This part of code is unclear to me

#1

// 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++;
}

#2

Hi @ssakshamprime,
This part suggest that if the character at the left pointer is not equal to ch then we would have already counted this character and increased c or count. hence decreasing a count because of left pointer and then increasing a count for the right pointer will not change the total number of swaps or count value hence this is a free swap as one is decreasing the count value and the other is increasing the count value…therefore count value remains the same. I hope this made it clear. If not then try to dry run the code once with paper and pen and you will understand the flow.

#3

Can you suggest any test case for this?

#4

Yeah got it, at this point j will be pointing to a character != ‘ch’ so we are just incrementing j to increase the size of window.

#5

Nice … If you still need test case then
2
aaabbbbbbaabbb
Take this as test case the answer is 11.

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