2516. Take K of Each Character From Left and Right - Detailed Explanation

Complexity Analysis (Brute Force):

• Time Complexity: O(n^2) in the worst case. We potentially check on the order of n^2 combinations of i and j. The counting of characters in each combination can be optimized using prefix/suffix counts, but even then the nested loops dominate. For n = 100k, this is astronomically large.

• Space Complexity: O(n) if we store prefix/suffix counts to speed up counting, or up to O(n) for building the taken string in the worst case. The brute force uses a lot of extra space and time and is not practical for large inputs.

Java Implementation

This brute force approach is useful to reason about the problem, but we need to optimize it further for real use.

Approach 2: Optimized Two-Pointer Sliding Window

Explanation:
To optimize, we need to avoid checking all combinations explicitly. A key insight is that any set of characters you take from the ends leaves behind a contiguous middle substring that you did not take. Instead of directly figuring out what to take, we can equivalently figure out what contiguous substring to leave in the middle. If we know which substring remains, then all characters outside that substring (i.e. from the prefix and suffix) are taken.

The requirement "at least k of each character taken" translates to a condition on the substring left behind:

• Let total_a, total_b, total_c be the total counts in the whole string. If you leave a certain substring in the middle, and you took the rest, then for each character:

  • Characters taken = total_count − (characters left in middle).

  • We need characters taken ≥ k.

  • Therefore, characters left in the middle (the contiguous substring) must be ≤ total_count - k for each character. In other words, the middle substring can contain at most total_a - k 'a's, at most total_b - k 'b's, and at most total_c - k 'c's. If the middle substring has more than total_x - k of some character x, then not enough of x were taken.

Using this idea, we can slide a window (two pointers) over the string to find the largest middle substring that satisfies those constraints. Once we find the longest valid substring we can leave, then the minimum characters to take is simply the rest of the string (i.e. total length - length_of_longest_valid_middle). We want the longest valid middle because leaving a longer middle means taking fewer characters.

How to implement with two pointers:

• Use two pointers left and right to maintain a window [left, right] (inclusive) that represents a candidate middle substring we are not taking.

• Maintain counts of 'a', 'b', and 'c' within this window.

• Initially, you can start with left = 0 and expand right step by step through the string (like a typical sliding window that grows).

• As you move right (include a new character in the window), update the window counts. If adding this character makes the window invalid (i.e., the window now has too many of one character such that not enough would remain outside), then move the left pointer to shrink the window until it becomes valid again.

  • "Window invalid" means for some char x, window_count[x] > total_x - k. Equivalently, the outside (taken) count of x = total_x - window_count[x] becomes < k. We can detect this by checking if any outside_count[x] < k.

• Keep track of the maximum window size that remained valid throughout this process.

By doing this, we effectively find the largest contiguous block we can leave such that taking the rest yields at least k of each character. The answer will be n - (max window length).

This approach is much more efficient because each character is considered at most twice (once when expanding right, once when moving left), making it linear time.

Sliding Window Algorithm in steps:

1. Compute total_a, total_b, total_c for the whole string. If any of these totals is less than k, immediately return -1 (not possible).

2. Set up a window with left = 0. Use an array or dictionary count to track counts of 'a','b','c' in the current window (middle substring). You can initialize it to { 'a': 0, 'b': 0, 'c': 0 }.

3. Initialize max_window_len = 0.

4. Iterate right from 0 to n-1 (indexing the string):

   • Include s[right] in the window: increment count[s[right]].
   • Now this means one less of s[right] is outside. Update the outside count for that character or simply check the condition:
     • While the current window is invalid (i.e., for any character x, count[x] > total_x - k), we need to shrink from the left:
       • This usually specifically happens when count[s[right]] exceeded the allowed limit for that char. We can check just that character, since only it increased.
       • While count[s[right]] > total_{s[right]} - k, we move left forward by 1 (remove s[left] from the window):

         • Decrement count[s[left]] (since we're no longer leaving that char in the middle; we are effectively taking it instead).

         • Increment left by 1.

       • Continue this until the window is valid again.
   • Now check the length of the current valid window: window_length = right - left + 1. If this is the largest seen so far, update max_window_len.

5. After expanding through the whole string, the answer (minimum characters to take) will be n - max_window_len. This is the minimum minutes needed.

Why this works: We are effectively finding the longest contiguous segment we can keep without violating the requirement that at least k of each char are taken. By always fixing violations by moving left, we ensure we never miss a potential valid window. This two-pointer technique ensures we explore all possible contiguous segments in linear time.

