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This approach utilizes a stack data structure to process the path components. By splitting the path on the '/' character, we collect the different components of the path. Using a stack helps efficiently manage directory movement due to '..'. For every directory name, it is pushed to the stack, for '..' we pop from the stack, and '.' is simply ignored. After processing all components, we join the names in the stack with '/' to form the simplified canonical path.
Time Complexity: O(N) where N is the length of the path string, as we perform operations linearly along the string.
Space Complexity: O(N) because we use a stack that can contain each part of the path in the worst case.
1function simplifyPath(path) {
2 const stack = [];
3 const components = path.split('/');
4
5 for (const component of components) {
6 if (component === '..') {
7 if (stack.length > 0) {
8 stack.pop();
9 }
10 } else if (component !== '.' && component !== '') {
11 stack.push(component);
12 }
13 }
14
15 return '/' + stack.join('/');
16}In JavaScript, we achieve the solution through an array used as a stack. By splitting the input path and processing each fragment, we dictate stack operations to navigate directories, crafting a succinct path at the end by concatenating stack elements with '/'.
This approach optimizes the simplification process using string handling techniques without explicitly using stack structures. It involves maintaining a result string directly, with an index pointer to simulate stack behavior. Iterating over path components allows addition, removal, or skipping of segments based on path rules, with a focus on reducing stack overhead. Finally, the result string is reconstructed as the simplified path.
Time Complexity: O(N) as parsing occurs once through entire path.
Space Complexity: O(N) utilized for the resultant path string (no explicit stack).
using namespace std;
string simplifyPath(string path) {
string result;
string temp;
int length = path.length();
for (int i = 0; i < length; ++i) {
if (path[i] == '/') continue;
int j = i;
while (j < length && path[j] != '/') ++j;
temp = path.substr(i, j - i);
if (temp == "..") {
if (!result.empty()) result.erase(result.find_last_of('/'));
} else if (temp != ".") {
result += '/' + temp;
}
i = j - 1;
}
return result.empty() ? "/" : result;
}Using C++, the path is processed segment by segment, storing result directly in a string. Addition or removal of directory paths into this result string is based on component interpretation, avoiding stack usage while deriving the simplified path.