Similar String Groups

Identifying Problem Isomorphism

“Similar String Groups” can be approximately mapped to “Number of Islands”.

Here’s the reasoning:

Both problems can be seen as counting the number of disconnected components in a graph. In “Similar String Groups”, each string can be seen as a node in a graph. There is an edge between two nodes if one string can be transformed into another by swapping any two characters. A similar string group, therefore, can be viewed as a connected component in this graph. The problem is then to count the number of these connected components.

Similarly, in “Number of Islands”, each cell in the grid can be seen as a node in a graph. There is an edge between two nodes if they are adjacent and both are land. An island can be viewed as a connected component in this graph, and the problem is to count the number of these connected components.

These problems operate in different domains: one in strings, the other in a 2-dimensional grid, which leads to differences in how adjacency is determined and how the graph is traversed.

“Number of Islands” is simpler as it involves a more straightforward adjacency relationship and typically uses basic depth-first search or breadth-first search to solve. “Similar String Groups” requires an understanding of how to construct the graph from the string transformations and may require more complex data structures and algorithms to efficiently solve.

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from collections import defaultdict

class Solution:
    def numSimilarGroups(self, strs):
        def is_similar(str1, str2):
            diff = sum([str1[i] != str2[i] for i in range(len(str1))])
            return diff == 2 or diff == 0

        def dfs(node):
            visited.add(node)
            for neigh in graph[node]:
                if neigh not in visited:
                    dfs(neigh)

        graph = defaultdict(list)
        for i, str1 in enumerate(strs):
            for j in range(i+1, len(strs)):
                if is_similar(str1, strs[j]):
                    graph[str1].append(strs[j])
                    graph[strs[j]].append(str1)

        visited = set()
        num_groups = 0

        for node in strs:
            if node not in visited:
                num_groups += 1
                dfs(node)

        return num_groups

Problem Classification

The problem lies in the realm of string processing and graph theory.

The ‘What’ components:

  1. List of strings: The problem statement provides us with a list of strings, all of which are anagrams of each other.
  2. Similarity of strings: The problem statement defines two strings as similar if we can make them identical by swapping at most two letters within one of the strings. We need to keep this similarity definition in mind while solving the problem.
  3. Grouping: The problem statement asks us to group similar strings. A group is formed if a string is similar to at least one other string in the group.
  4. Count of groups: The final goal is to return the number of such groups formed.

Classification:

  1. String Processing: As the problem deals with strings and their manipulations (specifically, swapping two letters), it falls under the category of string processing problems.
  2. Graph Theory: The problem can also be seen as a graph problem where each string is a node, and an edge exists between two nodes if the corresponding strings are similar. Therefore, we can solve the problem by finding the number of connected components in the graph.
  3. Union Find: The notion of grouping also leans towards a Union-Find approach. The groups can be considered as disjoint sets, and two strings that are similar should be in the same set.

This problem is a combination of string manipulation, graph theory, and disjoint set / union-find data structure. We need to find an approach that considers all these aspects to provide a solution.

Constraints

Given the nature of the problem and the constraints, there are a few points which could be exploited to devise an efficient solution:

  1. Anagrams: Since every string in strs is an anagram of every other string, this implies that any changes in the positioning of the letters do not affect the equality of two strings. This is an important characteristic that we can use to simplify the problem.

  2. Two-letter swap: The problem states that two strings are considered similar if at most two letters are swapped. This condition can be used to establish the relationship (edges) between different strings (nodes).

  3. Small Alphabet Size: As all strings consist of only lowercase letters, we have a maximum of 26 different characters to consider. This could be advantageous in certain string processing operations or when constructing the graph.

  4. Connected Component: The grouping operation described in the problem is essentially finding connected components in a graph where each node is a string and an edge exists between two nodes if the strings are similar.

  5. Size of input: The size of the strings (1 to 300) and the number of strings (1 to 300) is reasonably limited. This suggests that approaches with quadratic time complexity would still be feasible.

A combination of these insights could guide us towards a suitable strategy to tackle this problem. For example, we could use depth-first search (DFS) or breadth-first search (BFS) to find all connected components in the graph which would represent the groups of similar strings. Additionally, the characteristics of anagrams and the two-letter swap condition could simplify the process of determining if two strings are similar.

Thought Process

The problem statement describes a set of strings which form “groups” according to their similarity - namely, that two strings are “similar” if they can be made equivalent by swapping at most two letters. This suggests that we can treat the strings as nodes in a graph, with edges representing this “similarity” relationship.

This graph representation makes the problem a question about connected components in a graph - specifically, how many connected components exist in the graph. This is because each connected component corresponds to a group of “similar” strings, and the problem asks us to return the number of these groups.

Our thought process, then, can be divided into three main steps:

  1. Represent the problem as a graph: This will involve iterating over the list of strings, comparing each pair of strings to determine if they are “similar”, and if so, adding an edge between them in the graph.

  2. Identify the connected components in the graph: Here, we’ll need to implement a method for traversing the graph and identifying its connected components. A standard algorithm for this is Depth-First Search (DFS).

  3. Count the number of connected components: This is as simple as maintaining a count during the DFS in step 2.

Here’s the Python code for this approach:

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from collections import defaultdict

class Solution:
    def numSimilarGroups(self, strs):
        def is_similar(str1, str2):
            diff = sum([str1[i] != str2[i] for i in range(len(str1))])
            return diff == 2 or diff == 0

        def dfs(node):
            visited.add(node)
            for neigh in graph[node]:
                if neigh not in visited:
                    dfs(neigh)

        graph = defaultdict(list)
        for i, str1 in enumerate(strs):
            for j in range(i+1, len(strs)):
                if is_similar(str1, strs[j]):
                    graph[str1].append(strs[j])
                    graph[strs[j]].append(str1)

        visited = set()
        num_groups = 0

        for node in strs:
            if node not in visited:
                num_groups += 1
                dfs(node)

        return num_groups

In the function is_similar, we’re checking if two strings are similar by counting the number of positions where they differ. If this count is 2 or 0, the strings are similar according to the problem’s definition. In the dfs function, we’re performing a standard Depth-First Search traversal to identify the connected components of the graph.

