Solution For Similar String Groups
Problem:
You are given an array of strings words, where each word is unique. Find the number of differentiable groups of words.
A group is defined as a collection of two or more strings with the same length, where each string in the group differs from all other strings in the group by exactly one character.
For example, the group (red, ted, tad) has 3 strings and is a valid group since every string differs from the others by exactly one character. The group (red, bed) has 2 strings and is not valid since red and bed differ by more than one character.
Return the number of differentiable groups of words.
Solution:
To solve this problem, we can use the concept of disjoint sets. For each word in the input array, we can create a separate disjoint set with the word itself. Then, we can compare each pair of words to see if they differ by exactly one character. If they do, we can merge the sets to which they belong. Finally, we can count the number of disjoint sets remaining, as each set represents a differentiable group of words.
To compare two words, we can iterate over their characters and count the number of differences. If the number of differences is greater than one, we stop comparing and move on to the next pair of words. If the number of differences is exactly one, we can merge the sets to which the words belong.
We can use a hashmap to keep track of the disjoint sets. The keys of the hashmap will be the words in the input array, and the values will be the representative elements of the disjoint sets to which the words belong. Initially, each word is a representative element of its own set.
We can also use a function to find the representative element of a set. This function will take a word as input and follow the parent pointers until it reaches the representative element of the set.
Implementation:
Here is the Python code to solve the problem:
“`python
def numSimilarGroups(words: List[str]) -> int:
def find(word):
if word != sets[word]:
sets[word] = find(sets[word])
return sets[word]
def similar(word1, word2):
diffs = 0
for c1, c2 in zip(word1, word2):
if c1 != c2:
diffs += 1
if diffs > 1:
return False
return diffs == 1
sets = {}
for word in words:
sets[word] = word
for word1, word2 in combinations(words, 2):
if similar(word1, word2):
sets[find(word1)] = find(word2)
return len(set(find(word) for word in words))
“`
First, we define the function find to find the representative element of a set. This function uses path compression to reduce the height of the tree and speed up future lookups.
Next, we define the function similar to check if two words are similar. This function iterates over the characters of the words and counts the number of differences. If the number of differences is greater than one, the function returns False. Otherwise, the function returns True.
We create a hashmap sets to keep track of the disjoint sets. Initially, each word is a representative element of its own set.
We then iterate over all possible pairs of words using the combinations function from the itertools module. If a pair of words is similar, we merge the sets to which they belong by setting the representative element of one set to the representative element of the other set.
Finally, we count the number of disjoint sets left in the hashmap by converting the set of representative elements to a set and returning its length.
Time Complexity:
The time complexity of the algorithm is O(n^2 * m), where n is the number of words and m is the maximum length of a word. This is because we compare each pair of words, which takes O(m) time, and we may need to update the parent pointers of the disjoint sets, which takes O(log n) time in the worst case (if the sets form a chain). The total number of parent pointer updates is O(n^2 * m), which dominates the time complexity. However, in practice, the algorithm is much faster than O(n^2 * m) because most sets are merged quickly, and path compression reduces the height of the disjoint sets.
Step by Step Implementation For Similar String Groups
There are several ways to solve this problem. One approach would be to use a union-find data structure. We can initialize each string to be its own group. Then, for each pair of strings, we can check if they are similar. If they are, we can union their groups. After processing all pairs of strings, the number of groups will be the number of similar string groups. Another approach would be to use a graph data structure. We can initialize each string to be a node in the graph. Then, for each pair of strings, we can add an edge between them if they are similar. After processing all pairs of strings, the number of connected components in the graph will be the number of similar string groups.
This is a Python implementation of the Union-Find algorithm for solving the "Similar String Groups" problem on LeetCode.
Union-Find is a data structure that keeps track of a set of elements, partitioned into a number of disjoint (non-overlapping) subsets. It supports two operations:
Find: Determine which subset a particular element is in. This can be used for determining if two elements are in the same subset.
Union: Join two subsets into a single subset.
