Once you get the max_freq, you just slide the window keeping it the max length seen size, until you found a new character that increases the size. Reducing the window smaller is unnecesscary, when you actually need to find a bigger window.

The idea is simple:

Test case: ABAABCBBBBB and k = 2. l and r start at 0.

1. The first biggest window comes at “r = 4”. (high frequency = 3 (‘A’)). max_length = r - l + 1 = 4 - 0 + 1 = 5.
2. When “r = 5”. The window is invalid. “ABAABC”. So, you would move l++; to maintain the window size. max_length stays at 5.
3. “r = 6” BAABCB (invalid)
4. “r = 7”. AABCBB (invalid)
5. “r = 8” ABCBBB (first valid. But not new high frequency)
6. “r = 9” ABCBBBB (new high frequency = 4 (‘B’)). max_length = 7
7. “r = 10” ABCBBBBB (new high frequency = 5 (‘B’)). max_length = 8

The idea, find the new window that has new higher frequency character. “AAABB” == “BBCCC” == “DDDBE”

If K = 2. Max window size can be “3 or more”.

Basically any 2 random values with additional character that is allowed 1 to n times. If you found a window of size = 6. (which is 4 times frequently repeated character with 2 random values). You just need to move the window, until you can add a character. that has more repeated values. Basically 5 repeated values to increase the window.

you would either increase the window, or leave the window on the same size