File System - Amazon Top Interview Questions
Problem Statement :
Implement a data structure with the following methods: bool create(String path, int content) which creates a path at path if there's a parent directory and path doesn't exist yet, and sets its value as content. Returns whether it was newly created. Initially only the root directory at "/" exists. int get(String path) which returns the value associated with path. If there's no path, return -1. Constraints 0 ≤ n ≤ 100,000 where n is the number of calls to create and get Example 1 Input methods = ["constructor", "create", "create", "get", "get", "create", "get"] arguments = [[], ["/usr", 1], ["/usr/bin", 2], ["/usr"], ["/usr/bin"], ["/proc/1/exe", 3], ["/non/existent"]]` Output [None, True, True, 1, 2, False, -1] Explanation fs = FileSystem() # Only "/" exists fs.create("/usr", 1) == True # Now "/" and "/usr" exists fs.create("/usr/bin", 2) == True# Now "/", "/usr" and "/usr/bin" exists fs.get("/usr") == 1 fs.get("/usr/bin") == 2 fs.create("/proc/1/exe", 3) == False # Parent dir "/proc/1/" doesn't exist, so can't create here fs.get("/non/existent") == -1
Solution :
Solution in C++ :
struct Node {
string folderName;
int content;
unordered_map<string, Node*> children;
Node(string name, int c) {
folderName = name;
content = c;
children.clear();
}
};
class FileSystem {
private:
Node* root;
vector<string> getFolders(string path) {
path = path.substr(1);
vector<string> folders;
string temp = "";
for (int i = 0; i < path.size(); i++) {
if (path[i] != '/') {
temp.push_back(path[i]);
} else {
folders.push_back(temp);
temp = "";
}
}
if (temp != "") folders.push_back(temp);
return folders;
}
bool createHelper(string path, int content) {
vector<string> folders = getFolders(path);
bool newlyCreated = false;
// cout<<"CREATE"<<endl;
// for(auto x: folders)
// cout<<x<<" ";
// cout<<endl;
int ptr = 0;
Node* currentNode = root;
while (ptr < folders.size()) {
if (currentNode->children.find(folders[ptr]) != currentNode->children.end()) {
// cout<<"ALREADY HERE "<<folders[ptr]<<endl;
currentNode = currentNode->children[folders[ptr]];
} else {
// cout<<"NEW HERE "<<folders[ptr]<<endl;
if (ptr != folders.size() - 1) return false;
newlyCreated = true;
Node* node = new Node(folders[ptr], content);
currentNode->children[folders[ptr]] = node;
currentNode = node;
}
ptr++;
}
// cout<<endl<<endl;
return newlyCreated;
}
int getHelper(string path) {
vector<string> folders = getFolders(path);
Node* currentNode = root;
int ptr = 0;
while (ptr < folders.size()) {
if (currentNode->children.find(folders[ptr]) == currentNode->children.end()) return -1;
currentNode = currentNode->children[folders[ptr]];
ptr++;
}
return currentNode->content;
}
public:
FileSystem() {
root = new Node(".root", -1);
}
int get(string path) {
return getHelper(path);
}
bool create(string path, int content) {
return createHelper(path, content);
}
};
Solution in Python :
class FileTree:
def __init__(self):
self.children = {}
self.value = None
class FileSystem:
def __init__(self):
self.tree = FileTree()
self.tree.children[""] = FileTree()
def insert(self, pathList, content):
temp = self.tree
for idx, path in enumerate(pathList):
if path not in temp.children:
if idx < len(pathList) - 1:
return False
temp.children[path] = FileTree()
temp = temp.children[path]
if temp.value != None:
return False
temp.value = content
return True
def get(self, path):
temp = self.tree
for path in path.split("/"):
if path not in temp.children:
return -1
temp = temp.children[path]
return temp.value if temp.value != None else -1
def create(self, path, content):
return True if self.insert(path.split("/"), content) else False
View More Similar Problems
Game of Two Stacks
Alexa has two stacks of non-negative integers, stack A = [a0, a1, . . . , an-1 ] and stack B = [b0, b1, . . . , b m-1] where index 0 denotes the top of the stack. Alexa challenges Nick to play the following game: In each move, Nick can remove one integer from the top of either stack A or stack B. Nick keeps a running sum of the integers he removes from the two stacks. Nick is disqualified f
View Solution →Largest Rectangle
Skyline Real Estate Developers is planning to demolish a number of old, unoccupied buildings and construct a shopping mall in their place. Your task is to find the largest solid area in which the mall can be constructed. There are a number of buildings in a certain two-dimensional landscape. Each building has a height, given by . If you join adjacent buildings, they will form a solid rectangle
View Solution →Simple Text Editor
In this challenge, you must implement a simple text editor. Initially, your editor contains an empty string, S. You must perform Q operations of the following 4 types: 1. append(W) - Append W string to the end of S. 2 . delete( k ) - Delete the last k characters of S. 3 .print( k ) - Print the kth character of S. 4 . undo( ) - Undo the last (not previously undone) operation of type 1 or 2,
View Solution →Poisonous Plants
There are a number of plants in a garden. Each of the plants has been treated with some amount of pesticide. After each day, if any plant has more pesticide than the plant on its left, being weaker than the left one, it dies. You are given the initial values of the pesticide in each of the plants. Determine the number of days after which no plant dies, i.e. the time after which there is no plan
View Solution →AND xor OR
Given an array of distinct elements. Let and be the smallest and the next smallest element in the interval where . . where , are the bitwise operators , and respectively. Your task is to find the maximum possible value of . Input Format First line contains integer N. Second line contains N integers, representing elements of the array A[] . Output Format Print the value
View Solution →Waiter
You are a waiter at a party. There is a pile of numbered plates. Create an empty answers array. At each iteration, i, remove each plate from the top of the stack in order. Determine if the number on the plate is evenly divisible ith the prime number. If it is, stack it in pile Bi. Otherwise, stack it in stack Ai. Store the values Bi in from top to bottom in answers. In the next iteration, do the
View Solution →