Set - Google Top Interview Questions
Problem Statement :
Implement a set data structure with the following methods: CustomSet() constructs a new instance of a set add(int val) adds val to the set exists(int val) returns whether val exists in the set remove(int val) removes the val in the set This should be implemented without using built-in set. Constraints n ≤ 100,000 where n is the number of calls to add, exists and remove Example 1 Input methods = ["constructor", "add", "exists", "remove", "exists"] arguments = [[], [1], [1], [1], [1]]` Output [None, None, True, None, False] Explanation c = CustomSet() c.add(1) c.exists(1) == True c.remove(1) c.exists(1) == False
Solution :
Solution in C++ :
class CustomSet {
int S = 1e3;
vector<list<int>> v;
hash<int> hs;
public:
CustomSet() {
v.resize(S);
}
void add(int val) {
if (exists(val)) return;
int id = hs(val) % S;
v[id].push_back(val);
}
bool exists(int val) {
int id = hs(val) % S;
for (int x : v[id]) {
if (x == val) return true;
}
return false;
}
void remove(int val) {
int id = hs(val) % S;
for (auto it = v[id].begin(); it != v[id].end(); it++) {
if (*it == val) {
v[id].erase(it);
return;
}
}
}
};
Solution in Java :
import java.util.*;
class CustomSet {
ArrayList<Integer>[] buckets;
int T;
public CustomSet() {
T = 300;
buckets = new ArrayList[T];
}
public void add(int val) {
if (!exists(val)) {
buckets[map(val)].add(val);
}
}
public boolean exists(int val) {
int m = map(val);
if (buckets[m] == null)
buckets[m] = new ArrayList<Integer>();
return buckets[m].contains(val);
}
public void remove(int val) {
if (exists(val))
buckets[map(val)].remove(Integer.valueOf(val));
}
public int map(int val) {
return ((val % T) + T) % T;
}
}
Solution in Python :
class ListNode:
def __init__(self, val):
self.val = val
self.next = None
class CustomSet:
def __init__(self):
self.mod = 2069 # any large prime number
self.lst = [None] * self.mod
def add(self, val):
hashkey = val % self.mod
if not self.lst[hashkey]:
self.lst[hashkey] = ListNode(val)
else:
curr = self.lst[hashkey]
while curr.next:
if curr.val == val:
return
curr = curr.next
if curr.val == val:
return
curr.next = ListNode(val)
def exists(self, val):
hashkey = val % self.mod
if self.lst[hashkey]:
curr = self.lst[hashkey]
while curr:
if curr.val == val:
return True
curr = curr.next
return False
def remove(self, val):
hashkey = val % self.mod
if self.lst[hashkey]:
curr = self.lst[hashkey]
if curr.val == val:
self.lst[hashkey] = curr.next
return
while curr.next:
if curr.next.val == val:
curr.next = curr.next.next
return
curr = curr.next
View More Similar Problems
Jesse and Cookies
Jesse loves cookies. He wants the sweetness of all his cookies to be greater than value K. To do this, Jesse repeatedly mixes two cookies with the least sweetness. He creates a special combined cookie with: sweetness Least sweet cookie 2nd least sweet cookie). He repeats this procedure until all the cookies in his collection have a sweetness > = K. You are given Jesse's cookies. Print t
View Solution →Find the Running Median
The median of a set of integers is the midpoint value of the data set for which an equal number of integers are less than and greater than the value. To find the median, you must first sort your set of integers in non-decreasing order, then: If your set contains an odd number of elements, the median is the middle element of the sorted sample. In the sorted set { 1, 2, 3 } , 2 is the median.
View Solution →Minimum Average Waiting Time
Tieu owns a pizza restaurant and he manages it in his own way. While in a normal restaurant, a customer is served by following the first-come, first-served rule, Tieu simply minimizes the average waiting time of his customers. So he gets to decide who is served first, regardless of how sooner or later a person comes. Different kinds of pizzas take different amounts of time to cook. Also, once h
View Solution →Merging Communities
People connect with each other in a social network. A connection between Person I and Person J is represented as . When two persons belonging to different communities connect, the net effect is the merger of both communities which I and J belongs to. At the beginning, there are N people representing N communities. Suppose person 1 and 2 connected and later 2 and 3 connected, then ,1 , 2 and 3 w
View Solution →Components in a graph
There are 2 * N nodes in an undirected graph, and a number of edges connecting some nodes. In each edge, the first value will be between 1 and N, inclusive. The second node will be between N + 1 and , 2 * N inclusive. Given a list of edges, determine the size of the smallest and largest connected components that have or more nodes. A node can have any number of connections. The highest node valu
View Solution →Kundu and Tree
Kundu is true tree lover. Tree is a connected graph having N vertices and N-1 edges. Today when he got a tree, he colored each edge with one of either red(r) or black(b) color. He is interested in knowing how many triplets(a,b,c) of vertices are there , such that, there is atleast one edge having red color on all the three paths i.e. from vertex a to b, vertex b to c and vertex c to a . Note that
View Solution →