Solution For Simplify Path
Problem Statement:
Given a string path, which is an absolute path (starting with a slash ‘/’) to a file or directory in a Unix-style file system, convert it to the simplified canonical path.
In a Unix-style file system, a period ‘.’ refers to the current directory, a double period ‘..’ refers to the directory up a level, and any multiple consecutive slashes (i.e. ‘//’) are treated as a single slash ‘/’. For this problem, any other format of periods such as ‘…’ are treated as file/directory names.
The canonical path should have the following format:
- The path starts with a single slash ‘/’.
- Any two directories are separated by a single slash ‘/’.
- The path does not end with a trailing ‘/’.
- The path only contains the directories on the path from the root directory to the target file or directory (i.e., no period ‘.’ or double period ‘..’)
Return the simplified canonical path.
Solution:
This problem can be solved using a stack data structure. We can first split the given string path using the ‘/’ character as the delimiter. This will give us an array of directories, which we can then process one at a time.
We can use a stack to keep track of the directories that we want to keep in our simplified canonical path. We will iterate over the directories, and for each directory in the path, we will do the following:
- If the directory is a period ‘.’, we can ignore it as it refers to the current directory.
- If the directory is a double period ‘..’, we need to go up one level in the directory hierarchy. We can do this by popping the top element from the stack. If the stack is empty, we do not need to do anything as we are already at the root directory.
- If the directory is not a period or a double period, we can simply push it onto the stack.
After iterating over all the directories, we can pop the elements from the stack and append them to a string variable, forming the simplified canonical path.
Finally, we need to reverse the string as the stack order is in the reverse order of the path, and return the path by checking if the stack is empty or not.
Code:
class Solution:
def simplifyPath(self, path: str) -> str:
stack = []
for directory in path.split(‘/’):
if directory == ‘.’ or directory == ”:
continue
elif directory == ‘..’:
if stack:
stack.pop()
else:
stack.append(directory)
return ‘/’ + ‘/’.join(stack[::-1]) if stack else ‘/’
Step by Step Implementation For Simplify Path
/**
* Given an absolute path for a file (Unix-style), simplify it. Or in other words, convert it to the canonical path.
In a UNIX-style file system, a period . refers to the current directory. Furthermore, a double period .. moves the directory up a level. For more information, see: Absolute path vs relative path in Linux/Unix
Note that the returned canonical path must always begin with a slash /, and there must be only a single slash / between two directory names. The last directory name (if it exists) must not end with a trailing /. Also, the canonical path must be the shortest string representing the absolute path.
Example 1:
Input: "/home/"
Output: "/home"
Explanation: Note that there is no trailing slash after the last directory name.
Example 2:
Input: "/../"
Output: "/"
Explanation: Going one level up from the root directory is a no-op, as the root level is the highest level you can go.
Example 3:
Input: "/home//foo/"
Output: "/home/foo"
Explanation: In the canonical path, multiple consecutive slashes are replaced by a single one.
Example 4:
Input: "/a/./b/../../c/"
Output: "/c"
Example 5:
Input: "/a/../../b/../c//.//"
Output: "/c"
Example 6:
Input: "/a//b////c/d//././/.."
Output: "/a/b/c"
*/
import java.util.*;
class Solution {
public String simplifyPath(String path) {
// this will store the final simplified path
StringBuilder simplifiedPath = new StringBuilder();
// this is used as a stack to keep track of the directories
// we've visited so far
Deque stack = new ArrayDeque<>();
// this array will store all the individual directories in the path
String[] directories = path.split("/");
for (String dir : directories) {
// if the directory is empty or ".", we can simply continue
if (dir.isEmpty() || dir.equals(".")) {
continue;
}
// if the directory is "..", we need to backtrack
// by popping the last directory from the stack
if (dir.equals("..")) {
stack.pop();
continue;
}
// otherwise, we push the directory onto the stack
stack.push(dir);
}
// now, we need to construct the final simplified path
// by popping all the directories from the stack
while (!stack.isEmpty()) {
simplifiedPath.append("/");
simplifiedPath.append(stack.pollLast());
}
// if the simplified path is empty, return "/"
if (simplifiedPath.length() == 0) {
return "/";
}
return simplifiedPath.toString();
}
} def simplifyPath(path):
stack = []
for p in path.split("/"):
if p == "..":
if stack:
stack.pop()
elif p and p != ".":
stack.append(p)
return "/" + "/".join(stack)var simplifyPath = function(path) {
// your code here
};The code should be in C++. This is a C++ solution for the leetcode problem simplify-path: #include#include #include #include using namespace std; class Solution { public: string simplifyPath(string path) { // create a stack to store the simplified path stack stk; // create a variable to store the current element string curr; // split the path by '/' vector v = split(path, '/'); // iterate through the vector for (int i = 0; i < v.size(); i++) { // get the current element curr = v[i]; // if the current element is "..", pop from the stack if (curr == "..") { if (!stk.empty()) { stk.pop(); } } // if the current element is ".", do nothing else if (curr == ".") { continue; } // otherwise, push the element into the stack else { stk.push(curr); } } // create a variable to store the result string res; // if the stack is empty, return "/" if (stk.empty()) { res = "/"; } // otherwise, construct the path from the stack else { while (!stk.empty()) { res = "/" + stk.top() + res; stk.pop(); } } return res; } // a helper function to split a string by a given delimiter vector split(string s, char delim) { // create a vector to store the result vector res; // create a variable to store the current element string curr; // iterate through the string for (int i = 0; i < s.length(); i++) { // if the current character is the delimiter if (s[i] == delim) { // if the current element is not empty, push it into the vector if (curr != "") { res.push_back(curr); } // reset the current element curr = ""; } // otherwise, append the character to the current element else { curr += s[i]; } } // if the current element is not empty, push it into the vector if (curr != "") { res.push_back(curr); } return res; } };
public string SimplifyPath(string path) {
// create a stack to store the simplified path
Stack stack = new Stack();
// create a variable to store the current directory
string current = "";
// loop through each character in the path
for(int i = 0; i < path.Length; i++) {
// if the character is a slash, check if the current directory is empty
if(path[i] == '/') {
// if the current directory is not empty, push it onto the stack
if(current != "") {
stack.Push(current);
}
// reset the current directory
current = "";
}
// if the character is not a slash, append it to the current directory
else {
current += path[i];
}
}
// if the current directory is not empty, push it onto the stack
if(current != "") {
stack.Push(current);
}
// create a variable to store the simplified path
string simplified = "";
// while the stack is not empty, pop off the top directory and append it to the simplified path
while(stack.Count > 0) {
simplified += "/" + stack.Pop();
}
// return the simplified path
return simplified;
}