Working of second loop

#1

#include
#include
using namespace std;

//Function to count the length of window which can be made of char ch with <= k swaps
int countMaxWindowSize(const string &s, char ch, int k)
{
int i = 0; //Left pointer
int j = 0; //Right pointer

//First move the right pointer forward by k steps.
//If the character is already ch , do not count a swap and move freely

int c = 0; //Variable to count the swaps so far

int ans = 0; //Variable to store the final answer

for (j; c < k && j < s.size() - 1; j++)
{
    if (s[j] != ch)
    {
        //If s[j] is not ch then count it as a swap and move forward
        c++;
    }
    if (c == k)
    {
        //If no of swaps has reached k, stop moving j any more forward
        break;
    }
}

while (i < j)
{

    //Move j ahead if next element is ch as it doesn't count as a swap
    while (j < s.size() - 1 && s[j + 1] == ch)
    {
        j++;
    }

    //Store the maximum length of all windows
    int currentLength = j - i + 1;
    ans = max(ans, currentLength);

    //Move left pointer by one to slide the window
    i++;

    //If the char at previous position of left pointer was not ch, then that position must
    //have counted as a swap earlier. Now we have a free swap available.
    //Iterate right pointer forward to use that one free swap
    if (j < s.size() - 1 && s[i - 1] != ch)
    {
        j++;
    }
}

return ans;

}

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

if (k >= s.size())
{
    //If k is larger than s.size() then we can swap all the elements to either A or B
    //and obtain the answer equal to length of string
    cout << s.size();
    return 0;
}

//First let us check for longest perfect string of A's then we will find the same for B's and compare
int ansForA = countMaxWindowSize(s, 'a', k);

//Now we do the same for B's
int ansForB = countMaxWindowSize(s, 'b', k);

//Final answer is max of the two answers obtained above
cout << max(ansForA, ansForB);

return 0;

}

Can you please explain the working of second loop

and can you please refer any video for it