3223. Minimum Length of String After Operations - Detailed Explanation

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Problem Statement

Problem:
You are given a string s consisting only of the characters 'A', 'B', 'C', and 'D'. You can perform the following operations any number of times (in any order):

Remove one occurrence of the pair "AB" (i.e. remove one 'A' and one 'B').
Remove one occurrence of the pair "CD" (i.e. remove one 'C' and one 'D').

The operations do not require the letters to be adjacent in the original string (i.e. you can choose any 'A' and any 'B' to remove together, and similarly for 'C' and 'D'). Return the minimum possible length of s after performing any number of these operations.

Example Inputs & Outputs:

Example 1:
- Input: "AB"
- Operations: Remove 'A' and 'B' as a pair (using "AB" removal).
- Output: 0
- Explanation: Both letters can be removed, leaving an empty string.
Example 2:
- Input: "CAB"
- Operations: You can remove the 'A' and 'B' pair from the string. After removal, the string becomes "C".
- Output: 1
- Explanation: After removing one pair, only 'C' remains.
Example 3:
- Input: "ABCD"
- Operations:
  - Remove the pair "AB" to get "CD".
  - Then remove the pair "CD" to get an empty string.
- Output: 0
- Explanation: Both operations remove all characters.

Constraints:

$( 1 \leq s.length \leq 10^5 )$
$( s )$ consists only of the characters 'A', 'B', 'C', and 'D'.

Hints to Approach the Problem

Hint 1: Notice that each valid operation removes one 'A' with one 'B' or one 'C' with one 'D'. This means that every removal reduces the count of two characters by one each.
Hint 2: Think about the problem in terms of counts. Instead of simulating removals step by step, ask yourself: How many pairs can I form from the given counts?
This leads to a very efficient counting method.

Approaches

Approach 1: Brute Force Simulation

Idea:
Repeatedly search the string for any occurrence of "AB" or "CD". When found, remove that pair and concatenate the remaining parts. Continue until no such pair exists.
Steps:
1. While the string contains "AB" or "CD", remove one occurrence.
2. Return the length of the resulting string.
Why It’s Not Optimal:
This method may require scanning and reconstructing the string many times. In the worst case (with string length up to (10^5)) the time complexity can be very high (exponential or quadratic), making it inefficient.

Approach 2: Optimal Counting / Greedy Method

Observation:
Every operation removes one occurrence of 'A' and 'B' or one occurrence of 'C' and 'D'. Notice that the order in which you remove the pairs does not affect the final count.
- The maximum number of "AB" removals is the minimum of the count of 'A' and the count of 'B'.
- The maximum number of "CD" removals is the minimum of the count of 'C' and the count of 'D'.
Key Insight:
After removing all possible pairs, the remaining letters will be those that couldn’t be paired. Their total count is exactly:

$[ \text{remaining} = \text{abs(count('A') - count('B'))} + \text{abs(count('C') - count('D'))} ]$
Steps:
1. Count the occurrences of 'A', 'B', 'C', and 'D' in the string.
2. Compute the remaining letters for the 'A'–'B' pair as abs(countA - countB).
3. Compute the remaining letters for the 'C'–'D' pair as abs(countC - countD).
4. The final answer is the sum of these two values.
Why It’s Optimal:
This method only requires one pass through the string ((O(n)) time) and constant extra space ((O(1))). It efficiently leverages the problem’s structure without simulating removals.

Python Code

Python3

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Java Code

Java

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Complexity Analysis

Time Complexity:
The optimal approach makes a single pass through the string to count characters, which takes (O(n)) time where (n) is the length of s.
Space Complexity:
Only a constant amount of extra space is used for the counters, i.e., (O(1)).

Step-by-Step Walkthrough and Visual Examples

Walkthrough for Example 2: `"CAB"`

Initial String: "CAB"
Count Occurrences:
- 'A': 1
- 'B': 1
- 'C': 1
- 'D': 0
Calculate Removals:
- For "AB": $( \min(1, 1) = 1 )$ pair can be removed.
- For "CD": $( \min(1, 0) = 0 )$ pairs.
Remaining Letters Calculation:
- For A and B: $( \text{abs}(1 - 1) = 0 )$
- For C and D: $( \text{abs}(1 - 0) = 1 )$
Final Length: ( 0 + 1 = 1 )
The remaining string (conceptually) is just "C".

Visual Representation

Input:  C A B
         | | 
         V V   <-- Remove one A-B pair
Remaining: C
Final length = 1

Common Mistakes

Simulating the removals literally:
Attempting to repeatedly find and delete substrings (i.e., actually modifying the string) can lead to inefficient solutions, especially for large inputs.
Overcomplicating the pairing logic:
Remember that each operation only decreases two counts by one. The final result depends solely on the differences of these counts.
Not handling edge cases:
Failing to consider strings that do not have any possible pairs to remove (e.g., "CCCC") or strings that are already empty.

Edge Cases

Empty String:
If ( s ) is empty, the answer is ( 0 ).
No Valid Pairs:
For strings like "AAAA" or "DDDD", no removals are possible, so the answer is the length of the string.
Balanced but Non-removable (if removal were adjacent):
In this problem, since removals are based on counts (and not position), a balanced string like "ACBD" (with 1 of each) would yield (0) even though the letters might not be adjacent. (This works because the operation is defined to allow removal regardless of order.)

Adjacent Removals Variation:
A similar problem might restrict removals only to adjacent pairs. In that case, a stack-based simulation would be more appropriate. However, that would change the solution since the order would matter.
Different Pairing Rules:
Some variations might allow different pairs (for example, "BA" or "DC") or even allow removals of nonconsecutive characters. Always analyze how the allowed operation affects the count of characters.

Remove All Adjacent Duplicates in String:
Practice using a stack to remove pairs of duplicate characters.
Super Reduced String (HackerRank):
Remove adjacent pairs from a string repeatedly until no more removals can be made.
Valid Parentheses:
Use stack-based solutions to validate matching pairs in a string.
Longest Valid Parentheses:
Another string problem where stack or dynamic programming approaches are used.