839 Similar String Groups LeetCode: Python, C++, Java Solutions

The LeetCode problem 839 Similar String Groups challenges us to find the number of distinct groups of strings where two strings belong to the same group if they are "similar" (can be made identical by swapping exactly two characters) or if they are transitively similar. This comprehensive tutorial will guide you through solving the 839 Similar String Groups leetcode question in python, C++, java, detailing the underlying algorithms and providing complete code implementations. You'll learn how to leverage the Disjoint Set Union (DSU) data structure to efficiently tackle this grouping problem, understanding its core concepts, practical application, and performance considerations across multiple programming languages.

Understanding the Problem: 839 Similar String Groups

The problem 839 Similar String Groups asks us to count the number of distinct groups among a given list of strings. The core definition revolves around "similarity." Two strings are considered similar if we can swap exactly two characters in one of the strings to make it equal to the other. Importantly, similarity is transitive: if string A is similar to B, and B is similar to C, then A, B, and C all belong to the same group. All strings in the input array strs have the same length.

Let's break down the key aspects:

1. String Length: All strings in the input array strs will have the same length. This is a crucial constraint that simplifies comparisons.

2. Similarity Definition: Two strings, s1 and s2, are similar if:

   • They are identical (s1 == s2). In this case, zero swaps are needed, which can be interpreted as swapping two identical characters.
   • They differ by exactly two characters, and those two differing characters can be swapped to make the strings identical. For example, "abc" and "bac" are similar (swap 'a' and 'b'). "ab" and "ba" are similar. "abc" and "bca" are not similar because it would require more than one swap to transform "abc" into "bca" (e.g., 'a' to 'b', 'b' to 'c', 'c' to 'a' - a cyclic shift, not a single swap).
   • Consider "abac" and "aabc". Differing positions are index 1 ('b' vs 'a') and index 2 ('a' vs 'b'). If we swap 'b' and 'a' at these positions, "abac" becomes "aabc". So, they are similar.
   • Consider "abcde" and "abxye". Differing positions are index 2 ('c' vs 'x') and index 3 ('d' vs 'y'). If we swap 'c' and 'd' in "abcde", it becomes "abdce". This is not "abxye". These are not similar.

3. Transitivity: This is the hallmark of grouping problems. If A is similar to B, and B is similar to C, then A, B, and C all belong to the same group. This forms equivalence classes, which are perfectly suited for a Disjoint Set Union (DSU) data structure.

Example Scenarios:

Let's look at some examples to clarify similarity:

• s1 = "tars", s2 = "rats":
  • Differences: s1[0] ('t') vs s2[0] ('r'), s1[1] ('a') vs s2[1] ('a'), s1[2] ('r') vs s2[2] ('t'), s1[3] ('s') vs s2[3] ('s').
  • Differing characters are 't' at index 0 and 'r' at index 2 in s1, and 'r' at index 0 and 't' at index 2 in s2.
  • If we swap s1[0] and s1[2] ('t' and 'r'), s1 becomes "rats". They are similar.
• s1 = "love", s2 = "evil":
  • Differences: All four characters are different. This would require more than one swap to make them identical (or rather, the characters at the differing positions are not a simple two-way swap). Not similar.
• s1 = "abc", s2 = "abc":
  • No differences. They are identical, thus similar.

The goal is to count how many such independent "groups" or "connected components" exist after considering all possible similar pairs.

Prerequisites: Graph Theory and Disjoint Set Union (DSU)

This problem, at its core, is a graph problem. Each string can be thought of as a node in a graph. An edge exists between two nodes (strings) if they are similar. Since similarity is transitive, this problem asks us to count the number of connected components in this implicitly defined graph. The most efficient data structure for solving connected components problems, especially when edges are added dynamically, is the Disjoint Set Union (DSU), also known as Union-Find.

Graph Representation:

Vertices: Each unique string in the input array strs is a vertex. If there are N strings, there are N vertices.

Edges: An undirected edge exists between two vertices strs[i] and strs[j] if strs[i] and strs[j] are similar.

Goal: Count the number of connected components in this graph.

What is Disjoint Set Union (DSU)?

