2599 Take K Of Each Character From Left And Right
2599 Take K Of Each Character From Left And Right
Take K of Each Character From Left and Right 
You are given a string 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.*
Example 1:
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**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.
Example 2:
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**Input:** s = "a", k = 1
**Output:** -1
**Explanation:** It is not possible to take one 'b' or 'c' so return -1.
Constraints:
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1 <= s.length <= 105
s consists of only the letters 'a', 'b', and 'c'.
0 <= k <= s.length
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class Solution:
def takeCharacters(self, s: str, k: int) -> int:
count = [0] * 3
n = len(s)
# Count total occurrences
for c in s:
count[ord(c) - ord("a")] += 1
# Check if we have enough characters
for i in range(3):
if count[i] < k:
return -1
window = [0] * 3
left, max_window = 0, 0
# Find the longest window that leaves k of each character outside
for right in range(n):
window[ord(s[right]) - ord("a")] += 1
# Shrink window if we take too many characters
while left <= right and (
count[0] - window[0] < k
or count[1] - window[1] < k
or count[2] - window[2] < k
):
window[ord(s[left]) - ord("a")] -= 1
left += 1
max_window = max(max_window, right - left + 1)
return n - max_window
This post is licensed under CC BY 4.0 by the author.