Finally, in the numSimilarGroups function, we’re building our graph, and then performing a DFS from each node that hasn’t already been visited, incrementing our count of the number of groups each time we start a new DFS. This count is then returned as the result.

Solution Approach and Analysis

Here is a detailed breakdown of a potential approach to solving the problem:

  1. Create a Graph Representation: We treat each string as a node in a graph. A connection or edge is made between two nodes if the two strings are ‘similar’, i.e., they can become identical by swapping at most two letters. This forms the basis of our solution.

    Metaphorically, consider this as creating a social network where each person (string) forms a connection (edge) with another person if they have a similar hobby (can be made identical by swapping two letters).

  2. Find Connected Components: The next step is to find the connected components in the graph. A connected component of a graph is a subgraph in which any two vertices are connected to each other by paths and which is connected to no additional vertices. Each connected component would represent a group of ‘similar’ strings.

    Going back to the social network metaphor, this is like grouping people who are connected directly or through mutual friends because of their similar hobbies.

  3. Count the Connected Components: The number of such connected components would give us the number of groups of similar strings.

    This is like counting the number of different hobby groups in the social network.

The main operations in this approach are creating the graph and finding the connected components, which would primarily depend on the number of strings and their length. If there are more strings or the strings are longer, these operations would take more time.

Let’s go through an example:

Suppose strs = [“tars”,“rats”,“arts”,“star”]

Step 1: We create a graph where each node is a string. An edge is formed between “tars” and “rats”, “tars” and “arts”, “rats” and “arts”. “star” is a separate node with no edges because it is not similar to any of the other strings.

Step 2: We find connected components. {“tars”, “rats”, “arts”} form one component and {“star”} is another component.

Step 3: We count the number of connected components, which is 2 in this case.

So, the output is 2.

Language Agnostic Coding Drills

Here are the core concepts that the provided code solution breaks down into:

  1. Variables and Data Structures (Basic Level): The fundamental concepts of storing information and managing it in structured ways. This includes understanding of lists, dictionaries, sets, and how to manipulate these data structures.

  2. Flow Control (Basic Level): The basic understanding of loops and conditionals is crucial to navigate through the data and make decisions based on certain conditions. For loops and conditional statements (if statements) are used in the provided solution.

  3. Functions and Recursion (Intermediate Level): Functions are reusable pieces of code that perform a specific task. Recursion, a concept where a function calls itself, is an important concept used in Depth-First Search in the solution.

  4. Graph Theory (Advanced Level): A graph is a set of nodes (or vertices) connected by edges. Understanding of graph theory is critical in this problem, as it involves finding connected components in the graph.

  5. Depth-First Search (Advanced Level): Depth-First Search (DFS) is a fundamental algorithm in graph theory used to traverse or search through a graph in a depthward motion, exploring as far as possible along each branch before backtracking. It’s used here to find the connected components in the graph.

In order of increasing difficulty, here is how you could progressively learn and apply these concepts:

  1. Variables and Data Structures: Start by learning how to declare variables and understand the basic data structures. Practice creating and manipulating lists, sets, and dictionaries. Understand when and how to use each structure.

  2. Flow Control: Learn about if statements and for and while loops. Understand how to use these constructs to control the flow of your program, iterate over data structures and apply conditional logic.

  3. Functions and Recursion: Learn how to define your own functions to encapsulate pieces of code that perform a specific task. Understand the concept of recursion, and practice with problems that require a function to call itself.

  4. Graph Theory: Start learning about graphs and their properties. Understand how graphs can be represented in code, typically as adjacency lists or adjacency matrices.

  5. Depth-First Search: Once you have a grasp on graph theory, learn about depth-first search. Practice implementing DFS in different scenarios.

When presented with the problem statement, start by visualizing the problem in terms of a graph. Represent each string as a node and draw an edge between two nodes if the corresponding strings are similar. Implement a helper function to determine if two strings are similar. Once the graph is constructed, use Depth-First Search to find connected components, which correspond to groups of similar strings. For each unvisited node, perform a DFS and increment a count representing the total number of groups. This count is the final result.

Targeted Drills in Python

Sure, let’s break down the problem into several key coding drills:

  1. Variables and Data Structures:
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# Declare variables
count = 0
words = ["star", "rats", "arts", "tars"]

# Declare a set
unique_words = set(words)

# Declare a dictionary
word_connections = {}
  1. Flow Control:
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# Use a for loop to iterate over a list
for word in words:
    print(word)

# Use an if statement to check a condition
if "star" in words:
    print("'star' is in the list of words")
  1. Functions and Recursion:
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# Define a function
def hello(name):
    print(f"Hello, {name}!")

# Call a function
hello("World")
  1. Graph Theory & Depth-First Search:

To simplify this drill, let’s create a basic graph and implement a DFS to traverse it.

