plz explain its sol or give me an approach
#include <bits/stdc++.h>
#define ll long long int
using namespace std;
int main() {
// your code goes here
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() {
// your code goes here
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.