Problem Statement
Two strings, X and Y, are defined as similar if they are either identical or can be transformed into one another by swapping exactly two distinct letters in string X. For instance, the word "tars" can be transformed into "rats" by swapping the characters at positions 0 and 2. Additionally, words such as "rats" and "arts" are considered similar under the same conditions, but "star" cannot be transformed into "tars", "rats", or "arts" through any two-letter swap.
Importantly, these similarities allow strings to form connected groups. For example, "tars", "rats", and "arts" can be placed in a single group because they are either directly similar or can be linked through a series of similarities. However, "star" would constitute a separate group of its own, being dissimilar to the others in the context of just two swaps.
Given a list of strings named strs, where each string is an anagram of every other string in the list, the task is to determine the total number of such distinct similarity groups present.
Examples
Example 1
Input:
strs = ["tars","rats","arts","star"]
Output:
2
Example 2
Input:
strs = ["omv","ovm"]
Output:
1
Constraints
1 <= strs.length <= 3001 <= strs[i].length <= 300strs[i]consists of lowercase letters only.- All words in
strshave the same length and are anagrams of each other.
Approach and Intuition
Based on the problem details and constraints, we can approach our solution using the following steps:
Identify Similar Strings:
- For each string in
strs, compare it with every other string to check if they are similar based on the criteria of zero or exactly two letter swaps. - Utilize a helper function to check for similarity, where a comparison of character positions will confirm if no more than two characters need swapping.
- For each string in
Use Disjoint Set Union (DSU) for Grouping:
- Implement DSU to manage and group strings that are similar.
- Each string starts in its group, and as similarities are identified, groups are merged appropriately.
Iterate Through All String Comparisons:
- For each string, make comparisons with subsequent strings in the list, potentially merging their respective groups in DSU if they're similar.
Count and Return Distinct Groups:
- After processing all strings and performing necessary group merges in DSU, the count of unique parent groups (representing distinct similarity sets) will give us the desired result.
Using this approach ensures that all potential comparisons are made, while DSU efficiently manages the dynamic grouping of strings based on discovered similarities. The constraints ensure that this method remains computationally feasible within provided limits.
Solutions
- C++
class DisjointSet {
private:
vector<int> leader, height;
int total_components;
public:
DisjointSet(int capacity) {
leader.resize(capacity);
height.resize(capacity, 0);
total_components = capacity;
for (int index = 0; index < capacity; index++) {
leader[index] = index;
}
}
int resolve(int node) {
if (leader[node] != node) leader[node] = resolve(leader[node]);
return leader[node];
}
void merge(int node1, int node2) {
int root1 = resolve(node1), root2 = resolve(node2);
if (height[root1] < height[root2]) {
leader[root1] = root2;
} else if (height[root1] > height[root2]) {
leader[root2] = root1;
} else {
leader[root2] = root1;
height[root1]++;
}
}
};
class Solution {
public:
bool isSimilar(string& s1, string& s2) {
int differences = 0;
for (int index = 0; index < s1.size(); index++) {
if (s1[index] != s2[index]) {
differences++;
}
}
return differences == 0 || differences == 2;
}
int numSimilarGroups(vector<string>& strings) {
int size = strings.size();
DisjointSet ds(size);
int group_count = size;
for (int i = 0; i < size; i++) {
for (int j = i + 1; j < size; j++) {
if (isSimilar(strings[i], strings[j]) && ds.resolve(i) != ds.resolve(j)) {
group_count--;
ds.merge(i, j);
}
}
}
return group_count;
}
};
The Similar String Groups solution involves using the Union-Find (or Disjoint-Set) data structure to determine the number of groups of similar strings. In this C++ implementation:
A class
DisjointSetis defined to manage the union-find operations.- Initialize each string’s parent to itself and set heights to zero.
- The
resolvefunction uses path compression to find the root leader of a node. - The
mergefunction unites two sets using union by rank to optimize tree height.
The
Solutionclass comprises two primary functions:isSimilarchecks if two strings have exactly zero or two differing characters, significant for defining whether strings belong to the same group.numSimilarGroupsuses the union-find structure to count groups:- Iterate over each pair of strings and, if they are similar and not already united, merge their sets in the data structure.
As strings are processed, the number of groups is reduced whenever a valid merge operation occurs because of similar strings. The remaining count of separate sets when all strings are processed gives the total number of similar groups.
This method is efficient for problems involving relational grouping of elements and is implemented here to handle string similarity groups specifically.
- Java
class DisjointSet {
int[] leader;
int[] size;
public DisjointSet(int capacity) {
leader = new int[capacity];
for (int i = 0; i < capacity; i++)
leader[i] = i;
size = new int[capacity];
}
public int locate(int x) {
if (leader[x] != x)
leader[x] = locate(leader[x]);
return leader[x];
}
public void merge(int x, int y) {
int rootX = locate(x), rootY = locate(y);
if (rootX == rootY) {
return;
} else if (size[rootX] < size[rootY]) {
leader[rootX] = rootY;
} else if (size[rootX] > size[rootY]) {
leader[rootY] = rootX;
} else {
leader[rootY] = rootX;
size[rootX]++;
}
}
}
class Solution {
public boolean areSimilar(String s1, String s2) {
int differences = 0;
for (int i = 0; i < s1.length(); i++) {
if (s1.charAt(i) != s2.charAt(i)) {
differences++;
}
}
return differences == 0 || differences == 2;
}
public int countSimilarGroups(String[] strings) {
int n = strings.length;
DisjointSet ds = new DisjointSet(n);
int groupCount = n;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if (areSimilar(strings[i], strings[j]) && ds.locate(i) != ds.locate(j)) {
groupCount--;
ds.merge(i, j);
}
}
}
return groupCount;
}
}
The given Java solution addresses the problem of finding and counting similar string groups. It effectively utilizes a Disjoint Set (Union-Find) data structure to manage and merge groups of similar strings, each of which possibly differ by two characters or are exactly the same.
- The
DisjointSetclass maintains an arrayleaderto track the leader of each element, and an arraysizeto optimize merging processes based on the size of each group. - The
locatefunction recursively finds the root leader of an element and employs path compression for efficiency. - The
mergefunction joins two elements into a single set and decides which leader should represent the merged set based on the size of the sets being merged.
The Solution class defines methods to compare strings and to count the total number of distinct similar string groups:
- The
areSimilarmethod determines if two strings are similar under given conditions (no differences or exactly two different characters). - The
countSimilarGroupsmethod initializes a newDisjointSetfor the input strings, checks each pair of strings, merging sets in the disjoint set structure when they are similar.
This setup efficiently groups all similar strings and the count of these groups is maintained and returned as a final result, demonstrating a dynamic and optimal approach for grouping and counting with the Union-Find structure.
No comments yet.