Fastest Solution in C++

include

include

using namespace std;

int main(){ ios::sync_with_stdio(false); cin.tie(nullptr);

int n;
cin >> n;
cin.ignore();

string s;
cin >> s;

vector<int> freq(26); 

for(char c : s) ++freq[c - 'a'];

int max_len = 0;
for(char c1 = 'a'; c1 <= 'z'; ++c1) {
    if(freq[c1 - 'a'] == 0) continue;

    for(char c2 = c1 + 1; c2 <= 'z'; ++c2) {
        if(freq[c2 - 'a'] == 0) continue;
        if(abs(freq[c1 - 'a'] - freq[c2 - 'a']) > 1) continue;
        if(freq[c1 - 'a'] + freq[c2 - 'a'] < max_len) continue;

        char last = '\0'; int len = 0;

        for(char c : s) {
            if(c == last) { len = 0; break; }
            if(c != c1 && c != c2) continue;
            last = c; ++len;
        }

        max_len = max(max_len, len);
    }
}

cout << max_len << endl;

return 0;

}

return 0 and be surprised

public static int alternate(String str) {
    String[] uniqueCharacters = Arrays.stream(str.split("")).collect(Collectors.toSet()).toArray(new String[0]);
    int maxLength = 0;
    for(int i=0;i<uniqueCharacters.length-1; i++) {
        for(int j=i+1; j<uniqueCharacters.length; j++) {
            String firstChar = uniqueCharacters[i];
            String secondChar= uniqueCharacters[j];
            String newStr = str;
            List<String> lst = Arrays.stream(uniqueCharacters).filter(s -> !s.equals(firstChar) && !s.equals(secondChar)).collect(Collectors.toList());
            for(String l:lst) {
                newStr = newStr.replaceAll(l, "");
            }
            if(isAlternateCharacters(newStr) & newStr.length() > maxLength) {
                maxLength = newStr.length();
            }
        }
    }
    return maxLength;
}   


private static boolean isAlternateCharacters(String str) {
    if(str.length() == 1) return false;

    char firstchar = str.charAt(0);
    char secondchar = str.charAt(1);
    if(firstchar == secondchar) return false;

    for(int i=0; i<str.length(); i++) {
        if(i%2 == 0 && str.charAt(i) == firstchar) continue;
        if(i%2 != 0 && str.charAt(i) == secondchar) continue;
        return false;
    }
    return true;
}

Here is problem solution in python, java, c++ c and javascript - https://programmingoneonone.com/hackerrank-two-characters-problem-solution.html

This is my Python solution ✅ Explanation: 1. I first generate all unique character pairs from the input string. 2. For each pair, I build a filtered string that only includes those two characters. 3. I then check if the filtered string alternates properly (i.e., no repeated consecutive characters). 4. If it passes, I update the maximum length found so far. 5. In the end, return the longest valid alternating string length or 0 if none found.

📌 Note: Although this problem is tagged "Easy", it requires thoughtful implementation. I’d say it feels more like a Medium problem due to the character pairing and validation logic.

Hope this helps others! 😊

def alternate(s):
    # Write your code here
    def isConsecutive(st):
        if len(st) < 2:
            return True
        
        for i in range(1, len(st)):
            if st[i] == st[i-1]:
                return True
        
        return False
    
    chars = list(set(s))
    candidates = []
    maxalter = float('-inf')
    
    for i in range(len(chars) - 1):
        for j in range(i + 1, len(chars)):
            candidates.append([chars[i], chars[j]])
    
    for cand in candidates:
        temp = ''
        for c in s:
            if c in cand:
                temp += c
        if not isConsecutive(temp):
            maxalter = max(maxalter, len(temp))
    
    return maxalter if maxalter != float('-inf') else 0