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Leetcode 2506. Count Pairs Of Similar Strings

Problem Statement

You are given a 0-indexed string array words. Two strings are similar if they consist of the same characters.

• For example, “abca” and “cba” are similar since both consist of characters ‘a’, ‘b’, and ‘c’.
• However, “abca” and “cbad” are not similar since they contain different characters.

Return the number of pairs (i, j) such that 0 <= i < j < n and the two strings words[i] and words[j] are similar.

Clarifying Questions

1. Q: Are the strings case-sensitive?
   • A: The problem appears to focus on characters, assuming lowercase. Unless otherwise specified, we’ll assume the strings are case-sensitive.
2. Q: What is the maximum length of the words array?
   • A: The constraints are not specified, but typically LeetCode problems handle arrays up to a few thousands in length.
3. Q: If two words have the same characters but different frequencies, are they similar?
   • A: Yes, the frequency of characters does not matter, only the set of characters does.

Strategy

1. Set Representation: Convert each word into a set of characters. If two words have the same sets of characters, they are similar.

2. Map for Counting Sets: Use a hashmap to count the frequency of each unique set of characters.

3. Count Pairs: For each unique set, count the pairs using the formula for combinations. Specifically, for a set appearing ( k ) times, the number of ways to pick 2 out of ( k ) is ( \binom{k}{2} = \frac{k (k - 1)}{2} ).

Plan to Implement

1. Initialize a hashmap to store the frequency of each unique set.
2. Convert each word into a set of characters and update the hashmap.
3. Iterate over the hashmap to compute the number of pairs for each set that appears more than once.

#include <vector>
#include <string>
#include <unordered_map>
#include <set>

using namespace std;

class Solution {
public:
    int countPairs(vector<string>& words) {
        // Map to store the frequency of each unique set of characters
        unordered_map<string, int> charsetCount;

        // Helper function to get the sorted string representation of a set of characters
        auto getCharSet = [](const string& word) {
            set<char> charSet(word.begin(), word.end());
            string sortedSet(charSet.begin(), charSet.end());
            return sortedSet;
        };
        
        // Populate the charsetCount map
        for (const string& word : words) {
            string charSet = getCharSet(word);
            charsetCount[charSet]++;
        }
        
        // Calculate the number of similar pairs
        int count = 0;
        for (const auto& [charSet, freq] : charsetCount) {
            if (freq > 1) {
                count += (freq * (freq - 1)) / 2; // Combination nC2
            }
        }
        
        return count;
    }
};

Time Complexity

• Preprocessing each word: Converting a word into a set and then into a string takes ( O(L \log L) ), where ( L ) is the length of the word (set insertion cost).
• HashMap Operations: Inserting and looking up each set in the hash map takes ( O(1) ) (amortized average).
• Total Complexity: Let ( n ) be the number of words and ( L ) be the average length of the words. We do ( O(n \cdot L \log L) ) for preprocessing and ( O(n) ) for the hashmap operations, making the overall time complexity ( O(n \cdot L \log L) ).

This solution efficiently counts pairs of similar strings by leveraging the power of sets and hashmaps to group and count the unique character sets.

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