1. Iterate i from 0 to n.
2. Iterate j from 0 to n - i.
3. Construct the set of characters formed by s[0...i-1] and s[n-j...n-1].
4. Check if the counts of 'a', 'b', and 'c' in this combined set are all $\ge k$.
5. If valid, record i + j and update the minimum.

Pseudo-code:

python
min_len = infinity
for i from 0 to n:
    for j from 0 to n - i:
        current_selection = s[0:i] + s[n-j:n]
        if counts(current_selection) satisfy k:
            min_len = min(min_len, i + j)

Why it fails: The time complexity of this approach is $O(n^2)$ because of the nested loops. Given the constraint $n \le 10^5$, an $O(n^2)$ solution requires roughly $10^{10}$ operations, which far exceeds the typical limit of $10^8$ operations per second. This results in a Time Limit Exceeded (TLE) error.

Core Insight for Take K of Each Character From Left and Right

The key intuition for Take K of Each Character From Left and Right is essentially "Problem Inversion."

Directly deciding how many characters to take from the left and right is difficult because the two decisions are coupled; taking more from the left might allow taking fewer from the right, but they are separated by the middle of the string.

However, the characters we don't take form a single contiguous substring in the middle of the string. Let the total length of the string be $N$. Let the number of characters taken be $X$. Let the number of characters left in the middle be $Y$. We know that $X + Y = N$. To minimize $X$ (taken characters), we must maximize $Y$ (the middle substring).

The Constraint: For a middle substring to be valid, the characters remaining outside of it (the prefix and suffix) must contain at least k of 'a', 'b', and 'c'. Mathematically, if the total count of 'a' is $Total_A$ and the count of 'a' inside our middle window is $Window_A$, then we must maintain: $Total_A - Window_A \ge k$ (And similarly for 'b' and 'c').

This transformation converts the problem into finding the longest substring where the counts of characters inside the substring do not exceed $Total - k$.

Visual Description: Imagine the string as a fixed bar. We slide a window (the middle section) across this bar. As the window expands to the right (including more characters in the "ignore" pile), we check if we have "ignored" too many of a specific character. If Total - Window < k, it means we have kept too many inside the window and haven't left enough for the prefix/suffix. We then shrink the window from the left until the condition is restored.

Execution Flow

Let's trace s = "aabaaaacaabc", k = 2. Length $n=12$.

1. Total Counts: A: 8, B: 2, C: 2.

   • Required outside: 2 of each.
   • Allowed inside window: A: $8-2=6$, B: $2-2=0$, C: $2-2=0$.
   • This means our window cannot contain any 'b' or 'c', and at most 6 'a's.

2. Window Expansion:

   • right=0 ('a'): Window{'a':1}. Valid. Max len = 1.
   • right=1 ('a'): Window{'a':2}. Valid. Max len = 2.
   • right=2 ('b'): Window{'a':2, 'b':1}. Invalid (Limit for B is 0).
     • Shrink left. Remove 'a'. Window{'a':1, 'b':1}. Still Invalid.
     • Shrink left. Remove 'a'. Window{'b':1}. Still Invalid.
     • Shrink left. Remove 'b'. Window{}. Valid.
   • right=3 ('a'): Window{'a':1}. Valid. Max len = 2 (unchanged? No, current is 1. Max remains 2).
   • ... (Skipping ahead) ...
   • The window will struggle to grow because 'b' and 'c' limits are 0.
   • Actually, wait. The example output is 8. Let's re-read the trace carefully.

   Correction on Logic Application: The example output is 8, meaning the window length is $12 - 8 = 4$. Let's re-verify the "Allowed Inside" logic. Total: A=8, B=2, C=2. k=2. We need 2 of each outside. We have exactly 2 'b's and 2 'c's total. So, if we put 'b' or 'c' inside the window, the outside count drops to 1, which is . Therefore, the window can ONLY contain 'a's. The longest substring of only 'a's is "aaaa" (index 3 to 6). Length 4. Result: . Correct.

3. Algorithm Steps:

   • Initialize count array for current window.
   • ans = 0 (max window length).
   • Loop right from 0 to n:
     • Add s[right] to window count.
     • While count[s[right]] > total[s[right]] - k:
       • Remove s[left] from window count.
       • left++.
     • ans = max(ans, right - left + 1).
   • Return n - ans.