DSU is a data structure that keeps track of a set of elements partitioned into a number of disjoint (non-overlapping) subsets. It provides two primary operations:

1. find(element): Determines which subset a particular element belongs to. It returns a "representative" (or "root") of that element's subset.
2. union(element1, element2): Merges the subsets containing element1 and element2 into a single subset.

DSU is typically implemented using an array, parent, where parent[i] stores the parent of element i. If parent[i] == i, then i is the root of its set.

To optimize DSU for better performance, two heuristics are commonly used:

• Path Compression (for find operation): When find is called on an element i, it traverses up the parent array until it reaches the root. Path compression modifies the parent array during this traversal. As it goes up, it makes every node it visits point directly to the root. This flattens the tree structure, making future find operations on these nodes much faster.

  Algorithm for find(i) with Path Compression:

  1. If parent[i] == i, then i is the root; return i.
  2. Otherwise, set parent[i] = find(parent[i]) (recursively find the root and update the current node's parent).
  3. Return parent[i].

• Union by Rank or Size (for union operation): To keep the trees flat and balanced, when performing a union operation, we attach the root of the smaller tree to the root of the larger tree. This prevents skewed trees that could degrade performance.

  • Union by Rank: rank[i] stores an upper bound on the height of the tree rooted at i. When merging two trees, the tree with the smaller rank is attached to the root of the tree with the larger rank. If ranks are equal, one is arbitrarily chosen as the new root, and its rank is incremented.
  • Union by Size: size[i] stores the total number of nodes in the tree rooted at i. When merging, the tree with fewer nodes is attached to the root of the tree with more nodes. The size of the new root is updated by summing the sizes of the two merged trees. Union by size is often slightly simpler to implement and provides similar performance benefits.

  Algorithm for union(i, j) with Union by Size:

  1. Find the roots of i and j: root_i = find(i), root_j = find(j).
  2. If root_i != root_j (meaning they are in different sets):
     • If size[root_i] < size[root_j], swap root_i and root_j to ensure root_i always points to the larger set's root.
     • Set parent[root_j] = root_i.
     • Update size[root_i] += size[root_j].
     • Return true (sets were merged).
  3. Else (root_i == root_j): Return false (they were already in the same set).

Why DSU for this problem?

The problem asks for groups, which are essentially connected components. We iterate through all pairs of strings. If two strings are similar, we "union" their respective sets. After checking all pairs, the number of disjoint sets remaining will be the number of similar string groups. DSU efficiently manages these merges and queries for set membership. The optimized DSU operations (path compression and union by rank/size) provide nearly constant time complexity on average for find and union operations (specifically, O(alpha(N)) where alpha is the inverse Ackermann function, which grows extremely slowly, practically constant for any realistic N).

Core Algorithm Steps for 839 Similar String Groups

To solve the 839 Similar String Groups problem using DSU, we follow these steps:

1. Initialize Disjoint Set Union (DSU) Structure

We start by initializing a DSU structure for N strings, where N is the number of strings in the input array strs. Each string initially belongs to its own unique set.

• parent array: parent[i] = i for all i from 0 to N-1. This means each element is its own root.
• size array (for union by size optimization): size[i] = 1 for all i. Each set initially contains one element.
• num_groups: Initialize this counter to N, as initially, every string forms its own group. This counter will decrement each time a successful union operation merges two distinct groups.

2. Iterate Through All Pairs of Strings

We need to check every possible pair of strings to determine if they are similar. Since similarity is symmetric (s1 similar to s2 implies s2 similar to s1), we only need to check each pair once.

• Use nested loops: Outer loop i from 0 to N-1, inner loop j from i+1 to N-1. This ensures each unique pair (strs[i], strs[j]) is considered exactly once.