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graph = {
  'A' : ['B','C'],
  'B' : ['D', 'E'],
  'C' : ['F'],
  'D' : [],
  'E' : ['F'],
  'F' : []
}

visited = set()

def dfs(visited, graph, node):
    if node not in visited:
        print (node)
        visited.add(node)
        for neighbour in graph[node]:
            dfs(visited, graph, neighbour)

# Driver code
dfs(visited, graph, 'A')

Now, let’s implement problem-specific drills:

  1. String manipulation and similarity check:
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def is_similar(str1, str2):
    count_diff = sum([str1[i] != str2[i] for i in range(len(str1))])
    return count_diff == 2 or count_diff == 0

print(is_similar("tars", "rats"))  # Should return True
print(is_similar("tars", "star"))  # Should return False
  1. Graph construction from a list of strings:
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def build_graph(words):
    graph = {word: [] for word in words}
    for i in range(len(words)):
        for j in range(i + 1, len(words)):
            if is_similar(words[i], words[j]):
                graph[words[i]].append(words[j])
                graph[words[j]].append(words[i])
    return graph

print(build_graph(["tars", "rats", "arts", "star"]))  # Should return the graph
  1. Finding the number of connected components (groups) in the graph:
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def count_groups(graph):
    visited = set()
    count = 0
    for node in graph:
        if node not in visited:
            dfs(visited, graph, node)
            count += 1
    return count

print(count_groups(graph))  # Should return the number of groups

By mastering these drills, one would be able to understand and build the solution for the problem statement. The final solution would consist of using these drills and concepts in an integrated way: declaring necessary data structures, defining the similarity check function, building the graph, and finally running a DFS to count the number of connected components in the graph.

Q&A

For me is hard cause I have no experience with Union Find and I have no quick intuition to solve this. I thought at first I should look for anagrams but in constraints it says they are already anagrams. If I am thinking to swap chars manually to just match words, I really don’t wanna do that. Idk how DFS would check the words… Furthermore for me is hard, cause I have an empty mind approaching this.

It’s perfectly fine to feel overwhelmed when approaching a problem that requires a specific algorithm or data structure that you’re not familiar with. In this case, the problem requires an understanding of Union Find and Depth-First Search (DFS), which are popular techniques used in Graph Theory.

Here’s a more simplified explanation of how to approach this problem:

  1. Anagram Grouping: Given that all words in the input are anagrams of each other, any two words will have the same characters but in different orders. The problem hints that the words can be made similar by swapping at most two characters. It means the words are either already identical (no swaps needed) or we can make them identical by performing at most two swaps.

  2. Graph Representation: The problem can be visualized as a graph where each word is a node, and an edge connects two nodes if the corresponding words are similar. The challenge is to find the number of connected components in the graph. A connected component is a set of nodes in a graph that are connected to each other, but not connected to any other nodes outside that set.

  3. Union Find: Union Find is a data structure that keeps track of a partition of a set into disjoint (non-overlapping) subsets. It’s used to solve the problem of determining the connected components in a graph. Here, we would treat each word as a node of the graph. An edge would be drawn between two nodes if the corresponding words can be made identical by swapping at most two characters. The number of disjoint sets (connected components in the graph) would be the answer.

  4. Depth-First Search (DFS): In case you don’t want to use Union Find, DFS could be another approach to solve this problem. Using DFS, you could traverse the graph and mark all reachable nodes (similar words) as visited, count this as one group, and repeat this process until all words have been visited.

In both methods, the core idea is to build a graph based on the similarity of the words and find the connected components, which represent the groups of words that can be transformed into one another.

Remember that it’s perfectly okay if you don’t solve the problem at the first go. Learning algorithms and data structures is a process, and with practice, you’ll get better at recognizing which techniques to apply in a given situation. You’ve made a great start by asking questions and trying to understand the problem, so keep going!

Test case 60.

strs=[“kccomwcgcs”, “socgcmcwkc”, “sgckwcmcoc”, “coswcmcgkc”, “cowkccmsgc”, “cosgmccwkc”, “sgmkwcccoc”, “coswmccgkc”, “kowcccmsgc”, “kgcomwcccs”] I’m getting groups {kccomwcgcs, kgcomwcccs}, {socgcmcwkc}, {sgckwcmcoc, sgmkwcccoc}, {coswcmcgkc, coswmccgkc}, {cowkccmsgc, kowcccmsgc}, {cosgmccwkc}, which is 6 count, while testcase says there’s 5. What’s wrong?

This problem relies on the concept of “swapping at most two letters (in distinct positions) within the string X”. This is not a straightforward problem of grouping anagrams, but rather finding out how many groups of words can be transformed into each other with at most two swaps.

Looking at your test case, the strings “kccomwcgcs” and “kgcomwcccs” belong to the same group because they can be made identical by swapping at most two characters.

Similarly, the strings “sgckwcmcoc” and “sgmkwcccoc” belong to the same group for the same reason.

By this logic, it appears there are five distinct groups, which corresponds to the expected output of 5.

Here’s how you might group the strings:

  1. “kccomwcgcs”, “kgcomwcccs”
  2. “socgcmcwkc”, “cosgmccwkc”
  3. “sgckwcmcoc”, “sgmkwcccoc”
  4. “coswcmcgkc”, “coswmccgkc”
  5. “cowkccmsgc”, “kowcccmsgc”

Each group can be connected to each other by swapping at most two letters, hence they belong to the same group. And we can see that there are 5 groups here. This might be confusing at first but the key is understanding that we can create similar strings by swapping at most two letters.

Similar for me, I’ve counted many times over manually for this one & still get 6. Guess quickest way to verify is to check already working solution & compare what it does.

Your confusion might arise from how the swapping rule works. Let’s clarify this. Two words are similar if:

  1. They are identical, or
  2. We can swap at most two letters (in distinct positions) within one word to make them equivalent.

Here’s an important note: the swapping rule allows for indirect connections between words, as similarity is a transitive relation (if word A is similar to word B, and word B is similar to word C, then A is similar to C, even if A and C wouldn’t be considered similar on their own).

Considering your example, here are the groups of similar words:

  1. “kccomwcgcs”, “kgcomwcccs”
  2. “socgcmcwkc”, “cosgmccwkc”
  3. “sgckwcmcoc”, “sgmkwcccoc”
  4. “coswcmcgkc”, “coswmccgkc”, “cosgmccwkc”
  5. “cowkccmsgc”, “kowcccmsgc”

You can see that “coswmccgkc” and “cosgmccwkc” are similar by swapping the 4th and 5th characters, but “cosgmccwkc” is also similar to “socgcmcwkc”. Therefore, “socgcmcwkc”, “coswmccgkc”, and “cosgmccwkc” are all in the same group, even though “socgcmcwkc” and “coswmccgkc” themselves would not be considered similar. That’s why there are 5, not 6 groups.

