Dealing with constraints in strings can feel like a balancing act! 🧩 In LeetCode problem #2516: Take K of Each Character From Left and Right, the challenge is to efficiently remove characters from the edges of a string while ensuring at least k occurrences of 'a', 'b', and 'c' remain. Using the sliding window technique, we’ll determine the minimum time required to achieve this. Ready to dive into a smart and efficient solution? Let’s go! 🚀

🚀 Problem Description

You are given a string s containing only the characters 'a', 'b', and 'c', along with a non-negative integer k. The goal is to find the minimum number of minutes required to remove characters such that at least k instances of each character ('a', 'b', 'c') remain in the string.

You can only remove one character per minute, and the removal must be either from the leftmost or the rightmost side of the string. If it is impossible to meet the condition, return -1.

🛠️ Problem Details

• Difficulty: Medium
• Tags: Hash Table, String, Sliding Window
• LeetCode Link: Take K of Each Character From Left and Right

🧐 Key Intuition

To efficiently solve this problem, we use the sliding window technique. The main idea is to determine the smallest window that can be removed from the string while still ensuring at least k instances of 'a', 'b', and 'c' remain. This leads us to the answer by computing: [ \text{Answer} = \text{Length of } s - \text{Size of the largest valid window found.} ]

🔑 Steps to Solve

1. Character Counting: Use a frequency counter to track occurrences of 'a', 'b', and 'c'. If any character has fewer than k instances initially, the answer is immediately -1.

2. Sliding Window: Identify the largest contiguous substring (or “window”) that can be removed from s, leaving enough characters of 'a', 'b', and 'c' outside the window to satisfy the condition.

3. Optimize Window: Adjust the window boundaries dynamically using two pointers (i and j). If removing characters causes the count of any character to drop below k, move the left pointer (j) forward to restore balance.

4. Calculate Result: The minimum time required is the difference between the total string length and the size of the largest removable window.

🔧 Algorithm Implementation

Here’s a Python implementation of the above approach using the sliding window method:

from collections import Counter

class Solution:
    def takeCharacters(self, s: str, k: int) -> int:
        # Count the frequency of each character in the string
        char_count = Counter(s)
        
        # Check if it's impossible to satisfy the condition
        if any(char_count[char] < k for char in "abc"):
            return -1  # Return -1 if any character occurs less than k times
        
        # Initialize pointers and variables
        max_window_size = 0
        left = 0
        n = len(s)
        
        # Iterate over the string with the right pointer
        for right in range(n):
            # Decrement the count of the current character
            char_count[s[right]] -= 1
            
            # If any character count drops below k, adjust the left pointer
            while any(char_count[char] < k for char in "abc"):
                char_count[s[left]] += 1
                left += 1
            
            # Update the maximum window size that can be removed
            max_window_size = max(max_window_size, right - left + 1)
        
        # The minimum time required is the length of the string minus the largest window
        return n - max_window_size

🛠 Example Walkthrough

Let’s work through an example step-by-step to fully understand the solution.

Example 1:

Input:
s = "abbcabc"
k = 2

Steps:

1. Initial Frequency Count:

   {'a': 2, 'b': 3, 'c': 2}
   

   Each character occurs at least k times. Proceed to the sliding window.

2. Sliding Window Process:
   • Start with two pointers left = 0 and right = 0.
   • Iterate over the string:

   right	Removed	Adjust left	Updated Counts	Max Window
   0	'a'	No	{'a': 1, 'b': 3, 'c': 2}	1
   1	'b'	No	{'a': 1, 'b': 2, 'c': 2}	2
   2	'b'	Yes	{'a': 2, 'b': 2, 'c': 2}	2
   3	'c'	No	{'a': 1, 'b': 2, 'c': 1}	3
   …	…	…	…	…

3. Result Calculation:
   • Largest valid window found: “bca” (size 3).
   • Total string length: 7.
   • Answer: 7 - 3 = 4.

Output:
4

🔬 Complexity Analysis

1. Time Complexity:
   • The outer loop iterates over the string once (O(n)).
   • The inner loop adjusts the left pointer, but it only progresses forward, processing each character once (O(n)).
   • Total time complexity: O(n).
2. Space Complexity:
   • The space used by the Counter is constant (O(1)), as we only track counts for 'a', 'b', and 'c'.
   • Total space complexity: O(1).

🌟 Advantages of the Sliding Window Approach

• Efficiency: Linear time complexity ensures the solution scales well with longer strings.
• Dynamic Adjustment: The two-pointer technique dynamically adjusts the boundaries of the removable window, ensuring minimal removals.

🧮 Edge Cases

1. Insufficient Characters:
   Input: s = "aabbcc", k = 3
   Output: -1 (impossible to satisfy the condition).

2. Already Satisfied:
   Input: s = "aaa", k = 1
   Output: 0 (no removals needed).

3. Single Character Repeats:
   Input: s = "aaaaaa", k = 2
   Output: 4 (remove 4 characters).

💡 Conclusion

The sliding window method offers a clean and efficient solution to this problem, focusing on the key task of finding the largest valid removable window. With O(n) time complexity, it performs well on large inputs and handles edge cases gracefully. This approach demonstrates the power of dynamic window adjustment in solving substring problems efficiently.