3. Check for Similarity Between String Pairs

For each pair (strs[i], strs[j]), we need a helper function, let's call it areSimilar(s1, s2), that returns true if the strings are similar according to the problem's definition, and false otherwise.

areSimilar(s1, s2) Logic:

1. Initial check: If s1.length() != s2.length(), they cannot be similar. (Though the problem states all strings have the same length, it's good practice).
2. Count differences: Iterate through both strings, character by character. Store the indices where characters differ. Let's say these indices are stored in a list diff_indices.
3. Analyze diff_indices:
   • If diff_indices is empty: The strings are identical. They are similar. Return true.
   • If diff_indices.size() is 2: This is the critical case. Let the differing indices be idx1 and idx2.
     • Check if s1[idx1] == s2[idx2] AND s1[idx2] == s2[idx1]. This condition confirms that swapping s1[idx1] and s1[idx2] (or s2[idx1] and s2[idx2]) would make the strings identical. If this holds, they are similar. Return true.
   • In any other case (diff_indices.size() is 1 or greater than 2): They are not similar. Return false. A single difference means one character needs to change, not a swap. More than two differences means more than one swap is needed.

4. Perform Union Operation if Similar

If areSimilar(strs[i], strs[j]) returns true, it means strs[i] and strs[j] belong to the same group. We then perform a union(i, j) operation on our DSU structure.

• The union operation will merge the sets containing i and j. If they were already in the same set, union does nothing (or returns false).
• If they were in different sets and are now merged, the union operation effectively reduces the total number of disjoint groups by one. This is where num_groups is decremented.

5. Count Distinct Parents (or num_groups counter)

After iterating through all pairs and performing union operations where necessary, the num_groups counter will hold the final count of distinct similar string groups. This is because each union operation on previously distinct sets correctly decrements the count of total groups.

Alternatively, if not using a num_groups counter during union, you can iterate through the parent array at the end and count how many elements are their own parents (parent[i] == i). Each such element represents the root of a distinct group. This approach is slightly less efficient if N is very large, but conceptually equivalent. Using the num_groups counter within the union method is generally preferred for its simplicity and directness.

Implementing the Solution: Python

Let's put these steps into action with Python code. We'll define a DSU class and a helper function for similarity checking.

Helper Function: areSimilar(s1, s2)

This function checks if two strings are similar according to the problem's definition.

def areSimilar(s1: str, s2: str) -> bool:
    """
    Checks if two strings s1 and s2 are similar.
    Similar means they can be made identical by swapping exactly two characters.
    This also covers identical strings (0 differences).
    """
    diff = []
    for i in range(len(s1)):
        if s1[i] != s2[i]:
            diff.append(i)

    # Case 1: Strings are identical
    if not diff:
        return True

    # Case 2: Strings differ by exactly two characters
    # Check if swapping these two characters makes them identical
    if len(diff) == 2:
        idx1, idx2 = diff[0], diff[1]
        return s1[idx1] == s2[idx2] and s1[idx2] == s2[idx1]

    # Case 3: Strings differ by 1 or more than 2 characters
    # Not similar
    return False

DSU Class in Python

We'll implement a DSU class with find, union, and a count of groups.

class DSU:
    def __init__(self, n):
        """
        Initializes the DSU structure for n elements.
        Each element is initially its own parent and forms a group of size 1.
        """
        self.parent = list(range(n))
        self.size = [1] * n  # For union by size optimization
        self.num_groups = n  # Initialize with n groups

    def find(self, i):
        """
        Finds the representative (root) of the set containing element i.
        Applies path compression for optimization.
        """
        if self.parent[i] == i:
            return i
        self.parent[i] = self.find(self.parent[i]) # Path compression
        return self.parent[i]

    def union(self, i, j):
        """
        Unites the sets containing elements i and j.
        Applies union by size optimization.
        Returns True if a merge happened, False otherwise.
        """
        root_i = self.find(i)
        root_j = self.find(j)

        if root_i != root_j:
            # Union by size: attach smaller tree to root of larger tree
            if self.size[root_i] < self.size[root_j]:
                root_i, root_j = root_j, root_i  # Swap to ensure root_i is larger/equal

            self.parent[root_j] = root_i
            self.size[root_i] += self.size[root_j]
            self.num_groups -= 1 # Decrement group count as two groups merged
            return True
        return False

Main Solution Logic for 839 Similar String Groups

Now, combine the DSU and areSimilar function into the main numSimilarGroups function.

class Solution:
    def numSimilarGroups(self, strs: list[str]) -> int:
        n = len(strs)

