How do you solve Count Pairs Of Similar Strings on LeetCode?

Count Pairs Of Similar Strings (LeetCode #2506) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * k log k) time complexity and O(n) space complexity. Key insight: Two strings are similar if they have the same unique characters.

What is the brute force solution for Count Pairs Of Similar Strings?

The brute force approach for Count Pairs Of Similar Strings is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible pair of strings to see if they are similar. This is straightforward but can be slow for larger inputs. The optimal Optimal Solution approach improves this to O(n * k log k).

What algorithm or data structure is used to solve Count Pairs Of Similar Strings?

Count Pairs Of Similar Strings uses the following concepts: Array, Hash Table, String, Bit Manipulation, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Pairs Of Similar Strings?

Always think about how to reduce the number of comparisons. Consider using data structures like hash maps to store intermediate results. Be prepared to explain your thought process and why you chose a particular approach.

What are common mistakes when solving Count Pairs Of Similar Strings?

Not considering the order of characters when comparing sets. Failing to account for duplicate characters in the same string.

#2506

Count Pairs Of Similar Strings

Easy

ArrayHash TableStringBit ManipulationCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * k log k)
Space	O(1)	O(n)

💡

Intuition

Time O(n * k log k)Space O(n)

Instead of comparing each pair, we can use a hash map to group words by their unique characters. This allows us to count similar strings efficiently.

⚙️

Algorithm

4 steps

1Step 1: Create a hash map to store the frequency of each unique character set as a string.
2Step 2: For each word, convert it into a sorted string of unique characters and update its count in the hash map.
3Step 3: For each unique character set in the hash map, calculate the number of pairs using the formula count * (count - 1) / 2.
4Step 4: Sum all the pairs and return the result.

solution.py15 lines

1# Full working Python code
2
3def countSimilarPairs(words):
4    from collections import defaultdict
5    count_map = defaultdict(int)
6    for word in words:
7        unique_chars = ''.join(sorted(set(word)))
8        count_map[unique_chars] += 1
9    count = 0
10    for c in count_map.values():
11        count += c * (c - 1) // 2
12    return count
13
14# Example usage
15print(countSimilarPairs(["aba", "aabb", "abcd", "bac", "aabc"]))  # Output: 2

ℹ

Complexity note: The time complexity is O(n * k log k) where n is the number of words and k is the average length of the words due to sorting. The space complexity is O(n) for storing the hash map.

1Two strings are similar if they have the same unique characters.
2Using a hash map to group strings by their unique characters can significantly reduce the number of comparisons.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.