Sir , i am not able to understand your logic of solving string compression probel

#1

#include <bits/stdc++.h>
using namespace std;

string compress(string s)
{
if (s.size() == 0)
{
return “”;
}

char ch = s[0];
int i = 1;
while (i < s.size() && s[i] == ch)
{
    i++;
}

string ros = s.substr(i);
ros = compress(ros);

string charCount = to_string(i);
return ch + charCount + ros;

}

int main()
{
string s;
cin >> s;

cout << compress(s) << endl;

return 0;

}
please explain .

my code is
#include
#include
#include
using namespace std;
int main()
{
string s;
getline(cin,s);
int l=s.length();
int sum=1;
for(int i=0,j=1;i<l,j<=l;i++,j++)
{
if(s[i]==s[j])
{
sum++;
}
else if(s[i]!=s[j])
{
cout<<s[i];
cout<<sum;
sum=1;
}
else
{
cout<<s[i];
cout<<sum;
}
}
return 0;
}

#2

In this question you need to tell how many times a character appears in local clusters for example in aaabbbaaa answer would be a3b3a3, the logic is pretty simple j points to i+1, and if s[i] == s[j] we increment the counter and if s[i] != s[j] we know that the cluster has ended we print the character (s[i]) and its count.

#3

Sir, I know the logic but I am. Notable to understand the code of your editorial.

#4

again the code is very simple as explained above, it takes i and j where j points towards i+1, s[i] == s[j] we increment the counter and if s[i] != s[j] we know that the cluster has ended we print the character (s[i]) and its count.

#5

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