        # Initialize DSU for n strings
        dsu = DSU(n)

        # Iterate through all unique pairs of strings
        for i in range(n):
            for j in range(i + 1, n):
                # Check if strings strs[i] and strs[j] are similar
                if areSimilar(strs[i], strs[j]):
                    # If similar, unite their sets in DSU
                    dsu.union(i, j)

        # The number of groups is stored in dsu.num_groups
        return dsu.num_groups

# Helper function defined outside the Solution class or nested inside for competitive programming platforms
def areSimilar(s1: str, s2: str) -> bool:
    diff = []
    for i in range(len(s1)):
        if s1[i] != s2[i]:
            diff.append(i)

    if not diff:
        return True

    if len(diff) == 2:
        idx1, idx2 = diff[0], diff[1]
        return s1[idx1] == s2[idx2] and s1[idx2] == s2[idx1]

    return False

Python Time and Space Complexity Analysis

Time Complexity:

• Let N be the number of strings and L be the length of each string.
• areSimilar function: Takes O(L) time to compare two strings.
• Main loops: We iterate through N * (N - 1) / 2, which is O(N^2) pairs of strings.
• DSU operations: Each find and union operation, with path compression and union by size, takes nearly constant time, amortized O(alpha(N)), where alpha is the inverse Ackermann function.
• Total Time Complexity: O(N^2 * L + N * alpha(N)) which simplifies to O(N^2 * L) because alpha(N) is practically a constant, and L is typically much larger than alpha(N).

Space Complexity:

• DSU class: Stores parent and size arrays of size N. So, O(N) space.
• areSimilar function: diff list stores at most L indices. So, O(L) space.
• Total Space Complexity: O(N + L).

Implementing the Solution: C++

Next, let's look at the C++ implementation for the 839 Similar String Groups leetcode question. We'll use a class for DSU and a standalone helper function.

Helper Function: areSimilar(string& s1, string& s2)

#include <vector>
#include <string>
#include <numeric> // For iota
#include <algorithm> // For std::swap

// Helper function to check if two strings are similar
bool areSimilar(const std::string& s1, const std::string& s2) {
    std::vector<int> diff;
    for (int i = 0; i < s1.length(); ++i) {
        if (s1[i] != s2[i]) {
            diff.push_back(i);
        }
    }

    // Case 1: Strings are identical
    if (diff.empty()) {
        return true;
    }

    // Case 2: Strings differ by exactly two characters
    // Check if swapping these two characters makes them identical
    if (diff.size() == 2) {
        int idx1 = diff[0];
        int idx2 = diff[1];
        return s1[idx1] == s2[idx2] && s1[idx2] == s2[idx1];
    }

    // Case 3: Strings differ by 1 or more than 2 characters
    return false;
}

DSU Class in C++

class DSU {
public:
    std::vector<int> parent;
    std::vector<int> sz; // For union by size optimization
    int num_groups;

    DSU(int n) {
        parent.resize(n);
        std::iota(parent.begin(), parent.end(), 0); // Initialize parent[i] = i
        sz.assign(n, 1); // Initialize size of each group to 1
        num_groups = n; // Initially n groups
    }

    int find(int i) {
        if (parent[i] == i) {
            return i;
        }
        return parent[i] = find(parent[i]); // Path compression
    }

    bool unite(int i, int j) {
        int root_i = find(i);
        int root_j = find(j);

        if (root_i != root_j) {
            // Union by size: attach smaller tree to root of larger tree
            if (sz[root_i] < sz[root_j]) {
                std::swap(root_i, root_j); // Ensure root_i is the root of the larger/equal sized tree
            }

            parent[root_j] = root_i;
            sz[root_i] += sz[root_j];
            num_groups--; // Decrement group count
            return true;
        }
        return false;
    }
};

Main Solution Logic for 839 Similar String Groups

class Solution {
public:
    // Helper function (can be defined as a private member or a global function)
    bool areSimilar(const std::string& s1, const std::string& s2) {
        std::vector<int> diff;
        for (int i = 0; i < s1.length(); ++i) {
            if (s1[i] != s2[i]) {
                diff.push_back(i);
            }
        }

        if (diff.empty()) {
            return true;
        }

        if (diff.size() == 2) {
            int idx1 = diff[0];
            int idx2 = diff[1];
            return s1[idx1] == s2[idx2] && s1[idx2] == s2[idx1];
        }

        return false;
    }

    int numSimilarGroups(std::vector<std::string>& strs) {
        int n = strs.size();

