**Hash Table - Amazon Top Interview Questions**

### Problem Statement :

Implement a hash table with the following methods: HashTable() constructs a new instance of a hash table put(int key, int val) updates the hash table such that key maps to val get(int key) returns the value associated with key. If there is no such key, then return -1. remove(int key) removes both the key and the value associated with it in the hash table. This should be implemented without using built-in hash table. Constraints n ≤ 100,000 where n is the number of calls to put, get and remove\ Example 1 Input methods = ["constructor", "put", "get", "remove", "get"] arguments = [[], [1, 10], [1], [1], [1]]` Output [None, None, 10, None, -1]

### Solution :

` ````
Solution in C++ :
class HashTable {
public:
vector<vector<pair<int, int>>> hashMap;
HashTable() {
hashMap = vector<vector<pair<int, int>>>(1000, vector<pair<int, int>>());
}
int findHash(int key) {
return key % 1000;
}
void put(int key, int val) {
int hashVal = findHash(key);
for (int i = 0; i < hashMap[hashVal].size(); i++) {
if (hashMap[hashVal][i].first == key) {
hashMap[hashVal][i].second = val;
return;
}
}
hashMap[hashVal].push_back({key, val});
}
int get(int key) {
int hashVal = findHash(key);
for (int i = 0; i < hashMap[hashVal].size(); i++) {
if (hashMap[hashVal][i].first == key) return hashMap[hashVal][i].second;
}
return -1;
}
void remove(int key) {
int hashVal = findHash(key);
int n = hashMap[hashVal].size();
for (int i = 0; i < hashMap[hashVal].size(); i++) {
if (hashMap[hashVal][i].first == key) {
swap(hashMap[hashVal][i], hashMap[hashVal][n - 1]);
hashMap[hashVal].pop_back();
}
}
return;
}
};
```

` ````
Solution in Java :
import java.util.*;
class HashTable {
static class Pair {
int key;
int val;
Pair(int key, int val) {
this.key = key;
this.val = val;
}
}
int size;
ArrayList<Pair>[] buckets;
public HashTable() {
this.size = 10000;
this.buckets = new ArrayList[10000];
for (int i = 0; i < 10000; i++) buckets[i] = new ArrayList();
}
private int hash(int key) {
return key % size;
}
public void put(int key, int val) {
int hash_val = hash(key);
ArrayList<Pair> arr = buckets[hash_val];
if (contains(key, val)) {
for (Pair p : arr) {
if (p.key == key)
p.val = val;
}
} else {
arr.add(new Pair(key, val));
}
}
private boolean contains(int key, int val) {
int hash_val = hash(key);
ArrayList<Pair> arr = buckets[hash_val];
for (Pair p : arr) {
if (p.key == key)
return true;
}
return false;
}
public int get(int key) {
if (!contains(key, -1))
return -1;
int hash_val = hash(key);
ArrayList<Pair> arr = buckets[hash_val];
for (Pair p : arr) {
if (p.key == key)
return p.val;
}
return -1;
}
public void remove(int key) {
ArrayList<Pair> newArr = new ArrayList();
int hash_val = hash(key);
ArrayList<Pair> arr = buckets[hash_val];
for (Pair p : arr) {
if (p.key != key)
newArr.add(p);
}
buckets[hash_val] = newArr;
}
}
```

` ````
Solution in Python :
class Node:
def __init__(self, key=None, value=None):
self.key = key
self.value = value
self.next = None
class HashTable:
def __init__(self):
self.hashmap = [None for i in range(1001)]
def find_index(self, key):
return key % 1001
def find(self, key, node):
prev = None
cur = node
while cur and cur.key != key:
prev = cur
cur = cur.next
return prev
def put(self, key, value):
index = self.find_index(key)
if self.hashmap[index] is None:
self.hashmap[index] = Node()
node = self.hashmap[index]
node = self.find(key, node)
if node.next:
node.next.value = value
else:
node.next = Node(key, value)
def get(self, key):
index = self.find_index(key)
if self.hashmap[index] is None:
return -1
node = self.hashmap[index]
node = self.find(key, node)
if node.next:
return node.next.value
else:
return -1
def remove(self, key):
index = self.find_index(key)
if self.hashmap[index] is None:
return
node = self.hashmap[index]
node = self.find(key, node)
if node.next:
node.next = node.next.next
else:
return
```

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