Optimized Approach Example:
Consider s = "aabcccabba", k = 2. Total counts might be (just an example) total_a = 4, total_b = 3, total_c = 3. We require at least 2 of each char outside. That means the middle substring can have at most total_a - 2 a's, total_b - 2 b's, total_c - 2 c's (so at most 2 a's, 1 b, 1 c in this case). We use a sliding window to find the longest substring meeting these criteria:

• Expand right pointer, counting characters.

• If we ever have more than allowed (say the window has 2 c's but allowed 1 c), we move left pointer up until the count of that character is back within limit.

• Track the max window length that stayed valid. Suppose the max valid window length we find is 5.

• Then the answer = len(s) - 5 (we take the rest of the characters).

This approach avoids trying every combination explicitly.

Optimized Implementation (Python):

Complexity Analysis (Optimized Two-Pointer):

• Time Complexity: O(n). We traverse the string with the right pointer once. The left pointer also moves at most n steps in total. Each character is processed a constant number of times. Therefore, the time is linear in the length of the string. This is efficient for n up to 100k.

• Space Complexity: O(1). We use a fixed-size counter (for 'a', 'b', 'c') and a few integer variables. The space does not grow with n. (If we consider the input string itself, that's given and not extra space for the algorithm.)

Step-by-Step Walkthrough (Optimal Approach)

Let's walk through the optimized two-pointer approach with a detailed example, using a table to illustrate how the window expands and contracts. We will use Example 1: s = "aabaaaacaabc", k = 2. For reference, the total counts are total_a = 8, total_b = 2, total_c = 2 in this string. This means the middle substring we leave can contain at most total_a - k = 6 'a's, total_b - k = 0 'b's, and total_c - k = 0 'c's. In other words, the window cannot include any 'b' or 'c' (because all 'b's and 'c's must be taken to reach 2 each), and it can include at most 6 'a's.

We iterate right from 0 to 11 (indices of the string). The table below shows the state at each step after adjusting the window:

Explanation of the steps:

• By the time we finish, the longest valid window we saw had length 4 (from index 3 to 6, containing "aaaa"). This window had counts (a=4, b=0, c=0), which was within allowed limits (no b or c, and 4 a's ≤ 6 allowed). We could not expand this window further without including a 'c' at index 7, which made it invalid and forced us to reset the window.
• The final answer is the total length minus the max window length: 12 - 4 = 8. This matches the output we expected.

In summary, the two-pointer method found that leaving the substring "aaaa" (indices 3–6) in the middle is optimal. Removing everything else (8 characters: "aab" from the left and "caabc" from the right in some order) is the minimum required to satisfy the condition.

Python Implementation (Optimal Solution)

Complexity Analysis

• Brute Force Approach: Time complexity is O(n^2) (or worse if counting within loops), which is not feasible for large n (100k). Space complexity is O(n) or more due to storing counts or building substrings for checks. This approach will time out for large inputs and is only useful for understanding the problem in theory.
• Optimized Two-Pointer Approach: Time complexity is O(n). We use a sliding window that moves through the string with two pointers, each pointer moving at most n steps. Counting operations and comparisons are O(1). Space complexity is O(1), using a fixed number of counters (just 3 counters for 'a','b','c'). This approach efficiently handles the maximum input size.

Common Mistakes

• Not Checking Impossibility: Forgetting to check if the total count of 'a', 'b', or 'c' is less than k. If any character cannot reach k in total, you should immediately return -1. Skipping this check could lead to incorrect results or infinite loops in the two-pointer approach.
• Misinterpreting the Problem: Assuming you need to take exactly k characters from each side (left and right) separately. In reality, the requirement is k of each character in total from both ends combined, not per side. You can take them in any order from left/right as needed.
• Off-by-One Errors in Two-Pointer Loop: When adjusting the left pointer, one might incorrectly update the window length or miss updating the counts. It’s important to update the counts before moving the left index and ensure the window length is computed as right - left + 1.
• Allowing Overlap of Prefix/Suffix in Brute Force: In the brute-force approach, be careful not to double-count characters by taking more from left and right than the string’s length. Ensuring i + j <= n (or handling the case where the middle becomes empty) is important.
• Not Handling k = 0: If k = 0, the answer should be 0 (since you don’t need to take any characters to have “at least 0” of each). Some solutions might accidentally return -1 or some other value if this case isn’t handled, but according to constraints k can be 0 and the result should logically be 0 minutes (do nothing).

Edge Cases

• Not Enough of a Character: If s doesn’t contain at least k of one of the characters (e.g., s = "aaa", k = 1 for 'b' or 'c'), the function should return -1 immediately.