An algorithm to solve this problem could use the Union-Find data structure to join all similar words into the same group. It’s a common approach for problems that involve grouping items based on some form of transitive relation, such as this one.

I ran into this when doing DFS, and for me it was because I had essentially made a directed rather than undirected graph. Consider a case where you have 3 strings A, B, and C. A is similar to C but not B, but B is also similar to C. Thus the directed graph has the edges A->C, B->C. A, B, and C should all be in one group, but if you start from A on a directed graph, you will only reach C, count 1 group, then start at B and count another group.

Long story short, when constructing the graph, you want to make sure that you can go both ways between two nodes that are connected.

Correct, a depth-first search (DFS) approach to solving this problem requires an undirected graph to function correctly. This is because the “similarity” relationship is bidirectional, meaning that if word A is similar to word B, then word B is also similar to word A. Therefore, the graph should be undirected so that traversing from any node can reach all nodes within its connected component.

This issue could be addressed by constructing an undirected graph from the given strings. In this graph, each node would represent a string, and there would be an edge between two nodes if and only if their corresponding strings are similar. This can be done by iterating over each pair of strings and adding an edge if they are similar.

Now, when you perform a DFS starting from any node, it will visit all nodes in the connected component, ensuring that all similar strings are grouped together. This can be done by iterating over each node, performing a DFS if the node hasn’t been visited before, and incrementing the group count. The final group count will be the number of connected components in the graph, i.e., the number of groups of similar strings.

Here’s a Python skeleton code illustrating the idea:

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def numSimilarGroups(strs):
    def isSimilar(A, B):
        # Your implementation here

    def dfs(node):
        # Your implementation here

    n = len(strs)
    graph = [[] for _ in range(n)]
    visited = [0] * n
    groups = 0

    for i in range(n):
        for j in range(i+1, n):
            if isSimilar(strs[i], strs[j]):
                graph[i].append(j)
                graph[j].append(i)

    for i in range(n):
        if not visited[i]:
            dfs(i)
            groups += 1

    return groups

This approach will give you the correct number of groups of similar strings and ensure you visit all nodes in a group, no matter where you start.

{kccomwcgcs, kgcomwcccs} {socgcmcwkc} {sgckwcmcoc, sgmkwcccoc} {coswcmcgkc, *cosgmccwkc, *coswmccgkc} {cowkccmsgc, kowcccmsgc}

Thank you for providing the groups, it seems that your solution is almost correct. The strings “cosgmccwkc” and “coswmccgkc” are indeed similar, as they can be made equivalent by swapping two letters, and hence should be in the same group. However, it seems that “coswcmcgkc” is not similar to “cosgmccwkc” and “coswmccgkc” according to the rules defined in the problem statement, and it should not be grouped with them.

Based on your groups, we have:

  • Group 1: {“kccomwcgcs”, “kgcomwcccs”}
  • Group 2: {“socgcmcwkc”}
  • Group 3: {“sgckwcmcoc”, “sgmkwcccoc”}
  • Group 4: {“coswcmcgkc”}
  • Group 5: {“cosgmccwkc”, “coswmccgkc”}
  • Group 6: {“cowkccmsgc”, “kowcccmsgc”}

There are 6 distinct groups, which contradicts the given test case answer of 5. The discrepancy may be due to a mistake in manually grouping the strings or a misunderstanding of the similarity rules.

It would be helpful to carefully review the rules and make sure each group of strings indeed satisfies the conditions for similarity: they are identical or can be made equivalent by swapping at most two letters (in distinct positions) within the string.

Furthermore, ensure your algorithm correctly implements these rules, which should yield the correct number of groups. If necessary, go through the code step by step with this test case to identify where it may be going wrong.

For me, it was a case of aggressive optimization on my part, such that I wasn’t able to backtrack with my graph search implementation. Actually drawing the case in paper helped me find out the issue, maybe it will work for you as well… To expand upon my comment, assuming you start at 1, and weren’t able to backtrack, you’d get the sets (1,10), (2), (3,6,8), (4,8), (5,9), (7). 8 is being counted twice, because when you reached 8 for the first time, you didn’t go back and check the 4.

Yes, your observation is correct. Backtracking is a crucial part of Depth-First Search (DFS) when applied to this problem. When you encounter a node, you are not only exploring all possible paths going out of this node but also all the nodes that can reach this node. Therefore, not implementing backtracking might lead to a situation where you miss some of the relationships.

As you mentioned, drawing it out on paper can indeed help visualize the problem. For example, in this case, we can represent each string as a node in the graph, and a line between two nodes means they are similar. Then, the problem becomes finding all connected components in the graph, which is a standard application of DFS.

Here is a simple fix to your DFS process:

  1. Create a visited list to keep track of the nodes that have been visited.

  2. When you perform DFS on a node, mark it as visited.

  3. When you want to visit a new node from the current node, you first check whether it has been visited. If it has not been visited, you perform DFS on it.

This way, each node will be visited exactly once, ensuring all relationships are captured while avoiding duplications and infinite loops.

If your code is running pretty slow compared to other submissions and you are puzzled as to why optimizing union find (with path compression, union by rank) makes no difference, it is because the runtime is in fact dominated by checking the similarity of the strings (it runs in O(m) while your unionFind is ~O(1) if optimized).

So to reduce the runtime, it’s s the similarity check you should optimize. One way to do that is to break out of the loop as soon as you find more than 2 differences. It speeds things up considerably for the present testcases (by over a factor 10 in my case)

Absolutely, this is an important point. In this problem, the similarity checking between strings is indeed the operation that is most frequently performed, and therefore it’s likely to be the bottleneck in terms of time complexity.

