# Help me in solving LTB11121 problem

### My issue

plz explain its sol or give me an approach

### My code

``````#include <bits/stdc++.h>
#define ll long long int
using namespace std;

int main() {
ll t;
cin>>t;
while(t--){
ll n;
cin>>n;
string s,str;
cin>>s>>str;
ll diff=0;
for(ll i=0;i<n;i++){
if(s[i]!=str[i])diff++;
}
if()
}

return 0;
}

``````

Problem Link: LTB11121 Problem - CodeChef

``````//ladchat supremacy

#include <bits/stdc++.h>

using namespace std;

#define ll long long

#define MAX 1000000007

int main() {

ll t;

cin>>t;

while(t--){

ll n;

cin>>n;

string s,ans;

cin>>ans;

//string ans;

cin>>s;

vector<ll>need(n,0);

for(ll i=0;i<n;i++)if(s[i]!=ans[i])need[i]++;

ll first,last;

first=last=-1;

for(ll i=0;i<n;i++){

if(s[i]!=ans[i]){

first=i;

break;

}

}

for(ll i=n-1;i>=0;i--){

if(s[i]!=ans[i]){

last=i;

break;

}

}

ll seg=0;

ll it=0;

while(it<n){

if(s[it]!=ans[it]){

while(it<n && s[it]!=ans[it])it++;

seg++;

}

else it++;

}

//cout<<first<<" "<<last<<" ";

//cout<<seg<<" ";

if(seg==1){

ll fin=0;

fin=((first-0)+(n-1-last))*2+(last-first)*2; // first and last belong to same segment

cout<<fin<<endl;

}

else if(seg==0){

cout<<(n*(n+1)/2)<<endl;

}

else if(seg>2)cout<<0<<endl;

else{

cout<<6<<endl;

}

}

return 0;

}

``````

include <bits/stdc++.h>
define ll long long int
using namespace std;

int main() {
ll t;
cin >> t; // Read the number of test cases

``````while (t--) { // Iterate through each test case
ll n;
cin >> n; // Read the length of the strings

string s, str;
cin >> s >> str; // Read two strings of length n

ll diff = 0;
for (ll i = 0; i < n; i++) {
if (s[i] != str[i])
diff++; // Count the number of positions where the strings differ
}

// Now, you need to add some code here to determine the result based on the value of 'diff'.

// For example, you can print "YES" if diff is 0, indicating that the strings are identical.
// Or you can print "NO" if diff is not 0, indicating that the strings are different.

// Here's an example:
if (diff == 0) {
cout << "YES" << endl;
} else {
cout << "NO" << endl;
}
}

return 0;
``````

}
The code reads the number of test cases `t`, and for each test case, it reads two strings `s` and `str` of length `n`. It then calculates the number of positions where these strings differ and stores it in the `diff` variable. Finally, you need to add code to determine the result based on the value of `diff`.

In the example code I provided, it prints “YES” if `diff` is 0 (indicating the strings are identical) and “NO” otherwise. You can modify this part based on the specific problem requirements.