Proof of Correctness

The algorithm's correctness relies on the invariant maintained by the sliding window. At every step right, the while loop ensures that the window s[left...right] is valid. A valid window implies that for every character $c$, window_count[c] <= total_count[c] - k. Rearranging this inequality gives total_count[c] - window_count[c] >= k. Since total - window represents the count of characters outside the window (prefix + suffix), the invariant guarantees that the prefix and suffix combined satisfy the problem's condition. By finding the maximum length of such a window, we mathematically minimize the length of the prefix + suffix.

Reference Implementation

C++ Solution for LeetCode 2516

cpp
class Solution {
public:
    int takeCharacters(string s, int k) {
        int n = s.length();
        // Count total occurrences of a, b, c
        vector<int> total(3, 0);
        for (char c : s) {
            total[c - 'a']++;
        }

        // Check if solution is possible
        if (total[0] < k || total[1] < k || total[2] < k) {
            return -1;
        }

        // Sliding window to find max length of middle part
        int maxWindow = 0;
        int left = 0;
        vector<int> window(3, 0);

        for (int right = 0; right < n; right++) {
            int charIndex = s[right] - 'a';
            window[charIndex]++;

            // Shrink window if we took too many characters inside
            // (leaving too few outside)
            while (window[charIndex] > total[charIndex] - k) {
                window[s[left] - 'a']--;
                left++;
            }

            maxWindow = max(maxWindow, right - left + 1);
        }

        return n - maxWindow;
    }
};

Java Solution for LeetCode 2516

java
class Solution {
    public int takeCharacters(String s, int k) {
        int n = s.length();
        int[] total = new int[3];
        
        // Count total occurrences
        for (char c : s.toCharArray()) {
            total[c - 'a']++;
        }
        
        // Check feasibility
        if (total[0] < k || total[1] < k || total[2] < k) {
            return -1;
        }
        
        int[] window = new int[3];
        int left = 0;
        int maxWindow = 0;
        
        for (int right = 0; right < n; right++) {
            int charIndex = s.charAt(right) - 'a';
            window[charIndex]++;
            
            // If window contains too many of current char, shrink from left
            // "Too many" means remaining outside < k
            while (window[charIndex] > total[charIndex] - k) {
                window[s.charAt(left) - 'a']--;
                left++;
            }
            
            maxWindow = Math.max(maxWindow, right - left + 1);
        }
        
        return n - maxWindow;
    }
}

Python Solution for LeetCode 2516

python
class Solution:
    def takeCharacters(self, s: str, k: int) -> int:
        n = len(s)
        # Map 'a', 'b', 'c' to 0, 1, 2 for array indexing
        total = [0] * 3
        for char in s:
            total[ord(char) - ord('a')] += 1
            
        # If total counts are insufficient, return -1
        if any(count < k for count in total):
            return -1
            
        window = [0] * 3
        left = 0
        max_window = 0
        
        for right in range(n):
            idx = ord(s[right]) - ord('a')
            window[idx] += 1
            
            # Shrink window if remaining characters outside < k
            # Equivalent to checking if window[idx] > total[idx] - k
            while window[idx] > total[idx] - k:
                left_idx = ord(s[left]) - ord('a')
                window[left_idx] -= 1
                left += 1
            
            max_window = max(max_window, right - left + 1)
            
        return n - max_window

Q: Why does my code return -1 even when a solution exists? Mistake: Incorrectly initializing the feasibility check. Concept Gap: You must verify that total_count[char] >= k for all characters before starting the window logic. If the string doesn't even contain k 'a's initially, no amount of taking will help. Consequence: Correctness failure on edge cases.

Q: Why is my answer off by one? Mistake: Calculating window size as right - left instead of right - left + 1, or incorrectly subtracting from n. Concept Gap: The window length is inclusive. The number of characters removed is Total Length - Max Window Length. Consequence: Wrong answer.

Q: Why do I get negative indices or errors? Mistake: Not handling the case where total[char] - k is negative inside the loop logic (though the initial check should catch this) or confusing the logic of "count inside" vs "count outside". Concept Gap: The condition window[i] > total[i] - k is the precise mathematical inversion of total[i] - window[i] < k. Stick to one mental model.