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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.
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.