This is definitely not an easy question. One needs to know regexes and sets to solve this (at least, to solve it in a reasonable way). Please reconsider the points and the difficulty of this problem.

[Spoiler alert]

Here is an overview of my O(n) solution: I created a 26x26 matrix that kept track of each possible 2-letter alternation. If at some point I got an invalid alternation for a pair (which happens when a letter follows itself in that alternation), that combination was discarded.

It's easier to explain with an example. Let's simplify the matrix to just 3 letters:

Input: acbab

Matrix begins empty:
  a b c
a 
b
c

We begin with the first 'a'. We update the 'a' row and column with that letter as last:
  a b c
a a a a
b a
c a

  a b c
a a a c
b a   c
c c c c

  a b c
a a b c
b b b b
c c b c

  a b c
a a a a
b a b b
c a b c

Finally, here comes the last letter: 'b'. Note that there is already a 'b' in combination c-b, so that alternation won't work:
  a b c
a a b a
b b b b 
c a b c

To get the longest alternation, I simply kept track of the amount of updates for each combination in another matrix. When a combination was detected as invalid, I marked its count with -1 and never updated it again.

I hope it's useful for you.

well a long explanation but i can assure you guys this is complete explanation for this ques plus i have mentioned the methods you can use in between.

Explanation of solution for this problem: step 1. We can recognise all the unique characters in the string using a nested loop and stringname.charAt(index) function. Ex : if string is abcabfacb then unique characters would be : a,b,c,f

i advice you to take a pen and paper for the following steps.

step 2. construct a 2D matrix for the unique characters as following:

  a b c f
a
b
c
f

step 3. run through each letter of the string and update its row and column with the same letter. string = abcabfacb

 for letter a:   
   a b c f          
 a a a a a       
 b a               
 c a              
 f a          


 for letter b:
   a b c f
 a a b a a
 b b b b b
 c a b 
 f a b

while going through this process check if you are rewriting an alphabet in some other intersection oher than its diagonal intersection for ex : while updating a(string index =7) on the matrix , status before updating was:

   a b c f
 a a b a f
 b b b b f
 c a b c f
 f f f f f

you will notice that there is already a written on a-c intersection. "this means there has been no updation with c alphabet in matrix since previous a alphabet" this suggests that formation of t string is not possible with a-c combination. further going through this step you will find out that with a-c, a-f and b-f have also been ruled out. thus combinations: a-b,b-c and c-f are suitable to form the t string.

step4 : now to find the longest t string i suggest you to construct another similar matrix with those unique characters and update the intersection by adding 1 to that intersection which is being checked in the iteration and update -1 in matrix where the combination is not valid. i will write the final matrix for you guys: ruling out the same letter combinations(representing them by x)

    a  b  c  f
 a  x  6 -1 -1
 b  6  x  5 -1
 c -1  5  x  3
 f -1 -1  3  x

i want you guys to carefully observe the above matrix you will notice that the a-b combination row/column has been updated the most. hence obviously the longest t string will contain these two letters.

step 5: you can print directly the no. of updations on the screen or find out the t string by eliminating the other characters by using the method: s=s.replace("c",""); s=s.replace("f","");

i think you can understand what the method does by its name itself.

phew that was a very long explanation. surely this ques was not an easy one. i wrote this long explanation because this way of solving the ques was very interesting but a bit tricky and can be applied to various other questions. thank you guys for reading it thoroughly.

#include <bits/stdc++.h>
using namespace std;
string a = "abcdefghijklmnopqrstuvwxyz";
//dr0opy 
int main()
{
    int n;
    cin>>n;
    string s;
    cin>>s;
    int maxi=0;//to cal the max length of the possible string
    for(int i=0;i<26;i++){
        for(int j=i+1;j<26;j++){
            char x=a[i];
            char y=a[j];
            string t="";
            for(int k=0;k<n;k++){
                if(s[k]==x||s[k]==y){
                    t+=s[k];
                }
            }
            //cout<<t<<endl;
             bool flag=true;
           for(int f=0;f+1<t.size();f++)
           {
           	if(t[f]==t[f+1])
           	{
           		flag=false;break;
           	}
           }
           int ts=t.size();
           if((flag)&&(ts>1)) maxi = max(maxi,ts);
            
        }
    }
    cout<<maxi<<endl;
	return 0;
}

it took me 4 months and finally did it ;)

Why is this classified as 'easy'. Great question, but this is very misleading for newbies.