Similar String Groups

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StringsGraphs

Two strings, X and Y, are considered similar if either they are identical or we can make them equivalent by swapping at most two letters (in distinct positions) within the string X.

For example, "tars" and "rats" are similar (swapping at positions 0 and 2), and "rats" and "arts" are similar, but "star" is not similar to "tars", "rats", or "arts".

Together, these form two connected groups by similarity: {"tars", "rats", "arts"} and {"star"}. Notice that "tars" and "arts" are in the same group even though they are not similar. Formally, each group is such that a word is in the group if and only if it is similar to at least one other word in the group.

We are given a list strs of strings where every string in strs is an anagram of every other string in strs. How many groups are there?

Example 1:

Input: strs = ["tars","rats","arts","star"]
Output: 2

Example 2:

Input: strs = ["omv","ovm"]
Output: 1

Constraints:

1 <= strs.length <= 300
1 <= strs[i].length <= 300
strs[i] consists of lowercase letters only.
All words in strs have the same length and are anagrams of each other.

Try coding it in LeetCode

Similar String Groups

Hard

Asked by:

18 views

Topics:

StringsGraphs

Two strings, X and Y, are considered similar if either they are identical or we can make them equivalent by swapping at most two letters (in distinct positions) within the string X.

For example, "tars" and "rats" are similar (swapping at positions 0 and 2), and "rats" and "arts" are similar, but "star" is not similar to "tars", "rats", or "arts".

Together, these form two connected groups by similarity: {"tars", "rats", "arts"} and {"star"}. Notice that "tars" and "arts" are in the same group even though they are not similar. Formally, each group is such that a word is in the group if and only if it is similar to at least one other word in the group.

We are given a list strs of strings where every string in strs is an anagram of every other string in strs. How many groups are there?

Example 1:

Input: strs = ["tars","rats","arts","star"]
Output: 2

Example 2:

Input: strs = ["omv","ovm"]
Output: 1

Constraints:

1 <= strs.length <= 300
1 <= strs[i].length <= 300
strs[i] consists of lowercase letters only.
All words in strs have the same length and are anagrams of each other.

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 maximum length of a string in the input array, and what is the maximum number of strings in the array?
Are the strings in the input array guaranteed to be of the same length?
What characters can the strings contain (e.g., only lowercase English letters, or can they contain other characters)?
How should I handle an empty input array or an array containing null or empty strings?
If the input array is null, should I throw an exception or return a specific value (e.g., 0)?

Brute Force Solution

Approach

We want to group strings that are 'similar'. A brute force approach simply checks every possible pair of strings to see if they are similar, and keeps track of the resulting groups. It exhaustively compares each string with every other string.

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

Start by considering each string as being in its own separate group.
Pick the first string and compare it to every other string.
If two strings are similar, merge their groups together.
Repeat this process for the second string, the third string, and so on, comparing each string to all the strings that come after it.
After checking all possible pairs, count how many distinct groups you have. This count is the answer.

Code Implementation

def num_similar_groups_brute_force(string_list):
    number_of_strings = len(string_list)
    group_ids = list(range(number_of_strings))

    def find(index):
        if group_ids[index] != index:
            group_ids[index] = find(group_ids[index])
        return group_ids[index]

    def union(index_one, index_two):
        group_id_one = find(index_one)
        group_id_two = find(index_two)
        group_ids[group_id_one] = group_id_two

    def are_strings_similar(string_one, string_two):
        difference_count = 0
        for index in range(len(string_one)):
            if string_one[index] != string_two[index]:
                difference_count += 1
        return difference_count == 0 or difference_count == 2

    # Check every possible pair of strings.
    for first_index in range(number_of_strings):

        for second_index in range(first_index + 1, number_of_strings):
            # If the strings are similar, merge their groups.
            if are_strings_similar(string_list[first_index], string_list[second_index]):
                union(first_index, second_index)

    # Count the number of distinct groups.
    number_of_groups = 0
    for index in range(number_of_strings):
        if group_ids[index] == index:
            number_of_groups += 1

    return number_of_groups

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each of the n strings in the input list. For each string, it compares it with every other string that comes after it in the list. This pairwise comparison involves checking for similarity, which we'll assume takes constant time. Thus, the number of comparisons performed is proportional to n * (n-1) / 2. This simplifies to O(n²), representing the overall time complexity.

