You are given a string s consisting only of uppercase English letters.
You can apply some operations to this string where, in one operation, you can remove any occurrence of one of the substrings "AB" or "CD" from s.
Return the minimum possible length of the resulting string that you can obtain.
Note that the string concatenates after removing the substring and could produce new "AB" or "CD" substrings.
Example 1:
Input: s = "ABFCACDB" Output: 2 Explanation: We can do the following operations: - Remove the substring "ABFCACDB", so s = "FCACDB". - Remove the substring "FCACDB", so s = "FCAB". - Remove the substring "FCAB", so s = "FC". So the resulting length of the string is 2. It can be shown that it is the minimum length that we can obtain.
Example 2:
Input: s = "ACBBD" Output: 5 Explanation: We cannot do any operations on the string so the length remains the same.
Constraints:
1 <= s.length <= 100s consists only of uppercase English letters.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.
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.
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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.
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.
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.
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Time Complexity: O(n), dependent on examining each character once.
Space Complexity: O(1), as no additional data structures are used.
| Approach | Complexity |
|---|---|
| Stack-Based Approach | Time Complexity: O(n), where n is the length of the string. |
| Two-Pointer Iterative Method | Time Complexity: O(n), dependent on examining each character once. |
Minimum String Length After Removing Substrings - Leetcode 2696 - Python • NeetCodeIO • 6,398 views views
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