        parent_one = find(character_one)
        parent_two = find(character_two)

        # Ensure the smallest character becomes the parent.
        if parent_one != parent_two:
            if parent_one < parent_two:
                parent[parent_two] = parent_one
            else:
                parent[parent_one] = parent_two

    for index in range(len(string_one)):
        union(string_one[index], string_two[index])

    result = ''
    for character in base_string:
        # Find the equivalent representative for each character.
        result += find(character)

    return result

Big(O) Analysis

Time Complexity
O(n)The algorithm iterates through the input string s of length n once to find the representative of each character. Finding the representative involves traversing a path in the disjoint set data structure (union-find), which has a near-constant time complexity (amortized O(α(n)), where α(n) is the inverse Ackermann function, which grows extremely slowly and can be considered constant for practical input sizes). The union operations performed when merging equivalent characters are also near-constant time on average. Thus, the overall time complexity is dominated by the single pass through the string s, resulting in O(n).
Space Complexity
O(1)The algorithm uses a data structure to represent groups and their representative characters. In the worst-case scenario, each of the 26 lowercase English letters starts in its own group, requiring storage for 26 representative characters. Merging groups doesn't increase the amount of storage needed, it only updates existing references. Therefore, the auxiliary space is constant, independent of the length of the input strings, which are only read but not copied or significantly transformed in memory.

Edge Cases

CaseHow to Handle
Empty strings for s1, s2, or baseStrReturn baseStr immediately if s1 or s2 are empty, or if baseStr is empty, return an empty string.
s1 and s2 have different lengthsThrow an IllegalArgumentException since equivalent characters are only defined for strings of the same length.
s1 and s2 contain non-lowercase English lettersTreat these characters as distinct and independent, or throw an IllegalArgumentException depending on requirements.
Chains of equivalencies create cycles (e.g., a~b, b~c, c~a)The union-find algorithm correctly handles cycles by ensuring all nodes in a connected component point to the same root.
baseStr contains characters not present in s1 or s2Characters not linked to any other character are left unchanged in the baseStr.
Maximum string length for s1, s2 and baseStr stressing memory usageEnsure the union-find data structure has enough memory allocated to handle the maximum number of characters (26 for lowercase letters); the solution scales linearly with input length if path compression is used in union-find.
All characters in s1 and s2 are the same, creating maximum equivalenceThe union-find algorithm correctly collapses all characters to a single equivalence class.
Integer overflow during calculation of array indices based on character values.Characters can be directly used as array indices if the character set is small and controlled, thus avoiding the risk of overflow when doing calculations.