We can use Union-Find to solve the "Similar String Groups" problem as follows. First, we initialize a Union-Find data structure with one element (string) per group. Then, for each pair of strings, we check if they are in the same group. If they are not, we union their groups. Finally, we return the number of groups.
def find(parent, i):
if parent[i] == -1:
return i
return find(parent, parent[i])
def union(parent, i, j):
x = find(parent, i)
y = find(parent, j)
if x != y:
parent[x] = y
def numSimilarGroups(A):
parent = [-1] * len(A)
for i in range(len(A)):
for j in range(i+1, len(A)):
if areSimilar(A[i], A[j]):
union(parent, i, j)
return len([i for i in parent if i == -1])
def areSimilar(s1, s2):
count = 0
for i in range(len(s1)):
if s1[i] != s2[i]:
count += 1
if count > 2:
return False
return Truevar numSimilarGroups = function(A) {
// create a map of all unique letters to their corresponding indices
let map = {};
for (let i = 0; i < A.length; i++) {
for (let j = 0; j < A[i].length; j++) {
if (!(A[i][j] in map)) {
map[A[i][j]] = [i];
} else {
map[A[i][j]].push(i);
}
}
}
// create a set to track which indices have been visited
let visited = new Set();
// create a queue to keep track of which indices need to be visited
let queue = [];
// keep track of the number of groups
let numGroups = 0;
for (let i = 0; i < A.length; i++) {
// if the index has already been visited, continue
if (visited.has(i)) continue;
// add the index to the queue and mark it as visited
queue.push(i);
visited.add(i);
// while the queue is not empty...
while (queue.length > 0) {
// get the first index from the queue
let idx = queue.shift();
// get all the indices that are one letter different from the current index
let oneLetterDifferent = getOneLetterDifferent(A[idx], map);
// for each of those indices...
for (let j = 0; j < oneLetterDifferent.length; j++) {
// if the index has not been visited yet, add it to the queue and mark it as visited
if (!visited.has(oneLetterDifferent[j])) {
queue.push(oneLetterDifferent[j]);
visited.add(oneLetterDifferent[j]);
}
}
}
// increment the number of groups
numGroups++;
}
return numGroups;
};
// this function returns an array of indices that are one letter different from the given index
var getOneLetterDifferent = function(word, map) {
let oneLetterDifferent = [];
// for each letter in the word...
for (let i = 0; i < word.length; i++) {
// get all the indices that are associated with that letter
let indices = map[word[i]];
// for each of those indices...
for (let j = 0; j < indices.length; j++) {
// if the index is not the same as the given index and the words are one letter different, add it to the array
if (indices[j] !== i && isOneLetterDifferent(word, A[indices[j]])) {
oneLetterDifferent.push(indices[j]);
}
}
}
return oneLetterDifferent;
};
// this function returns true if the two words are one letter different, false otherwise
var isOneLetterDifferent = function(word1, word2) {
let diff = 0;
for (let i = 0; i < word1.length; i++) {
if (word1[i] !== word2[i]) {
diff++;
}
if (diff > 1) {
return false;
}
}
return true;
};There are several ways to solve this problem. One approach would be to use a graph data structure to store the strings and their connections. Then, you could use a depth-first search or a breadth-first search to find all of the connected strings. Another approach would be to use a union-find data structure. You would store the strings in the union-find data structure, and then use the union-find algorithm to determine which strings are in the same group.
using System;
using System.Collections.Generic;
using System.Linq;
public class Solution {
public IList> GroupAnagrams(string[] strs) {
// create a list to store the final results
List> results = new List>();
// create a dictionary to store the intermediate results
Dictionary> dict = new Dictionary>();
// iterate over the input array
foreach(string s in strs) {
// create a key by sorting the string
string key = String.Concat(s.OrderBy(c => c));
// check if the key is already present in the dictionary
if(dict.ContainsKey(key)) {
// if yes, then add the current string to the list of values corresponding to that key
dict[key].Add(s);
}
else {
// if not, then create a new key-value pair in the dictionary
dict.Add(key, new List(){s});
}
}
// finally, add all the values in the dictionary to the results list
foreach(var kvp in dict) {
results.Add(kvp.Value);
}
// return the results
return results;
}
}