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This approach revolves around using a stack to efficiently identify and remove the substrings "AB" and "CD" from the given string. By traversing each character of the string:
The characters that remain on the stack are the ones that could not form any removable patterns, thus forming the result string.
Time Complexity: O(n), where n is the length of the string.
Space Complexity: O(n), to store the stack elements, based on the worst case where no pairs are removed.
1using System;
2using System.Collections.Generic;
3
4public class MinLengthString {
5 public static int MinLengthAfterRemovals(string s) {
6 Stack<char> stack = new Stack<char>();
7 foreach (char c in s) {
8 if (stack.Count > 0 && ((c == 'B' && stack.Peek() == 'A') || (c == 'D' && stack.Peek() == 'C'))) {
9 stack.Pop();
10 } else {
11 stack.Push(c);
12 }
13 }
return stack.Count;
}
public static void Main(string[] args) {
string s = "ABFCACDB";
Console.WriteLine(MinLengthAfterRemovals(s));
}
}This C# solution utilizes a stack of characters to capture and eliminate removable patterns. The stack is processed using standard push and pop operations while iterating through the input string.
This technique uses a two-pointer approach to reduce the string by altering it in place. The core idea is to selectively overwrite positions in the string:
After reaching the end of the string, the length of the string from the start to the write pointer represents the reduced string length.
Time Complexity: O(n), dependent on examining each character once.
Space Complexity: O(1), as no additional data structures are used.
This Java solution translates the two-pointer logic into Java by deconstructing the input to a character array. This allows direct manipulation similar to C-style string approaches, achieving an in-place modification.