The similarity checking involves iterating through all the characters in a pair of strings, and counting how many positions have different characters. The worst case scenario is when we iterate through all the characters in both strings, which takes O(m) time if m is the length of the strings.

Given that the operation is performed for each pair of strings, the overall time complexity could go up to O(n^2 * m) in the worst case scenario, where n is the number of strings and m is the length of each string. So any optimization in the similarity checking could significantly speed up the algorithm.

Your suggestion of early termination when more than 2 differences are found is an excellent one. This makes use of the problem’s specific condition that two strings are considered similar if we can make them equivalent by swapping at most two letters. Once we find a third difference, we can conclude that the two strings are not similar and break out of the loop early. This will not change the worst case time complexity, but it will significantly improve the average case, which is what we typically care about in practice.

Hardest part is to conveince yourself that cubic solution is actually feasible…

Indeed, this problem might appear to have a cubic time complexity at first glance. It might seem that you need to compare every pair of strings for similarity (which would be quadratic time complexity) and then for each pair, iterate over all the characters in the strings (which would add another factor, making it cubic time complexity).

However, a cubic solution isn’t really necessary for this problem, nor is it likely to be feasible given the problem constraints (strings of length up to 300 and up to 300 strings). The key insight is to recognize that the problem can be solved using Union Find data structure, a powerful tool for handling disjoint-set queries, which typically has nearly linear time complexity for large inputs.

So, while it’s important to convince yourself that a cubic solution might not be feasible, it’s even more crucial to convince yourself that a more efficient solution is possible and work towards finding it. This often involves identifying the underlying data structures or algorithms that can help, and applying problem-specific optimizations.

Input strs = [“koqnn”,“knnqo”,“noqnk”,“nqkon”] Output 1 Expected 3

can anyone please explain this, how the expected output can be 3

The output should be 1, not 3, for the provided input.

Here’s why:

  • “koqnn” and “knnqo” are similar because we can swap ‘o’ at position 1 and ’n’ at position 3 in “koqnn” to get “knnqo”.
  • “koqnn” and “noqnk” are similar because we can swap ‘k’ at position 0 and ‘o’ at position 1 in “koqnn” to get “noqnk”.
  • “koqnn” and “nqkon” are similar because we can swap ‘k’ at position 0 and ‘o’ at position 1 in “koqnn” to get “nqkon”.

So, all these words are connected and form one group. Hence, the answer should be 1. If the expected output of this testcase in your source is 3, it seems there is a mistake in that source.

Many existing solutions can not pass the following test cases efficiently (usually runs > 200ms for single test case).

2000 strings, all are ‘aaaaaaaaaa’ 2000 strings, each 5 ‘a’ + ‘bcdef’.shuffle() 100 strings, each 195 ‘a’ + ‘bcdef’.shuffle() 20 strings, each 995 ‘a’ + ‘bcdef’.shuffle()

These test cases are specifically designed to stress test the algorithms that solve this problem. They do so by creating inputs that either have a high number of strings, long lengths for each string, or both.

For the first case, all the strings are the same, so the program must not waste time trying to group identical strings, a fast solution would simply identify all strings are similar without extra computation.

For the second and third cases, they are composed of a prefix of ‘a’ followed by a shuffled ‘bcdef’. Only the strings with identical ‘bcdef’ after ‘a’s are similar. An efficient solution would need to reduce unnecessary comparison, for instance, by using a hashing mechanism to group strings with same ‘bcdef’.

The last case is a stress test for string length. Even though there are only 20 strings, each one is very long. Efficient handling of string manipulation and comparison will be crucial here.

An optimal solution should be able to handle these edge cases efficiently, demonstrating a good balance between handling both a large number of strings and long strings. This might involve techniques such as pre-processing, using efficient data structures, or optimizing the comparison mechanism.

Line 1034: Char 9: runtime error: reference binding to null pointer of type ‘int’ (stl_vector.h) SUMMARY: UndefinedBehaviorSanitizer: undefined-behavior /usr/bin/../lib/gcc/x86_64-linux-gnu/9/../../../../include/c++/9/bits/stl_vector.h:1043:9 there is a runtime error in my code how to resolve it

this is my code vector parent; vector rank; int count; void makeset(vector& parent,vector& rank, int n){ for(int i =0;i<n;i++){ parent[i]=i; rank[i]=0; } count = n; } int find(int i) { if (parent[i] != i) { parent[i] = find(parent[i]); } return parent[i]; } void merge(int i, int j) { int pi = find(i); int pj = find(j); if (pi != pj) { if (pi < pj) { parent[pj] = pi; } else { parent[pi] = pj; } count–; } } bool isSimilar(const string& s1, const string& s2) { int diff = 0; for (int i = 0; i < s1.size(); ++i) { if (s1[i] != s2[i]) { diff++; if (diff > 2) { return false; } } } return diff == 2; }

int numSimilarGroups(vector& strs) {

int n = strs.size();
vector<int> parent(n), rank(n);
makeset(parent,rank,n);

for (int i = 0; i < n; ++i) {
    for (int j = i+1; j < n; ++j) {
        if (strs[i] == strs[j] || isSimilar(strs[i], strs[j])) {
            merge(i, j);
        }
    }
}   
return count;

}

Your makeset() function is trying to initialize your parent and rank vectors without properly allocating the space for them first. Even though you passed n to your makeset() function, you haven’t actually resized your vectors to be able to hold n elements.

Your makeset() function should be like this:

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void makeset(vector<int>& parent,vector<int>& rank, int n){
    parent.resize(n);
    rank.resize(n);
    for(int i = 0; i < n; i++){
        parent[i]=i;
        rank[i]=0;
    }
    count = n;
}

This will properly initialize your parent and rank vectors to the correct size, and then you can safely set the elements to their respective values.

