Sponsored
Sponsored
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.
1#include <stdio.h>
2#include <string.h>
3
4int minLengthAfterRemovals(char *s) {
5 int n = strlen(s);
6 int write = 0;
7 for (int read = 0; read < n; ++read) {
8 if (write && ((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 char s[] = "ABFCACDB";
19 printf("%d\n", minLengthAfterRemovals(s));
20 return 0;
21}This method involves editing the string in place. With two pointers ('read' and 'write'), we adjust the string as we progress. The write pointer helps track the new end of the string by overwriting unnecessary characters.