## Similar Problems

Similar Problems not available

# Similar String Groups - Leetcode Solution

## Companies:

LeetCode: Similar String Groups Leetcode Solution

Difficulty: Hard

Topics: hash-table depth-first-search union-find string breadth-first-search array

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:

```
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.

## Similar String Groups Solution Code

`1`