In addition, please ensure to define your vectors with a type. For example, you should declare vector<int> parent; instead of vector parent;. This is necessary in C++, as the type information is required at compile time.

how come that [“blw”,“bwl”,“wlb”] should have an output of 1 not 2??

The output should be 1 because all three strings (“blw”, “bwl”, “wlb”) are considered similar according to the problem’s definition of similarity, and therefore belong to the same group. Here’s why:

The problem statement allows us to consider two strings as similar if we can swap at most two letters within one of the strings to make them identical. Let’s see how this applies to our case:

  1. “blw” and “bwl” are similar because we can swap the ’l’ and ‘w’ in “blw” to make it identical to “bwl”.
  2. “bwl” and “wlb” are similar because we can swap the ‘b’ and ‘w’ in “bwl” to make it identical to “wlb”.

Since “blw” is similar to “bwl”, and “bwl” is similar to “wlb”, all three strings are part of the same similarity group, hence the output is 1.

Remember, the problem statement specifies that a word is in the group if it is similar to at least one other word in the group, not necessarily all of them. The relationship of similarity does not have to be directly between each pair of words in the group; it can be indirect, like in this case.

If, like me, you’re too lazy to implement Union-Find, the constraints are small enough to allow you to get away with a naive brute-force (N^2) DFS :V

Indeed, given the problem’s constraints of 1 <= strs.length <= 300 and the string length up to 300 characters, a brute-force depth-first search (DFS) approach can indeed solve this problem within acceptable time limits. This is especially true because we’re allowed to make at most two swaps to determine similarity, which can be checked relatively quickly.

Here is an example of a simple DFS implementation in Python:

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class Solution:
    def numSimilarGroups(self, A):
        def is_similar(a, b):
            return sum([x!=y for x, y in zip(a, b)]) <= 2

        def dfs(i):
            for j in range(len(A)):
                if j not in visited and is_similar(A[i], A[j]):
                    visited.add(j)
                    dfs(j)

        visited = set()
        res = 0
        for i in range(len(A)):
            if i not in visited:
                res += 1
                visited.add(i)
                dfs(i)
        return res

In this implementation, the function is_similar checks if two strings are similar (i.e., they can be made the same by at most two swaps). The function dfs performs a depth-first search starting at a given node (string). The set visited keeps track of the nodes that have already been visited. The main loop goes through all nodes (strings in A) and, if a node has not been visited yet, it increments the result counter and starts a DFS from that node.

While this solution works for this problem, it’s worth noting that it’s not generally advisable to resort to brute-force solutions. They can be inefficient and slow for larger inputs. In many real-world cases, more efficient algorithms like Union-Find can significantly speed up the program. Also, having knowledge of such algorithms and being able to implement them is important for technical interviews and competitive programming.

Since ‘being in the same group’ is an equivalence relation, we can build a graph that represents the relation. Then, the problem is equivalent to finding the number of connected components, which is a graph traversal problem.

Absolutely correct! The problem is indeed about finding the number of connected components in a graph. Each word is a node and an edge exists between two nodes if they are ‘similar’.

Finding connected components is a common problem in graph theory and can be solved with a standard Depth-First Search (DFS) or Breadth-First Search (BFS) algorithm.

Here’s a step-by-step approach on how to do this:

  1. Create the Graph: Create a graph where each node represents a string in the given list. If two strings are similar, connect them with an edge. This step could be optimized, since the comparison of each pair of strings might be computationally expensive.

  2. Initialize Visited Set/Data Structure: We need to keep track of the nodes we have visited to avoid revisiting them. This is crucial to prevent infinite loops in our DFS/BFS.

  3. DFS/BFS to Find Connected Components: Start a DFS/BFS from every unvisited node. This will visit all nodes in the current connected component. The number of times you have to start a new DFS/BFS is equal to the number of connected components.

  4. Count the Connected Components: Count the number of times DFS/BFS is invoked in step 3. This is the answer to the problem.

The DFS/BFS for each node runs in O(N), and we do this for every node, so the total time complexity is O(N^2), where N is the number of strings. This is feasible because N ≤ 300 as per the problem’s constraints.

However, keep in mind that the creation of the graph in step 1 could be optimized since we don’t necessarily need to compare each pair of strings. Instead, we could directly start DFS/BFS and in the process, while visiting a node, determine the next nodes to visit. This could reduce the time complexity significantly, especially when the length of the strings is large.

What is an equivalence relation?

An equivalence relation is a type of relation that satisfies three fundamental properties:

  1. Reflexivity: For all elements ‘a’, ‘a’ is related to itself. Mathematically, it’s expressed as aRa for all ‘a’ in the set. This means every element is similar to itself.

  2. Symmetry: If ‘a’ is related to ‘b’, then ‘b’ is also related to ‘a’. Mathematically, if aRb, then bRa. This means if string ‘a’ is similar to string ‘b’, then ‘b’ is also similar to ‘a’.

  3. Transitivity: If ‘a’ is related to ‘b’ and ‘b’ is related to ‘c’, then ‘a’ is related to ‘c’. Mathematically, if aRb and bRc, then aRc. This means if string ‘a’ is similar to string ‘b’ and ‘b’ is similar to string ‘c’, then ‘a’ is also similar to ‘c’.

In the context of this problem, two strings are considered “related” if they are similar. The similarity relation between strings is an equivalence relation because it is reflexive (a string is similar to itself), symmetric (if string ‘a’ is similar to ‘b’, then ‘b’ is similar to ‘a’), and transitive (if string ‘a’ is similar to ‘b’ and ‘b’ is similar to ‘c’, then ‘a’ is similar to ‘c’).

In other words, the equivalence relation groups the strings into equivalence classes where each class consists of strings that are all similar to each other. The problem asks for the number of such classes.

How is being in the same group considered as an equivalence relation?