        // Initialize DSU for n strings
        DSU dsu(n);

        // Iterate through all unique pairs of strings
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                // Check if strings strs[i] and strs[j] are similar
                if (areSimilar(strs[i], strs[j])) {
                    // If similar, unite their sets in DSU
                    dsu.unite(i, j);
                }
            }
        }

        // The number of groups is stored in dsu.num_groups
        return dsu.num_groups;
    }
};

C++ Time and Space Complexity Analysis

Time Complexity:

• Let N be the number of strings and L be the length of each string.
• areSimilar function: Takes O(L) time.
• Main loops: O(N^2) pairs.
• DSU operations: Amortized O(alpha(N)).
• Total Time Complexity: O(N^2 * L).

Space Complexity:

• DSU class: parent and sz vectors of size N. So, O(N) space.
• areSimilar function: diff vector stores at most L indices. So, O(L) space.
• Total Space Complexity: O(N + L).

Implementing the Solution: Java

Finally, let's implement the solution for the 839 Similar String Groups leetcode question in Java. We'll use an inner class for DSU within the Solution class.

Helper Function: areSimilar(String s1, String s2)

import java.util.ArrayList;
import java.util.List;

class Solution {
    // Helper function to check if two strings are similar
    private boolean areSimilar(String s1, String s2) {
        List<Integer> diff = new ArrayList<>();
        for (int i = 0; i < s1.length(); ++i) {
            if (s1.charAt(i) != s2.charAt(i)) {
                diff.add(i);
            }
        }

        // Case 1: Strings are identical
        if (diff.isEmpty()) {
            return true;
        }

        // Case 2: Strings differ by exactly two characters
        // Check if swapping these two characters makes them identical
        if (diff.size() == 2) {
            int idx1 = diff.get(0);
            int idx2 = diff.get(1);
            return s1.charAt(idx1) == s2.charAt(idx2) && s1.charAt(idx2) == s2.charAt(idx1);
        }

        // Case 3: Strings differ by 1 or more than 2 characters
        return false;
    }

    // ... DSU class and numSimilarGroups method will go here
}

DSU Class in Java

import java.util.ArrayList;
import java.util.List;
import java.util.Arrays; // For Arrays.fill and Arrays.setAll

class Solution {
    // Inner DSU class
    class DSU {
        int[] parent;
        int[] size; // For union by size optimization
        int numGroups;

        DSU(int n) {
            parent = new int[n];
            // Initialize parent[i] = i
            for (int i = 0; i < n; i++) {
                parent[i] = i;
            }
            size = new int[n];
            Arrays.fill(size, 1); // Initialize size of each group to 1
            numGroups = n; // Initially n groups
        }

        int find(int i) {
            if (parent[i] == i) {
                return i;
            }
            parent[i] = find(parent[i]); // Path compression
            return parent[i];
        }

        boolean unionSets(int i, int j) {
            int rootI = find(i);
            int rootJ = find(j);

            if (rootI != rootJ) {
                // Union by size: attach smaller tree to root of larger tree
                if (size[rootI] < size[rootJ]) {
                    int temp = rootI;
                    rootI = rootJ;
                    rootJ = temp; // Swap to ensure rootI is the root of the larger/equal sized tree
                }

                parent[rootJ] = rootI;
                size[rootI] += size[rootJ];
                numGroups--; // Decrement group count
                return true;
            }
            return false;
        }
    }

    // ... areSimilar method and numSimilarGroups method will go here
}

Main Solution Logic for 839 Similar String Groups

import java.util.ArrayList;
import java.util.List;
import java.util.Arrays;

class Solution {
    // Helper function to check if two strings are similar
    private boolean areSimilar(String s1, String s2) {
        List<Integer> diff = new ArrayList<>();
        for (int i = 0; i < s1.length(); ++i) {
            if (s1.charAt(i) != s2.charAt(i)) {
                diff.add(i);
            }
        }

        if (diff.isEmpty()) {
            return true;
        }

        if (diff.size() == 2) {
            int idx1 = diff.get(0);
            int idx2 = diff.get(1);
            return s1.charAt(idx1) == s2.charAt(idx2) && s1.charAt(idx2) == s2.charAt(idx1);
        }

        return false;
    }

    // Inner DSU class
    class DSU {
        int[] parent;
        int[] size;
        int numGroups;

