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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.
1function minLengthAfterRemovals(s) {
2 let arr = s.split(''); // Convert string to array for direct manipulation
3 let write = 0;
4 for (let read = 0; read < arr.length; read++) {
5 if (write > 0 && ((arr[read] === 'B' && arr[write - 1] === 'A') || (arr[read] === 'D' && arr[write - 1] === 'C'))) {
6 write--;
7 } else {
8 arr[write++] = arr[read];
9 }
10 }
11 return write;
12}
13
14const s = "ABFCACDB";
15console.log(minLengthAfterRemovals(s));JavaScript's string-to-array conversion allows us to directly alter the sequence with a write pointer to evaluate and bypass pairs.