**Two Characters**

### Problem Statement :

Given a string, remove characters until the string is made up of any two alternating characters. When you choose a character to remove, all instances of that character must be removed. Determine the longest string possible that contains just two alternating letters. Example s = 'abcacdabd' Delete a, to leave bcdbd. Now, remove the character c to leave the valid string bdbd with a length of 4. Removing either b or d at any point would not result in a valid string. Return 4. Given a string s, convert it to the longest possible string t made up only of alternating characters. Return the length of string t. If no string t can be formed, return 0. Function Description Complete the alternate function in the editor below. alternate has the following parameter(s): string s: a string Returns. int: the length of the longest valid string, or 0 if there are none Input Format The first line contains a single integer that denotes the length of s. The second line contains string s. Constraints 1 <= length of s <= 1000

### Solution :

` ````
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)
```

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