“Being in the same group” is considered an equivalence relation because it satisfies the three key properties of an equivalence relation: reflexivity, symmetry, and transitivity.

Here’s how these properties apply to the problem at hand:

  1. Reflexivity: A string is always in the same group as itself. In other words, every string is “similar” to itself. This means that the relationship is reflexive.

  2. Symmetry: If string A is in the same group as string B, then string B is also in the same group as string A. This means that the relationship is symmetric.

  3. Transitivity: If string A is in the same group as string B, and string B is in the same group as string C, then string A is also in the same group as string C. This means that the relationship is transitive.

Therefore, the relationship “being in the same group” is an equivalence relation.

In terms of the problem, two strings are considered to be in the same group if they are similar (either identical or can be made equivalent by swapping at most two letters). The relation of “being in the same group” groups the strings into classes, where each class consists of strings that are all similar to each other. This is why we can construct a graph with the strings as vertices and similarity relations as edges. The connected components of this graph represent the groups of similar strings.

How do you make the logical connection from equivalence relation to representing that relation in a graph?

The concept of an equivalence relation naturally lends itself to representation as a graph.

Let’s break this down:

  1. Vertices: In the context of an equivalence relation, the elements being compared (in this case, the strings) form the vertices (or nodes) of the graph.

  2. Edges: The relation between the elements – in this case, whether two strings are similar – determines whether there’s an edge connecting the corresponding vertices in the graph. If two strings are similar (that is, they are in the same equivalence class), we draw an edge between the vertices representing those strings.

This representation works because the properties of an equivalence relation mirror the properties of connected components in a graph:

  1. Reflexivity: Each vertex is connected to itself (often implicitly, as we usually don’t draw loops in such a graph).

  2. Symmetry: If there’s an edge from vertex A to vertex B, there’s also an edge from B to A. This is the definition of an undirected graph.

  3. Transitivity: If vertex A is connected to vertex B and vertex B is connected to vertex C, then vertex A is also connected to vertex C. In graph terms, all three vertices belong to the same connected component.

So, once we’ve drawn the graph, each connected component represents an equivalence class – in this case, a group of similar strings. The problem is then reduced to finding the number of connected components in the graph.

So any equivalence relation can be represented using graph?

Yes, any equivalence relation can be represented using a graph. This is due to the inherent properties of an equivalence relation: reflexivity, symmetry, and transitivity.

These properties align perfectly with a graph representation:

  • Reflexivity (each element is related to itself) can be represented by self-loops in the graph, though these are often left implicit since they don’t add much useful information.

  • Symmetry (if a is related to b, then b is related to a) corresponds to the graph being undirected: if there’s an edge from node a to node b, then there’s an edge from node b to node a.

  • Transitivity (if a is related to b and b is related to c, then a is related to c) means that all nodes that are connected, directly or indirectly, form a connected component in the graph.

In this way, each connected component of the graph represents an equivalence class under the relation.

This conceptual framework can be a powerful tool for solving problems involving equivalence relations, as it allows you to apply graph algorithms and techniques. However, it’s also worth noting that the specifics of the problem may mean that there’s a more efficient or simpler approach than explicitly constructing the graph.

How to generate the insights?

  1. Understand the problem: The problem involves determining the number of groups of similar strings. Two strings are similar if they are identical or if they can be made identical by swapping two letters. All words in the list are anagrams of each other. The main objective is to find the total number of groups.

  2. Identify relationships: The similarity relation between two strings is key here. If two strings are similar, they belong to the same group. Also, if string A is similar to string B, and string B is similar to string C, then strings A, B, and C all belong to the same group, even if A and C are not directly similar.

  3. Model the problem: The problem can be modeled as a graph, where each node is a string and an edge exists between two nodes if the corresponding strings are similar. Each connected component in this graph represents a group of similar strings.

  4. Identify potential strategies: Strategies for solving this problem might include constructing the graph and then traversing it to count the number of connected components. This could be done using various graph traversal algorithms such as Depth-First Search (DFS) or Breadth-First Search (BFS). Alternatively, a Union-Find data structure could be used to efficiently keep track of the connected components.

What’s wrong with testcase 76/77? I’ve got a Wrong Answer (3 instead of 1 expected), not TLE.

Testcase 76 is a list of one hundred 200 characters strings as “aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaebdfc”

Anybody got the same problem?

The provided test case consists of a list of one hundred 200-character strings that are all anagrams of each other, with only the positions of the last five characters differing among them.

In the problem statement, it is mentioned that two strings X and Y are similar if and only if we can swap two characters in X exactly once to get Y.

Given that all strings in this test case are anagrams, they all have exactly the same characters but possibly in a different order. To make two of these strings identical, we need at most two swaps.

  1. If a character is in the wrong position in string X, we can swap it with the character that is supposed to be in that position.

  2. If that character from step 1 was also in the wrong position in the original string X, it’s now in its correct place in string Y.

  3. If it was in the correct position in X, then it’s now in the wrong position in Y, but we can fix this with one more swap.

Therefore, any two strings in this test case are similar to each other, which means they all belong to the same group. So, the correct answer should indeed be 1, and not 3.

If you are getting 3, there could be an issue with your implementation. Check to make sure you’re identifying similar strings correctly and grouping all similar strings together.

You should be able to swap 2 letters in star to get one of the strings “tars”, “rats”, or “arts”, which is not possible. So, it is not similar. Easy way to understand is, two strings are similar if Hamming Distance <= 2

Yes, that’s a great way to put it! The Hamming distance is a measure used to quantify the difference between two strings of equal length. In this problem’s context, it corresponds to the number of positions at which the corresponding symbols (characters in a string) are different.

As per the problem statement, two strings are similar if we can swap two characters in one string exactly once to obtain the other string. If the Hamming distance is 1 or 2, we can make the strings identical with at most one swap, hence they are similar. But if the Hamming distance is greater than 2, we’d need more than one swap, so the strings are not considered similar.

