# PROBLEM LINK:

*Author:* Shriram Chandran

*Editorialist:* Shriram Chandran

# DIFFICULTY:

Medium

# PREREQUISITES:

Constructive Algorithms, strings.

# PROBLEM:

Given two binary strings, make both of them palindromes or report that it is not possible.

# EXPLANATION:

First, since both strings are of even length, it is not possible to make both of them palindromic if frequency of each character is odd.

Now, consider the case when both occur an even number of times. This time, the answer definitely exists.

Consider the following process: let us consider all characters in String 1 from index 1 to n/2: and if s_{1,i} \neq s_{1,n+1-i}, we find a character such that s_{1,i}=s_{1,j} and i<j<n+1-i. If such a character exists, we do operation 1 on indices j and n+1-i. If such a character doesnâ€™t exist, we search for it throughout string 2 and find s{2,j} such that s_{1,i}=s_{2,j} and perform operation 2 on indices n+1-i and j.

Convince yourself that string 1 is now a palindrome.

Now, we start string 2 from index 1 to n/2-1 (convince yourself why we donâ€™t do this till position n/2): if s_{2,i} \neq s_{2,n+1-i}, we find a character such that s_{2,i}=s_{2,j} and i<j<n+1-i. This time, the character exists definitely since the answer has to exist. Hence do operation 3 on j and n+1-i.

We are doing at most 1 operation in each iteration and we are doing n-1 iterations, hence we are doing at most n-1 operations.

This is one of multiple similar solutions. All correct answers shall be given AC verdict.

# SOLUTION:

## Setter's Solution

```
#include<bits/stdc++.h>
#define ll long long
#define max(a,b) ((a>b)?a:b)
using namespace std;
void solve()
{
ll n;
cin>>n;
string s1,s2;
cin>>s1>>s2;
ll co=0;
for(int i=0;i<n;i++)
{
if(s1[i]=='a')
co++;
if(s2[i]=='a')
co++;
}
if(co%2==1)
{
cout<<-1<<endl;
return;
}
vector<tuple<ll,ll,ll>>> ans;
for(ll i=0;i<n/2;i++)
if(s1[i]!=s1[n-1-i])
{
bool flag=false;
for(int j=i+1;j<n-i-1;j++)
{
if(s1[j]==s1[i])
{
flag=true;
char temp=s1[j];
s1[j]=s1[n-1-i];
s1[n-1-i]=temp;
ans.push_back(make_tuple(1,i,j));
break;
}
}
if(flag)
continue;
for(int j=0;j<n;j++)
if(s2[j]==s1[i])
{
char temp=s2[j];
s2[j]=s1[n-1-i];
s1[n-1-i]=temp;
ans.push_back(make_tuple(2,i,j));
break;
}
}
for(ll i=0;i<n/2;i++)
if(s2[i]!=s2[n-1-i])
for(int j=i+1;j<n-i-1;j++)
if(s2[j]==s2[i])
{
char temp=s2[j];
s2[j]=s2[n-1-i];
s2[n-1-i]=temp;
ans.push_back(make_tuple(3,i,j));
break;
}
cout<<ans.size()<<endl;
for(ll i=0;i<ans.size();i++)
cout<<get<0>(ans[i])<<" "<<get<1>(ans[i])<<" "<<get<2>(ans[i])<<endl;
}
int main()
{
int t;
cin>>t;
while(t--)
solve();
return 0;
}
```