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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.
1def minLengthAfterRemovals(s):
2 stack = []
3 for char in s:
4 if stack and ((char == 'B' and stack[
This Python implementation uses a list to serve as a stack, taking advantage of Python's list pop and append to simulate stack operations effectively.
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.
1#include <iostream>
2#include <string>
3using namespace std;
4
5int minLengthAfterRemovals(string &s) {
6 int write = 0;
7 for (int read = 0; read < s.size(); ++read) {
8 if (write > 0 && ((s[read] == 'B' && s[write - 1] == 'A') || (s[read] == 'D' && s[write - 1] == 'C'))) {
9 --write;
10 } else {
11 s[write++] = s[read];
12 }
13 }
14 return write;
15}
16
17int main() {
18 string s = "ABFCACDB";
19 cout << minLengthAfterRemovals(s) << endl;
20 return 0;
21}This approach efficiently alters the string 's' by condensing it in place. The write index adjusts the length of the final string as patterns are popped and skipped during traversal.