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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.
1function minLengthAfterRemovals(s) {
2 let stack = [];
3 for (let char of s) {
4 if (stack.
The JavaScript implementation mimics stack operations using an array. The solution processes the input string, leveraging the array's push and pop methods to remove patterns as identified.
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.
1using System;
2
3public class MinLengthString {
4 public static int MinLengthAfterRemovals(string s) {
5 char[] arr = s.ToCharArray();
6 int write = 0;
7 for (int read = 0; read < arr.Length; read++) {
8 if (write > 0 && ((arr[read] == 'B' && arr[write - 1] == 'A') || (arr[read] == 'D' && arr[write - 1] == 'C'))) {
9 write--;
10 } else {
11 arr[write++] = arr[read];
12 }
13 }
14 return write;
15 }
16
17 public static void Main(string[] args) {
18 string s = "ABFCACDB";
19 Console.WriteLine(MinLengthAfterRemovals(s));
20 }
21}Using character arrays in C#, this solution directly modifies the sequence of characters to simulate a condensed version of the input after removing specified patterns.