 # BTSTRT4 - Editorial

Setter: Rahul Kumar
Tester: Aniket Kumar
Editorialist: Rahul Kumar

DIFFICULTY:
SIMPLE - EASY

PROBLEM:
Given a string of length N which contains only A or B characters. A good string is the one in which all the characters are same. You need to find the longest substring that is a good string by doing the one of the following operation only once.

• Choose a good substring and subtract 1 from all the characters of this substring.
• or add 1 to all the characters of this substring.

Subtracting 1 from a character means replacing it with a letter below it for e.g. B-1=A, D-1=C. And adding 1 means replacing it with letter above it for e.g. B+1=C, X+1=Y.
Print the max length of good substring after the operation.
Print minimum XOR.

EXPLANATION:
We will store the frequency of letter for every continuous characters in an array.
Then we will traverse the array checking maximum value of A[i-1] + A[i] + A[i+1] .
So that the operation can be performed in the A[i] part to make the characters same.

TIME COMPLEXITY:
O(N)

SOLUTION:

``````#include<bits/stdc++.h>
using namespace std;
#define fio ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define int long long int
#define vi vector<int>
#define w(x) int x; cin>>x; while(x--)
#define pb push_back
#define print(v) for(int i=0;i<v.size();i++) cout<<v[i]<<" ";

int32_t main() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
fio
w(t)
{
int n;
cin>>n;
vi p;
string v;
cin>>v;
char c=v; /*storing any of the two letters*/
int o=0,z=0;
for(int i=0;i<n;i++)
{
if(v[i]==c)
{
z++;
if(o>0)
{
p.pb(o); /*storing no. of 2nd letter when transition from 2nd  to 1st*/
o=0;
}
}
else
{
o++;
if(z>0)
{
p.pb(z); /*storing no. of 1st letters when transition from 1st to 2nd */
z=0;
}
}
/*When reaches last index*/
if(i==n-1)
{
if(z>0)
{
p.pb(z); z=0;
}
else if(o>0)
{
p.pb(o); o=0;
}
}
}
int m=0;
for(int i=0;i<p.size()-1;i++)
{
if(i==0)
{
m=max(p[i]+p[i+1],m);
}
else if(i==n-1)
{
m=max(p[i]+p[i-1],m);
}
else
m=max(p[i-1]+p[i]+p[i+1],m);
}
cout<<m<<"\n";
}
}
``````

Please comment below if you have any questions, alternate solutions, or suggestions. 