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

Here is my c++ easy solution, explantion here : https://youtu.be/WAPtXpj-PSU

int validSize(string s, char first, char second){
    string ans = "";
    for(int i = 0; i < s.size(); i++){
        if(s[i] == first || s[i] == second){
            if(ans.size() > 0 && ans[ans.size() - 1] == s[i]) return 0;
            else ans+=s[i];
        }
    }
    if(ans.size() < 2) return 0;
    return ans.size();
}

int alternate(string s) {
    int ans = 0;
    for(char i = 'a'; i < 'z'; i++){
        for(char j = i + 1; j <= 'z'; j++){
           int r = validSize(s, i, j);
           if(r > ans) ans = r;
        }
    }
    return ans;
}