# 2516. Take K of Each Character From Left and Right

• Time: $O(n)$
• Space: $O(1)$
  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 class Solution { public: int takeCharacters(string s, int k) { const int n = s.length(); int ans = n; vector count(3); for (const char c : s) ++count[c - 'a']; if (count[0] < k || count[1] < k || count[2] < k) return -1; for (int l = 0, r = 0; r < n; ++r) { --count[s[r] - 'a']; while (count[s[r] - 'a'] < k) ++count[s[l++] - 'a']; ans = min(ans, n - (r - l + 1)); } return ans; } }; 
  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 class Solution { public int takeCharacters(String s, int k) { final int n = s.length(); int ans = n; int[] count = new int[3]; for (final char c : s.toCharArray()) ++count[c - 'a']; if (count[0] < k || count[1] < k || count[2] < k) return -1; for (int l = 0, r = 0; r < n; ++r) { --count[s.charAt(r) - 'a']; while (count[s.charAt(r) - 'a'] < k) ++count[s.charAt(l++) - 'a']; ans = Math.min(ans, n - (r - l + 1)); } return ans; } } 
  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 class Solution: def takeCharacters(self, s: str, k: int) -> int: n = len(s) ans = n count = collections.Counter(s) if any(count[c] < k for c in 'abc'): return -1 l = 0 for r, c in enumerate(s): count[c] -= 1 while count[c] < k: count[s[l]] += 1 l += 1 ans = min(ans, n - (r - l + 1)) return ans