        DSU(int n) {
            parent = new int[n];
            for (int i = 0; i < n; i++) {
                parent[i] = i;
            }
            size = new int[n];
            Arrays.fill(size, 1);
            numGroups = n;
        }

        int find(int i) {
            if (parent[i] == i) {
                return i;
            }
            parent[i] = find(parent[i]);
            return parent[i];
        }

        boolean unionSets(int i, int j) {
            int rootI = find(i);
            int rootJ = find(j);

            if (rootI != rootJ) {
                if (size[rootI] < size[rootJ]) {
                    int temp = rootI;
                    rootI = rootJ;
                    rootJ = temp;
                }

                parent[rootJ] = rootI;
                size[rootI] += size[rootJ];
                numGroups--;
                return true;
            }
            return false;
        }
    }

    public int numSimilarGroups(String[] strs) {
        int n = strs.length;

        // Initialize DSU for n strings
        DSU dsu = new DSU(n);

        // Iterate through all unique pairs of strings
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                // Check if strings strs[i] and strs[j] are similar
                if (areSimilar(strs[i], strs[j])) {
                    // If similar, unite their sets in DSU
                    dsu.unionSets(i, j);
                }
            }
        }

        // The number of groups is stored in dsu.numGroups
        return dsu.numGroups;
    }
}

Java Time and Space Complexity Analysis

Time Complexity:

• Let N be the number of strings and L be the length of each string.
• areSimilar function: Takes O(L) time.
• Main loops: O(N^2) pairs.
• DSU operations: Amortized O(alpha(N)).
• Total Time Complexity: O(N^2 * L).

Space Complexity:

• DSU class: parent and size arrays of size N. So, O(N) space.
• areSimilar function: diff list stores at most L indices. So, O(L) space.
• Total Space Complexity: O(N + L).

Common Pitfalls and Considerations

While the DSU approach is straightforward, a few common mistakes can lead to incorrect solutions or Time Limit Exceeded (TLE).

• Incorrect areSimilar Logic: The most crucial part is correctly implementing the similarity check. Many candidates might misinterpret "swap exactly two characters." Forgetting to handle identical strings (zero differences) or mistakenly allowing strings with more than two differences will lead to errors. Ensure the check for s1[idx1] == s2[idx2] && s1[idx2] == s2[idx1] is correct for the two-difference case.
• DSU Implementation Errors:
  • Missing Path Compression: Without path compression in the find operation, the DSU trees can become very deep, degrading find and union operations to O(N) in the worst case. This would increase overall complexity significantly, often resulting in TLE for larger inputs.
  • Missing Union by Rank/Size: Similar to path compression, not using union by rank or size can lead to skewed trees, also causing O(N) worst-case performance for DSU operations.
  • Incorrect num_groups Decrement: If you're using a num_groups counter, ensure it's decremented only when union successfully merges two different sets. If root_i == root_j, no merge happens, and the count should not change.
• Time Limit Exceeded (TLE) - O(N^2 * L): This complexity is inherent to the problem's constraints (comparing all pairs).
  • For N up to 100, N^2 is 100^2 = 10,000. If L is up to 60, N^2 * L is 10,000 * 60 = 600,000, which is perfectly acceptable.
  • However, if N is larger (e.g., 2000), N^2 is 4 * 10^6. With L=60, this becomes 2.4 * 10^8, which might be too slow for a typical 1-second time limit. For this specific problem (N up to 100), O(N^2 * L) is expected. Be aware of the constraints for future problems.
• String Hashing Optimization: For problems with larger N but smaller L, or if many duplicate strings are expected, you might consider string hashing to group identical strings initially and then map them to integer IDs to reduce the N for DSU. However, for LeetCode 839, the provided constraints (N up to 100) mean the O(N^2 * L) approach is perfectly fine and preferred for its simplicity.

