Sanket and Strings Problem: What is wrong in my approach? It's giving wrong answers

#1

import java.util.Scanner;
class Main{
public static Scanner scn = new Scanner(System.in);
public static void main(String[] args){
sanketStrings();
}

public static void sanketStrings(){
    int k = scn.nextInt();
    String str = scn.next();
    int count_a = 0;
    int count_b = 0;

    for(int i = 0;i< str.length();i++){
        if(str.charAt(i)=='a'){
            count_a++;
        }else{
            count_b++;
        }
    }
    int perfect = 0;
    if(count_a>count_b){
        perfect=count_a;
        if(count_b<=k){
            perfect = perfect + count_b;
        }else{
            perfect = perfect + k;
        }
    }else{
        perfect = count_b;
        if(count_a<=k){
            perfect = perfect + count_a;
        }else{
            perfect = perfect + k;
        }
    }

    System.out.println(perfect);

}

}

#2

1
ababababaaba
its correct output is:4
And Your Code’s output is: 8

#3

logic is not correct.
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’.

#4


you can see this

#5

Oh ,I got the problem with my logic

#6

Question is asking for max length substring…Sorry :frowning:

#7

Hey, can you once again share your code? It’s code replaced by a tree code now

#8