Sanket and strings

#1

I want the left, right and mid walla solution. I’m unable to find it please help.
Did this question on my own but getting TLE


link to my code

#2

hi @Yuvraj_Luhach
This problem can be solved with help of two pointers. Let the first pointer is l and the second pointer is r. Then for every position l we will move right end r until on the substring si.si + 1… sr it is possible to make no more than k swaps to make this substring beautiful. Then we need to update the answer with length of this substring and move l to the right.

#3

Hi, Sir I’m unable to understand this explanation could you send me the code so that I can look over and understand it better with your explanation .

#4

you are using brute force approach try optimize your code through the upon explanation its better for you to solve it own your own.

#5

Sir , I can’t understand the explanation could you please provide the code.

#6

ok
import java.util.Scanner;

public class SanketAndString {

public static void main(String[] args) {
    // TODO Auto-generated method stub
    Scanner scn = new Scanner(System.in);
    int k=scn.nextInt();
    String str = scn.next();
    int left=0, ans=0;
    int[] count = {0,0};
    for (char c: str.toCharArray()) {
        count[c-'a']++;
        if(Math.min(count[0], count[1]) > k) {
            count[str.charAt(left)-'a']--;
            left++;
        } else {
            ans++;
        }
    }
    System.out.println(ans);
}

}

#7

@Yuvraj_Luhach
please mark your doubt as resolved and rate me as well.

#8

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