Leetcode #2516: Take K of Each Character From Left and Right
In this guide, we solve Leetcode #2516 Take K of Each Character From Left and Right in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Hash Table, String, Sliding Window
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
class Solution:
def takeCharacters(self, s: str, k: int) -> int:
cnt = Counter(s)
if any(cnt[c] < k for c in "abc"):
return -1
mx = j = 0
for i, c in enumerate(s):
cnt[c] -= 1
while cnt[c] < k:
cnt[s[j]] += 1
j += 1
mx = max(mx, i - j + 1)
return len(s) - mx
Complexity
The time complexity is , where is the length of string . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.