#2516

Medium

auto_awesomeSliding window with running state updates

LeetCode Problem Workspace

Take K of Each Character From Left and Right

Find the minimum number of minutes needed to take at least k of each character from both ends of a string.

Problem Statement

You are given a string s containing only the characters 'a', 'b', and 'c', and a non-negative integer k. Each minute, you may take either the leftmost or the rightmost character from s. Your goal is to determine the minimum number of minutes required to take at least k of each character, or return -1 if it is not possible.

For example, with s = "aabaaaacaabc" and k = 2, the correct output is 8, since you need to take characters from both ends. You must ensure that you count at least k occurrences of 'a', 'b', and 'c'.

Examples

Example 1

Input: s = "aabaaaacaabc", k = 2

Output: 8

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

Input: s = "a", k = 1

Output: -1

It is not possible to take one 'b' or 'c' so return -1.

Constraints

1 <= s.length <= 105
s consists of only the letters 'a', 'b', and 'c'.
0 <= k <= s.length

Solution Approach

Frequency Counting

First, count the frequency of each character in the string to determine if it's possible to take at least k occurrences of each character. If any character has a frequency less than k, return -1 immediately.

Sliding Window Technique

Use a sliding window to check the minimum number of characters you need to take from both the left and right sides. The window should grow and shrink while maintaining the count of characters and ensuring at least k occurrences of each character.

Efficient State Updates

As the window moves, update the state efficiently by tracking the number of characters on the left and right. This dynamic approach will help you find the minimum time needed to fulfill the requirement.

Complexity Analysis

Metric	Value
Time	O(n)
Space	O(3) = O(1)

The time complexity is O(n) due to the linear pass through the string, with each character being processed at most twice (once when entering and once when exiting the window). The space complexity is O(1), since the space used is constant for tracking character frequencies, regardless of input size.

What Interviewers Usually Probe

Look for understanding of sliding window technique and its relationship to character frequency counting.
Evaluate the candidate’s ability to efficiently update the window state and maintain character counts.
Check if the candidate can handle edge cases, like when k is greater than the number of occurrences of a character.

Common Pitfalls or Variants

Common pitfalls

Failing to check if k is feasible before attempting to process the string, which may lead to unnecessary computation.
Incorrectly managing the sliding window and failing to maintain accurate character counts across both ends of the string.
Overcomplicating the problem by using unnecessary data structures instead of the efficient sliding window with simple character counting.

Follow-up variants

Implementing this approach for strings with more than three characters.
Extending this solution to handle varying values of k for different characters simultaneously.
Optimizing for larger strings or strings with non-trivial distributions of characters.

FAQ

What is the sliding window technique?

The sliding window technique is a method used to maintain a subset of elements in a sequence, allowing efficient updates as the window moves through the data, often used for substring or subarray problems.

How do I know if it’s possible to take k of each character?

First, count the frequency of each character in the string. If any character’s frequency is less than k, it’s impossible to take k occurrences, and you should return -1.

What if the string contains characters other than 'a', 'b', and 'c'?

This problem specifically assumes the string only contains 'a', 'b', and 'c'. For strings containing other characters, the problem would need to be modified.

How does GhostInterview assist with the sliding window technique?

GhostInterview provides guidance on how to maintain the sliding window’s state, ensuring efficient character counting and helping you solve the problem step by step.

What is the optimal time complexity for this problem?

The optimal time complexity is O(n), where n is the length of the string. This is achieved by using the sliding window technique with efficient state management.

terminal

Solution

Solution 1: Sliding Window

First, we use a hash table or an array of length $3$, denoted as $cnt$, to count the number of each character in string $s$. If any character appears less than $k$ times, it cannot be obtained, so we return $-1$ in advance.

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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

Continue Practicing

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