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This approach involves using a stack data structure to handle the character removals efficiently.
Iterate through each character in the string s. Use the stack to maintain a record of the characters processed so far. For each new character, compare it with the character at the top of the stack. If they form a bad pair (for example, if one is a lower-case and the other is the same letter in upper-case), pop the stack. Otherwise, push the current character onto the stack.
The stack will ensure that all bad pairs are removed as you process the string, and the remaining elements in the stack will form the good string.
Time Complexity: O(n), where n is the length of the string. We make a single pass through the string.
Space Complexity: O(n), because we may store all characters of the string in the worst case (if no removals are made).
1#include <iostream>
2#include <string>
3
4std::string makeGood(const std::string& s) {
5 std::string stack;
6 for (char c : s) {
7 if (!stack.empty() && (stack.back() == c + 32 || stack.back() == c - 32)) {
8 stack.pop_back();
9 } else {
10 stack.push_back(c);
11 }
12 }
13 return stack;
14}
int main() {
std::string s = "leEeetcode";
std::string result = makeGood(s);
std::cout << result << std::endl;
return 0;
}In C++, the STL provides a string structure that allows us to use it as a stack, thanks to methods like push_back and pop_back. We go through each character, checking if it forms a bad pair with the character at the end of the stack and take corresponding actions.
The two-pointer technique allows us to traverse the string using two indices to find and remove bad pairs directly. This could potentially minimize contiguous sections of bad pairs before further adjusting the string.
By initially placing one pointer at the start and the other incrementing through the string, check adjacent pairs for bad sequences directly, shifting strings as necessary, allowing further analysis.
While effectively a simulation of stack behavior, avoiding using additional storage structures explicitly, this approach may involve more complex logic to ensure the traversal maintains correct indices.
Time Complexity: O(n), iterating through input once.
Space Complexity: O(1) and no auxiliary data structures are used.
Java allows room optimization by leveraging character arrays and remapping to refine the original input by dismissing offending pairs while securing the efficient retention of necessary elements.