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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.
1#include <iostream>
2#include <stack>
3using namespace std;
4
5int minLengthAfterRemovals(string s) {
6 stack<char> st;
7 for (char c : s) {
8 if (!st.empty() && ((c == 'B' && st.top() == 'A') || (c == 'D' && st.top() == 'C'))) {
9 st.pop();
10 } else {
11 st.push(c);
12 }
13 }
14 return st.size();
}
int main() {
string s = "ABFCACDB";
cout << minLengthAfterRemovals(s) << endl;
return 0;
}This C++ solution uses the standard stack from the STL to manage the stack operations. We iterate through the string and apply push/pop operations based on matching conditions.
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.
1def minLengthAfterRemovals(s):
2 s = list(s)
3 write = 0
4 for read in range(len(s)):
5 if write > 0 and ((s[read] == 'B' and s[write - 1] == 'A') or (s[read] == 'D' and s[write - 1] == 'C')):
6 write -= 1
7 else:
8 s[write] = s[read]
9 write += 1
10 return write
11
12s = "ABFCACDB"
13print(minLengthAfterRemovals(s))In Python, the string is transformed into a mutable list of characters. Using two indices, characters are compared and overwritten as needed, ensuring patterns are skipped and not translated into the resulting sequence.