Two Characters

                            Solution in C :

In  C++  :







#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> ii;

int valid(string x) {
    const int n = x.size();
    for (int i = 1; i < n; ++i)
        if (x[i] == x[i-1])
            return false;
    return true;
}

int main() {
    int asd;
    cin>>asd;
    string s;
    cin>>s;
    int ans = 0;
    for (char a = 'a'; a <= 'z'; ++a)
    for (char b = 'a'; b <= 'z'; ++b)
    if (a != b)
    {
        if (s.find(a) == string::npos) continue;
        if (s.find(b) == string::npos) continue;
        string x;
        for (const char ch : s)
            if (ch == a || ch == b)
                x.push_back(ch);
        if (valid(x))
            ans = max(ans, (int)x.size());
    }
    printf("%d\n", ans);
}








In   Java  :







import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        String str = in.next();
        HashSet<Character> set = new HashSet<Character>();
        char[] ar = str.toCharArray();
        for (char c : ar) set.add(c);
        int ans = 0;
        Character[] keys = set.toArray(new Character[set.size()]);
        for (int i = 0 ; i < set.size() ; i++) {
            for (int j = i+1 ; j < set.size() ; j++) {
                String b = str;
                for (char c : set) {
                    if (c != keys[i] && c != keys[j]) b = b.replace(c + "", "");
                }
                
                char[] ts = b.toCharArray();
                boolean valid = ts.length > 0;
                for (int k = 0 ; valid && k < ts.length - 1 ; k++) {
                    if (ts[k] == ts[k+1]) valid = false; 
                }
                if (valid) ans = Math.max(ans, ts.length);
            }
        }
        System.out.println(ans);
    }
}








In   C :






#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main() {
	int len, you[26] = { 0 }, c = 0, i, j, k, bv = 0, stat;
	scanf("%d", &len);
	char s[1010];
	scanf("%s", s);
	for (i = 0; i < len; i++)you[s[i] = s[i] - 'a']++;
	for (i = 0; i < 26; i++)if (you[i])c++;
	if (c <= 1) {
		puts("0");
		return 0;
	}
	for (i = 0; i < 26; i++)for (j = i + 1; j < 26; j++)if (you[i] && you[j]) {
		stat = -1;
		c = 0;
		for (k = 0; k < len; k++)if (s[k] == i || s[k] == j) {
			if (stat == -1)stat = (s[k] == j);
			else if (stat == (s[k] == j))goto nxt;
			else stat = 1 - stat;
			c++;
		}
		if (c > bv)bv = c;
	nxt:;
	}
	printf("%d\n", bv);
	return 0;
}








In   Python3  :






def char_range(c1, c2):
    for c in range(ord(c1), ord(c2)+1):
        yield chr(c)
        
def is_alternating(s):
    for i in range(len(s) - 1):
        if s[i] == s[i+1]:
            return False
    return True
        
len_s = int(input())

s = input()

max_len = 0
        
for c1 in char_range('a', 'z'):
    for c2 in char_range('a', 'z'):
        if c1 == c2:
            continue
        new_string = [c for c in s if c == c1 or c == c2]
        new_string_len = len(new_string)
        if is_alternating(new_string) and new_string_len > 1:
            if new_string_len > max_len:
                max_len = new_string_len
                
print(max_len)