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Tested Python solution for LeetCode 2516 with 20 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 2516 # by problem number
lcpy gen -s take_k_of_each_character_from_left_and_right # by problem name
s consisting of the characters 'a', 'b', and 'c' and a non-negative integer k. Each minute, you may take either the leftmost character of s, or the rightmost character of s.
Return the minimum number of minutes needed for you to take at least k of each character, or return -1 if it is not possible to take k of each character.
Input: s = "aabaaaacaabc", k = 2
Output: 8
Explanation:
Take three characters from the left of s. You now have two 'a' characters, and one 'b' character.
Take five characters from the right of s. You now have four 'a' characters, two 'b' characters, and two 'c' characters.
A total of 3 + 5 = 8 minutes is needed.
It can be proven that 8 is the minimum number of minutes needed.
Input: s = "a", k = 1
Output: -1
Explanation: It is not possible to take one 'b' or 'c' so return -1.
1 <= s.length <= 10^5s consists of only the letters 'a', 'b', and 'c'.0 <= k <= s.lengthclass Solution:
# Time: O(n)
# Space: O(1)
def take_characters(self, s: str, k: int) -> int:
n = len(s)
count = [0, 0, 0]
for ch in s:
count[ord(ch) - ord("a")] += 1
if any(c < k for c in count):
return -1
# Keep the longest middle window whose removal leaves >= k of each char.
best = 0
left = 0
for right, ch in enumerate(s):
count[ord(ch) - ord("a")] -= 1
while count[ord(ch) - ord("a")] < k:
count[ord(s[left]) - ord("a")] += 1
left += 1
best = max(best, right - left + 1)
return n - best
| Time | Space |
|---|---|
| O(n) | O(1) |