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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 <stdio.h>
2#include <string.h>
3
4int minLengthAfterRemovals(const char *s) {
5 char stack[100
This implementation carries out a linear traversal of the string and uses a simple array to simulate stack operations. The stack's top is managed using an integer (top) that tracks the index of the last element.
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.
#include <string>
using namespace std;
int minLengthAfterRemovals(string &s) {
int write = 0;
for (int read = 0; read < s.size(); ++read) {
if (write > 0 && ((s[read] == 'B' && s[write - 1] == 'A') || (s[read] == 'D' && s[write - 1] == 'C'))) {
--write;
} else {
s[write++] = s[read];
}
}
return write;
}
int main() {
string s = "ABFCACDB";
cout << minLengthAfterRemovals(s) << endl;
return 0;
}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.