So, you are absolutely correct. Two strings in this problem are similar if and only if their Hamming distance is less than or equal to 2. Therefore, “star” and “tars” (or “rats”, “arts”) are not similar because their Hamming distance is 4.

Its not that hard, I got 2 TLE and then I realised, there are only lower case letters, You don’t have to swap every index and then count. Just swap and check for those index where s[ i ] != s[ j ]

Yes, you’re correct. The strategy you outlined is a great optimization for this problem. Since the strings consist of only lowercase letters, and we’re allowed to perform at most two swaps to make the strings similar, we can focus only on the positions where the strings differ.

Here’s the optimized approach with these considerations:

  1. Loop through every pair of strings in the input list. For each pair:

    • Initialize a counter for differences at 0.
    • Compare the strings character by character. If the characters at a position differ, increment the counter.
    • If the counter exceeds 2 (meaning more than two swaps are required), stop the comparison and move on to the next pair.
    • If the comparison completes with the counter at 2 or less, union the pair as they are in the same group.

  2. After all pairs are processed, the number of disjoint sets in the union-find structure represents the number of groups.

This solution benefits from short-circuiting the comparison when more than two swaps are needed, reducing unnecessary computation and enhancing the performance. Remember that this is an optimization over the basic approach and it’s crucial to understand the fundamental concept of union-find and how it applies to the problem first.

This problem involves finding the number of groups of similar strings, where two strings are considered similar if they can be transformed into each other by swapping two letters. The strings in the input are anagrams of each other, meaning they have the same length and contain the same letters but in a different order.

Sure, let’s break this down!

  1. The problem is asking us to find how many different groups of strings we can make from a given list. Here, a group consists of strings that are similar.

  2. But what does it mean for two strings to be similar? In this context, if you can turn one string into another by swapping at most two letters within the string, those two strings are considered to be similar.

  3. For example, let’s take the strings “star” and “rats”. We can swap ’s’ with ‘r’ and ’t’ with ‘a’ in the string “star” to get the string “rats”. So, “star” and “rats” are similar.

  4. Here’s another detail: all strings in the list are anagrams of each other. An anagram is a word formed by rearranging the letters of a different word. For example, “listen” is an anagram of “silent”. This means all strings in our list have the same set of characters, just in a different order.

  5. So, we need to group these anagram strings in such a way that each group consists of strings that can become each other by swapping at most two letters.

In simple terms, imagine you have a box of differently arranged LEGO structures made from the same pieces. Your task is to group these structures by the number of small adjustments (two piece swaps) you need to make to transform one structure into another within the group. This problem is a similar task, but with strings of characters instead of LEGO structures.

Does this problem require path compression?

Path compression is not strictly necessary to solve this problem, but it can help optimize the solution.

In the context of this problem, the Union Find algorithm is used to group the strings into sets, or “groups,” of similar strings. The basic Union Find algorithm doesn’t require path compression: it can group elements into sets and check which set an element belongs to using separate union and find operations.

However, without path compression, the find operation can take time proportional to the depth of the tree representing the sets, which in the worst case can be O(N) where N is the total number of strings.

Path compression is an optimization to the find operation. Every time find is called on an element, path compression changes the parent of that element and all its ancestors in the tree to be the root of the tree. This can help flatten the tree, making future find operations faster. On average, the time complexity of the find operation with path compression can be nearly constant, O(1).

In short, while path compression is not absolutely necessary to solve this problem, it can significantly speed up the Union Find operations in certain cases, making the solution more efficient.

In editorials the method which checks for similar words, why its just looking for difference of 2 characters . Should it not swap and compare both words if they are same. For Example abcd azcy diff is 2 but they are not similar. Please correct me if I am wrong ?

You are correct. For two words to be similar, they not only need to have exactly two positions with different characters, but swapping the characters at these positions in one word should make it the same as the other word.

In the example you gave, ‘abcd’ and ‘azcy’, the characters in the 3rd and 4th positions are different. However, if we swap ‘c’ and ’d’ in ‘abcd’, we get ‘abdc’, not ‘azcy’. Therefore, ‘abcd’ and ‘azcy’ are not similar.

The condition of having exactly two positions with different characters is a necessary but not sufficient condition for two words to be similar. In addition to this condition, we also need to check that swapping the characters at these positions makes the two words the same.

This means that in the function that checks for similarity between two words, we need to add a check to ensure that swapping the different characters results in the two words being identical.

To correct the method that checks for similar words, it could be modified like this:

  1. If two words are identical, they are similar.
  2. If two words have more than two positions with different characters, they are not similar.
  3. If two words have exactly two positions with different characters, check if swapping the characters at these positions in one word makes it the same as the other word.

This way, the method will correctly determine whether two words are similar according to the problem’s requirements.

this is a bit of a dirty and unnecessary shortcut they use. The problem states that all the words are garanteed to be anagrams. Under those conditions, if there are 2 differences, it necessarily means that the swap is possible, because the set of letters is the same for both words. Note that your example doesn’t satisfy the anagram condition I don’t understand why they made this assumption tho. It is an unnecessary loss of generality because explicitly checking the swap is not really more complicated

Yes, you’re absolutely right. The requirement that all input strings are anagrams of each other is a critical detail that simplifies the problem.

Because all strings are anagrams (i.e., they contain the exact same letters, just arranged differently), if two strings have exactly two differing characters at the same positions, you can always swap those characters in one string to match the other string. This condition allows the problem to be solved more efficiently, since we don’t have to actually perform the swap and comparison; we can simply count the differences.

However, you’re also correct that this is a specific constraint of this problem. In a more general problem where the strings aren’t guaranteed to be anagrams, we would indeed need to verify that a swap results in identical strings.

As for why this constraint was included, it’s a design choice by the problem’s authors, likely aimed at keeping the problem solvable within a certain level of difficulty and performance requirements.