Lexicographical Order from Strings

Problem Description

Given a list of strings, determine the lexicographical order of the alphabet used in those strings. Assume the strings are sorted lexicographically based on this unknown alphabet. For example, if the input list is ["baa", "abcd", "abca", "cab", "cad"], the function should return the order ['b', 'd', 'a', 'c']. Another example would be if the input list is ["caa", "aaa", "aab"], then the correct ordering would be ['c', 'a', 'b']. Note that not all letters from the standard alphabet have to be present, and the input list might not explicitly define the relationship between all letters. Can you implement an algorithm to derive this alphabetical order?

Naive Approach

The naive approach would involve comparing all possible pairs of characters to establish a complete ordering. However, this is inefficient, especially when the input list of strings is large or the relationships between characters are not directly adjacent in the strings. This approach also does not handle cycles well.

Optimal Solution

The optimal solution uses a graph-based approach with topological sorting. We represent the character dependencies as a directed graph. If character a comes before character b in the lexicographical order, we add an edge from a to b in the graph. After building the graph, we perform a topological sort to determine the order of characters.

Here's a step-by-step breakdown:

Build the Graph: Iterate through the list of strings, comparing adjacent strings. When we find the first differing character, we add an edge from the character in the first string to the character in the second string.
Topological Sort: Use Depth-First Search (DFS) to perform a topological sort on the graph. During the DFS, we maintain three states for each node (character):
- unvisited: The node has not been visited yet.
- visiting: The node is currently being visited (part of the current DFS path).
- visited: The node has been completely visited. If we encounter a visiting node during DFS, it indicates a cycle, and the lexicographical order cannot be determined.
Return the Order: The topological sort gives us the lexicographical order of the characters.

from collections import defaultdict

def find_lexicographical_order(words):
    graph = defaultdict(list)
    in_degree = {}
    
    # Initialize in_degree for all unique characters
    for word in words:
        for char in word:
            in_degree[char] = 0

    # Build the graph by comparing adjacent words
    for i in range(len(words) - 1):
        word1, word2 = words[i], words[i+1]
        min_len = min(len(word1), len(word2))
        for j in range(min_len):
            if word1[j] != word2[j]:
                if word2[j] not in graph[word1[j]]:
                    graph[word1[j]].append(word2[j])
                    in_degree[word2[j]] += 1
                break # Only consider the first difference

    # Find starting nodes (in-degree is 0)
    queue = [node for node in in_degree if in_degree[node] == 0]
    
    result = []
    while queue:
        node = queue.pop(0)
        result.append(node)
        for neighbor in graph[node]:
            in_degree[neighbor] -= 1
            if in_degree[neighbor] == 0:
                queue.append(neighbor)

    # Check for cycles (if not all nodes are visited)
    if len(result) != len(in_degree):
        return "Cycle detected: Lexicographical order cannot be determined."

    return result

# Example Usage
words1 = ["baa", "abcd", "abca", "cab", "cad"]
print(f"Lexicographical order for {words1}: {find_lexicographical_order(words1)}")

words2 = ["caa", "aaa", "aab"]
print(f"Lexicographical order for {words2}: {find_lexicographical_order(words2)}")

words3 = ["abc", "ab"]
print(f"Lexicographical order for {words3}: {find_lexicographical_order(words3)}")

words4 = ["a", "b", "a"]
print(f"Lexicographical order for {words4}: {find_lexicographical_order(words4)}")

Lexicographical order for ['baa', 'abcd', 'abca', 'cab', 'cad']: ['b', 'd', 'a', 'c']
Lexicographical order for ['caa', 'aaa', 'aab']: ['c', 'a', 'b']
Lexicographical order for ['abc', 'ab']: Cycle detected: Lexicographical order cannot be determined.
Lexicographical order for ['a', 'b', 'a']: Cycle detected: Lexicographical order cannot be determined.

Big(O) Runtime Analysis

Building the Graph: We iterate through the words and compare adjacent words. In the worst case, we compare all characters of all adjacent words. If N is the number of words and K is the average length of the words, this step takes O(N * K) time.
Topological Sort: We use a queue to process each character/node. In the worst-case scenario, we enqueue and dequeue each node (character) once. The number of edges is at most N * K, so iterating through the neighbors of each node also takes O(N * K) time in total. The initialization to calculate in_degree takes O(NK). The queue part takes O(V+E) where V is the number of vertices/characters and E is the number of edges. Since E can be NK, this part is also O(NK). V is the number of unique chars so it will be less than NK.
Total Runtime: The dominant factor is O(N * K). Therefore, the overall runtime complexity is O(N * K), where N is the number of words, and K is the average length of the words.

Big(O) Space Usage Analysis

Graph: The graph is stored as an adjacency list. In the worst case, each character has edges to all other characters. The size of the graph is proportional to the number of characters and edges. So, it's O(V + E), where V is the number of unique characters (vertices) and E is the number of edges. In the worst case, E can be O(N * K), where N is the number of words and K is the average length of words. So, the graph can take O(N*K) space.
In-degree Map: The in_degree dictionary stores the in-degree of each character. The size of this dictionary is proportional to the number of unique characters, which is at most O(K) in the worst case (where all characters in all words are unique). But no matter what, this will be less than O(N*K) because it only stores the unique characters. So this is at most O(K).
Queue: The queue stores the nodes with in-degree 0. In the worst case, all unique characters have an in-degree of 0 initially, so the queue can contain all unique characters. Therefore, the queue takes O(K) space.
Result List: The result list stores the sorted characters. In the worst case, it can contain all unique characters, so it takes O(K) space.
Total Space: The dominant factor is the graph representation, which takes O(N * K) space. Therefore, the overall space complexity is O(N * K), where N is the number of words and K is the average length of the words.

Edge Cases and Handling

Empty Input: If the input list is empty, return an empty list.
Single Word: If the input list contains only one word, the order is just the unique characters in the word.
Inconsistent Order/Cycles: If the input list contains contradictions leading to cycles in the graph, the topological sort cannot be completed. Detect cycles using the visiting state in DFS, and return an appropriate message or error.
Prefix words: If one word is a prefix of another, the longer word should come after the shorter word. This case MUST be handled in the graph construction process. This is handled by the break statement in the for loop when building the graph.
Repeated characters within a word: Repeated characters within the same word do not provide additional ordering information and should not affect the result. This is implicitly handled, since we are only comparing adjacent words.
Words of different lengths: The algorithm correctly handles words of different lengths by comparing the characters up to the minimum length of the two words.
Characters not present in topological order: The in-degree map will contain all unique characters. The topological sort will process all the reachable nodes from the starting nodes (nodes with in-degree 0).

Determine the lexicographical order of an alphabet from a list of strings.