Lecture 158:-Simplify Path

The "Simplify Path" problem is a common coding challenge where you are given an absolute file path and you need to simplify it by removing redundant slashes and unnecessary directories. For example, given the input "/a/./b/../../c/", the simplified path should be "/c".

Here's an approach to solve the "Simplify Path" problem using a stack:

1. Split the input path by slashes ("/") to get individual components.
2. Iterate through the components: a. If the component is ".", ignore it as it refers to the current directory. b. If the component is ".." and the stack is not empty, pop the top element from the stack as it refers to the parent directory. c. Otherwise, push the component onto the stack.
3. After processing all components, construct the simplified path by joining the stack elements with slashes ("/").

Here's a C++ implementation of the "Simplify Path" problem:

cppCopy code

#include <iostream> #include <stack> #include <vector> #include <sstream> std::string simplifyPath(const std::string& path) { std::vector<std::string> components; std::stringstream ss(path); std::string component; // Split the input path by slashes and store components in a vector while (getline(ss, component, '/')) { if (component == "" || component == ".") { continue; } else if (component == "..") { if (!components.empty()) { components.pop_back(); } } else { components.push_back(component); } } // Construct the simplified path std::stack<std::string> stack; for (const std::string& comp : components) { stack.push(comp); } std::string simplifiedPath; while (!stack.empty()) { simplifiedPath = "/" + stack.top() + simplifiedPath; stack.pop(); } return simplifiedPath.empty() ? "/" : simplifiedPath; } int main() { std::string input = "/a/./b/../../c/"; std::string simplified = simplifyPath(input); std::cout << "Input path: " << input << std::endl; std::cout << "Simplified path: " << simplified << std::endl; return 0; }

In this implementation, the simplifyPath function takes the input path and follows the steps mentioned earlier to simplify it. The resulting simplified path is returned.

For the input "/a/./b/../../c/", the output will be "/c", which is the simplified path after removing redundant and unnecessary components.