Space Complexity

O(N) – The described brute force approach utilizes a data structure to keep track of the groups. In the worst-case scenario where no strings are similar, each string initially forms its own distinct group. To represent these groups, a data structure like a disjoint set (union-find) data structure, implicitly used for merging, might be employed, requiring space proportional to the number of strings, N. Therefore, the auxiliary space complexity is O(N), where N is the number of strings.

Optimal Solution

Approach

We need to figure out how many groups of strings are similar. The optimal approach is to connect similar strings together, forming groups, and then count how many separate groups exist. We can efficiently find these connected groups using a technique that avoids redundant checks.

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

Imagine each string as a point. We need to figure out which points are connected.
Two strings are connected if they are similar (differ by only two characters).
Start with the first string. Find all other strings that are similar to it and connect them together.
Now, for each string you just connected, find any other unconnected strings that are similar to it, and connect those as well.
Keep doing this until you can't connect any more strings to the current group.
This process creates one complete group of similar strings.
If there are any strings left that are not in a group, start a new group with one of those strings and repeat the process.
Once all the strings are in a group, count how many different groups you have. That's the number of similar string groups.

Code Implementation

def similar_string_groups(strings):
    number_of_strings = len(strings)
    visited = [False] * number_of_strings
    number_of_groups = 0

    def are_strings_similar(string1, string2):
        difference_count = 0
        for i in range(len(string1)):
            if string1[i] != string2[i]:
                difference_count += 1
        return difference_count == 0 or difference_count == 2

    def depth_first_search(string_index):
        visited[string_index] = True
        
        # Check other strings if they are similar
        for next_string_index in range(number_of_strings):
            if not visited[next_string_index] and are_strings_similar(strings[string_index], strings[next_string_index]):

                # Recursively explore the connected component
                depth_first_search(next_string_index)

    # Iterate to find all disjoint components
    for string_index in range(number_of_strings):
        if not visited[string_index]:

            # Start a DFS to discover group members
            depth_first_search(string_index)
            number_of_groups += 1

    return number_of_groups

Big(O) Analysis

Time Complexity

O(n² * m) – The algorithm iterates through each of the n strings in the input array. For each string, it compares it with potentially all other n-1 strings to check for similarity. The similarity check itself involves comparing each character, which takes O(m) time, where m is the length of the string. Thus, the overall time complexity becomes O(n * n * m), which simplifies to O(n² * m).

Space Complexity

O(N) – The algorithm implicitly uses a data structure to keep track of which strings have been visited and assigned to a group. In the worst case, we might need to mark each of the N strings as visited before all groups are identified. Therefore, the space needed to store the visited status of each string will be proportional to the number of strings, N. This results in a space complexity of O(N).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 as there are no strings to form groups.
Array with only one string	Return 1 as a single string forms its own group.
All strings in the input are identical	The algorithm should correctly identify all strings as belonging to the same group, resulting in a count of 1.
Maximum string length exceeded causing performance issues	Implement optimizations like early exit if difference between strings exceeds the defined maximum (e.g., 2 for 'similar').
Large number of strings with small length	Ensure the Union-Find or DFS algorithm scales well with a large number of small strings, optimizing memory usage and avoiding stack overflow.
Strings containing characters outside the expected alphabet (e.g., special characters, numbers)	Validate input strings to only contain expected characters and handle invalid characters gracefully (e.g., throw an error or ignore them).
Integer overflow when calculating string differences or comparing string lengths	Use appropriate data types (e.g., long) to prevent overflow during calculations involving string lengths or character differences.
Highly skewed distribution of string similarities, leading to a deeply unbalanced Union-Find tree	Employ path compression and union by rank in the Union-Find implementation to maintain a balanced tree structure and improve performance.