Problem of The Day: Determine if Two Strings Are Close

Problem Statement

leetcode problem link

My Solution [Accepted]

class Solution:
    def closeStrings(self, word1: str, word2: str) -> bool:
        if len(word1) != len(word2):
            return False

        counter1 = Counter(word1)
        counter2 = Counter(word2)

        s1 = counter1.keys()
        s2 = counter2.keys()

        return sorted(counter1.values()) == sorted(counter2.values()) and sorted(s1) == sorted(s2)

public class Solution {
    public bool CloseStrings(string word1, string word2) {
        if (word1.Length != word2.Length)
            return false;

        var dict1 = new Dictionary<char, int>();
        var dict2 = new Dictionary<char, int>();
        var hashSet1 = new HashSet<char>(word1);
        var hashSet2 = new HashSet<char>(word2);
        var sortedList1 = hashSet1.ToList();
        var sortedList2 = hashSet2.ToList();

        sortedList1.Sort();
        sortedList2.Sort();

        foreach(var ch in word1) {
            if (dict1.ContainsKey(ch))
                dict1[ch]++;
            else
                dict1[ch] = 1;
        }

        foreach(var ch in word2) {
            if (dict2.ContainsKey(ch))
                dict2[ch]++;
            else
                dict2[ch] = 1;
        }

        var values1 = dict1.Values.ToList();
        values1.Sort();
        var values2 = dict2.Values.ToList();
        values2.Sort();
        return sortedList1.SequenceEqual(sortedList2) && values1.SequenceEqual(values2);
    }
}

Improve CSharp Solution

public class Solution
{
    public bool CloseStrings(string word1, string word2)
    {
        if (word1.Length != word2.Length)
            return false;

        int[] freq1 = new int[26];
        int[] freq2 = new int[26];

        foreach (char c in word1)
            freq1[c - 'a']++;

        foreach (char c in word2)
            freq2[c - 'a']++;

        // Check if both words use the exact same characters
        for (int i = 0; i < 26; i++)
        {
            if ((freq1[i] == 0) != (freq2[i] == 0))
                return false;
        }

        // Compare sorted frequency distributions
        Array.Sort(freq1);
        Array.Sort(freq2);

        return freq1.SequenceEqual(freq2);
    }
}

Editorial

Approach 1: Using HashMap

class Solution:
    def closeStrings(self, word1: str, word2: str) -> bool:
        if len(word1) != len(word2):
            return False

        word1_elem_freq = Counter(word1)
        word2_elem_freq = Counter(word2)

        return set(word1_elem_freq) == set(word2_elem_freq) and sorted(word1_elem_freq.values()) == sorted(word2_elem_freq.values())

Approach 2: Using Frequency Array Map

class Solution {
public:
    bool closeStrings(string word1, string word2) {
        if (word1.size() != word2.size()) return false;
        vector<int> word1Map(26, 0);
        vector<int> word2Map(26, 0);
        for (auto c : word1) {
            word1Map[c - 'a']++;
        }
        for (auto c : word2) {
            word2Map[c - 'a']++;
        }
        for (int i = 0; i < 26; i++) {
            if ((word1Map[i] == 0 && word2Map[i] > 0) ||
                (word2Map[i] == 0 && word1Map[i] > 0)) {
                return false;
            }
        }
        sort(word1Map.begin(), word1Map.end());
        sort(word2Map.begin(), word2Map.end());
        return (word1Map == word2Map);
    }
};