TO MAKE A WINDOW ALL ONE LETTER, YOU REPLACE EVERY TILE THAT IS NOT ITS MOST COMMON ONE. SO A WINDOW COSTS ITS LENGTH MINUS THE TOP LETTER COUNT, AND IT IS ALLOWED WHEN THAT COST IS AT MOST K. GROW WHILE IT FITS, AND WHEN THE COST WOULD EXCEED K, SLIDE THE LEFT EDGE RIGHT BY ONE. THE WINDOW NEVER SHRINKS, SO ITS LARGEST LENGTH IS THE ANSWER.

GOLD TILES ARE THE LETTER YOU KEEP, THE MOST COMMON IN THE WINDOW. BLUE TILES ARE THE UP TO K YOU REPLACE. L AND R MARK THE WINDOW ENDS, AND THE NEED LINE READS LENGTH MINUS THE TOP COUNT AGAINST K. WHEN A NEW LETTER PUSHES THE COST OVER K, L SLIDES RIGHT AND THE WINDOW HOLDS ITS SIZE.

1def character_replacement(s, k):2    char_freq = {}3    max_l = 04    l = 05    for r in range(len(s)):6        c = s[r]7        char_freq[c] = 1 + char_freq.get(c, 0)8        if (r - l + 1) - max(char_freq.values()) > k:9            char_freq[s[l]] -= 110            l += 111        max_l = max(max_l, r - l + 1)12    return max_l