Need Some Hint how to approach this problem

#1

Cant make the logic and how to start this problem.

#2

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’.

refer this code --> https://ide.codingblocks.com/s/642677

#3

hi @mohit26900
i hope its clear now… is there anything else??

#4

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#5

@mohit26900 we can easily solve this problem using 2 pointer sliding window approach
refer to below code for more clarity

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

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

int n=s.length();
int l=0,r=0;
int ans=0;
int counta=0,countb=0;
while(l<=r && l<n && r<n)
{
	if(counta>k || countb>k)
	{
		while(counta>k || countb>k)
		{
			if(s[l]=='a')
			counta--;
			if(s[l]=='b')
			countb--;
			l++;
		}
	}
	else
	{
		ans=max(ans,r-l+1);
		if(s[r]=='a')
		counta++;
		else
		countb++;
		
		r++;
	}
}
cout<<ans<<endl;
return 0;

}

Ip: string = “abba” k=2
op: 4

#6

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