String First Questions cannot understand how to solve it

#1

String First Questions cannot understand how to solve it

#2

@ashishsoni1412 You this approach :
You can solve this problem in O(n) time using the two pointer approach.

  • Make two variabes , say i and j .
  • i defines the beginning of a window and j defines its end.
  • Start i from 0 and j from k.
  • Let’s talk about the singular case when we are considering the max window for only 'a’s and consider only the swapping of b-> a. If we are able to get the answer for max window of consecutive 'a’s , we can simply implement the same algo for the max ‘b’ window as well.
  • So we started i from 0 and j from k.
  • Move j ahead freely as long as there are ‘a’ characters at s[ j ] position.
  • Maintain a count variable which counts the number of swaps made or the number of 'b’s in our A window.
  • If you encounter a ‘b’ char at s[ j ] position , increment the count variable. Count should never exceed k .
  • Take the size of the window at every point using length = j - i + 1;
  • Compute the max size window this way and do the same for ‘b’ as well.
  • Output the maximum size window of ‘a’ and ‘b’.
#3

plz provide a solution sir

#4

import java.util.*; public class Main { public static void main(String args[]) { Scanner sc =new Scanner(System.in); String S=sc.nextLine(); int L=0;int R=0; HashSet seen=new HashSet(); int max=0; while(R<S.length()){ char c=S.charAt®; if(seen.add©){ max=Math.max(max,R-L+1); R++; }else{ while(S.charAt(L)==R){ seen.remove(S.charAt(L)); L++; }seen.remove©; L++; } } return max; } }

#5

@ashishsoni1412 You can refer my code :
``

import java.util.Scanner;

public class Main {

    // Function to count the length of window which can be made of char ch with <= k
    // swaps
    static int countMaxWindowSize(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 (; c < k && j < s.length() - 1; j++) {
            if (s.charAt(j) != ch) {
                // If s.charAt(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.length() - 1 && s.charAt(j + 1) == ch) {
                j++;
            }

            // Store the maximum length of all windows
            int currentLength = j - i + 1;
            ans = Math.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.length() - 1 && s.charAt(i - 1) != ch) {
                j++;
            }
        }

        return ans;
    }

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        int k = sc.nextInt();
        String s = sc.next();

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

        // 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
        System.out.println(Math.max(ansForA, ansForB));
    }
}
#6

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