Further Optimizations and Advanced Topics

While the O(N^2 * L) DSU solution is optimal for the given constraints of LeetCode 839, it's worth briefly touching on scenarios where further optimization might be considered.

Handling Larger N (Number of Strings)

If N were significantly larger (e.g., 10^5) but L (string length) remained relatively small, the O(N^2) string comparisons would be too slow. In such advanced cases, the approach would shift towards using hashing and graph-traversal algorithms instead of direct DSU on all pairs:

1. Generate Permutations/Neighbors: For each string s, generate all possible "similar" strings by trying to swap every pair of characters. This generates O(L^2) potential neighbors.
2. Hashing and Adjacency List: Store all input strings in a hash set (or unordered_map in C++, HashSet in Java) for O(1) average-time lookups. For each input string s, iterate through its O(L^2) generated similar strings s'. If s' exists in the hash set, s and s' are similar. Build an adjacency list graph[s] containing s'.
3. Graph Traversal (BFS/DFS): Once the graph (or a partial graph for similar strings) is built, perform a graph traversal (BFS or DFS) to find connected components. Count the number of times you start a new traversal from an unvisited node. This would typically bring the complexity closer to O(N * L^2 * alpha(N)) or O(N * L^2 + N + E) where E is number of edges (which could be N * L^2 in worst case).

However, this O(N * L^2) approach only works if L^2 is small enough. For LeetCode 839, L is up to 60, so L^2 is 3600. N * L^2 would be 100 * 3600 = 3.6 * 10^5, which is faster than N^2 * L = 100^2 * 60 = 6 * 10^5 in this specific scenario, but the constant factors and overhead of generating permutations can make it slower in practice for the given constraints. The N^2 * L approach is simpler to implement and often performs better for N up to around 100-200.

Trie-based Optimization

Another highly complex optimization for very specific string similarity definitions (like Hamming distance 1 or 2) can involve Tries. If you need to find all strings "similar" to a given string very quickly across a large dictionary of strings, a Trie could be used to efficiently search for neighbors. This is generally overkill for a "swap two characters" similarity and the given constraints.

For the purpose of the 839 Similar String Groups leetcode question, sticking with the O(N^2 * L) DSU approach is the standard and most effective solution. The primary takeaway should be the correct application of DSU and the precise definition of "similar strings."

Conclusion

The 839 Similar String Groups LeetCode problem is an excellent exercise in applying the Disjoint Set Union (DSU) data structure to solve connectivity problems. By modeling the strings as nodes in a graph and defining similarity as an edge, we effectively transform the problem into finding the number of connected components. The efficiency of DSU, particularly with path compression and union by size optimizations, makes it the ideal tool for handling the transitive nature of string similarity.

This tutorial has guided you through a comprehensive understanding of the problem, the theoretical underpinnings of DSU, and detailed implementations in Python, C++, and Java. Mastering these concepts and their practical application is crucial for competitive programming and algorithm design, much like understanding Dijkstra's Algorithm or other fundamental graph traversals. Continue to practice similar problems to solidify your understanding of DSU and its versatility in solving complex grouping and connectivity challenges.

Frequently Asked Questions

Q: What defines "similar" strings in LeetCode 839?

A: Two strings are similar if they are identical (zero differences), or if they can be made identical by swapping exactly two characters. If they differ by only one character or by more than two characters, they are not considered similar.

Q: Why is Disjoint Set Union (DSU) suitable for this problem?

A: DSU is ideal because string similarity is transitive: if A is similar to B and B to C, then A, B, and C are all in the same group. DSU efficiently manages these equivalence classes, allowing quick merging of related groups and counting the final distinct sets.

Q: What is the time complexity of the solution?

A: The overall time complexity of this solution is O(N^2 * L), where N is the number of strings in the input array and L is the length of each string. This accounts for iterating through all N^2 pairs and performing an O(L) similarity check for each.

Further Reading & Resources

• LeetCode Problem 839: Similar String Groups
• Disjoint Set Union (Union-Find) Data Structure Explained
• Introduction to Graph Theory