• Exactly k Characters Total: If one of the characters appears exactly k times in total, it means you must take all occurrences of that character from the ends. For example, in "aab" with k=2 for 'b', since there are exactly 2 'b's, any middle substring cannot contain a 'b'. The two-pointer window logic should handle this by never allowing 'b' inside.

• k = 0: As mentioned, this should return 0 because you don’t need to take any characters to satisfy “at least 0 of each”. The algorithm naturally handles this, but it’s good to note separately.

• All Characters Present in Abundance: If the string has plenty of each character (for instance, s = "aaabbbccc", k = 1), the minimum minutes might just be k*3 = 3 if you can take one of each from somewhere without needing extra removals. The algorithm will find a large middle (possibly the whole string if already satisfied, resulting in n - n = 0 if no removal needed). Ensure the logic handles cases where the best solution might be not removing much or any at all (if k is 0).

• String Length Equals k: If k equals the length of the string, then you need to take the entire string to get k of each character. This is only possible if the string’s length is composed of at least k of each char (which in practice means the string must have exactly k of each, since length = k and need k of each of 3 chars implies n must be ≥ 3k, which it isn't if n=k except k=0 case). Typically, if k = n, the only way to have k of each is if n is 0 or it's impossible (for n > 0, you can't have n of each of three chars if total length is n). So usually k won’t be equal to n for valid cases unless two of the characters require 0 (k=0 scenario). But it’s an edge parameter to think about.

Alternative Variations

• More Character Types: This problem specifically had only 'a', 'b', and 'c'. A variation could involve more types of characters (for example, a string with letters 'a' through 'z' and you need to take k of each distinct letter present). The approach would generalize by maintaining counts for each required character. The two-pointer technique would still apply, although the complexity of checking all characters in the window grows with the number of character types.

• Different k for Each Character: Another variation could ask for taking at least k1 of 'a', k2 of 'b', and k3 of 'c', where the k values might differ for each character. The solution approach would be similar: check totals and use a sliding window ensuring the window doesn’t contain more than total_a - k1 of 'a', etc., for each character.

• Exact Collection vs At Least: A twist on the problem could require taking exactly k of each character from the ends. This would be more complex, as you would have to stop once you have exactly k of each, possibly leaving some required characters in the middle which wouldn't satisfy "exactly". The strategy and problem definition would differ (and likely require a different approach, maybe backtracking or BFS on states, because the flexibility of "at least" allows a greedy window strategy).

• Weighted Characters or Different Removal Costs: Imagine if taking certain characters cost different amounts of "time" or had different benefits. The problem could become an optimization with weighted values, which might then be tackled with dynamic programming or greedy strategies depending on the setup.

• Taking Characters from One End Only: A simpler variation is if you were only allowed to take from one end (say only from the left). Then the problem reduces to just checking a prefix of the string – you would take the smallest prefix that contains k of each character (if possible). This is much simpler than the two-ended version.

In all these variations, the key concept of ensuring counts of characters meets requirements remains, but the approach to achieve the goal might change.

Related Problems

Here are some related problems and exercises that involve similar techniques or ideas, which can further solidify your understanding:

• LeetCode 1423 – Maximum Points You Can Obtain from Cards: You have an array of card points and you can take some cards from the beginning and some from the end. You want to maximize the score. This is similar in that it involves taking from two ends; a common solution uses a complementary subarray (or sliding window) approach, analogous to leaving a middle segment unpicked.

• LeetCode 1658 – Minimum Operations to Reduce X to 0: Given an array, you can remove numbers from either end to reduce the sum to X. You want the minimum number of removals. This is another problem that can be transformed into finding a subarray (in the middle) with a certain sum, using two-pointer or sliding window. It’s conceptually similar to this problem’s approach of finding a middle segment to leave behind.

• LeetCode 76 – Minimum Window Substring: This classic problem asks for the smallest substring of s that contains at least one of each character of another string t. While it’s not about taking from ends, it uses a similar sliding window technique to maintain character counts and ensure conditions (having enough of each required char). It’s good practice for managing counts in a window and adjusting two pointers.

• LeetCode 1358 – Number of Substrings Containing All Three Characters: This problem involves strings with 'a', 'b', 'c' and asks for the count of substrings containing all three. It uses two-pointer techniques to count valid substrings. Although it's counting substrings rather than removing from ends, the focus on 'a', 'b', 'c' and satisfying a condition with a sliding window makes it related.

• General Sliding Window Practice: Problems like "Longest Substring Without Repeating Characters", "Longest Substring with At Most K Distinct Characters", or "Max Consecutive Ones with at most K zeros flipped" – while different in subject, all hone your skills with the two-pointer/sliding window pattern which is used in the optimal solution here. Practicing those will make the approach used in this problem more intuitive.