Hash Map in Python

Hash maps are built-in in Python, they’re called dictionaries:

streetno = {}                        #create a dictionary called streetno
streetno["1"] = "Sachin Tendulkar"   #assign value to key "1"

Usage:

"1" in streetno                      #check if key "1" is in streetno
streetno["1"]                        #get the value from key "1"

See the documentation for more information, e.g. built-in methods and so on. They’re great, and very common in Python programs (unsurprisingly).

All you wanted (at the time the question was originally asked) was a hint. Here’s a hint: In Python, you can use dictionaries.

It’s built-in for Python. See dictionaries.

Based on your example:

streetno = {"1": "Sachine Tendulkar",
            "2": "Dravid",
            "3": "Sehwag",
            "4": "Laxman",
            "5": "Kohli" }

You could then access it like so:

sachine = streetno["1"]

Also worth mentioning: it can use any non-mutable data type as a key. That is, it can use a tuple, boolean, or string as a key.

streetno = { 1 : "Sachin Tendulkar",
            2 : "Dravid",
            3 : "Sehwag",
            4 : "Laxman",
            5 : "Kohli" }

And to retrieve values:

name = streetno.get(3, "default value")

Or

name = streetno[3]

That’s using number as keys, put quotes around the numbers to use strings as keys.

Here is the implementation of the Hash Map using python
For the simplicity hash map is of a fixed size 16.
This can be changed easily.
Rehashing is out of scope of this code.

class Node:
    def __init__(self, key, value):
        self.key = key
        self.value = value
        self.next = None

class HashMap:
    def __init__(self):
        self.store = [None for _ in range(16)]
    def get(self, key):
        index = hash(key) & 15
        if self.store[index] is None:
            return None
        n = self.store[index]
        while True:
            if n.key == key:
                return n.value
            else:
                if n.next:
                    n = n.next
                else:
                    return None
    def put(self, key, value):
        nd = Node(key, value)
        index = hash(key) & 15
        n = self.store[index]
        if n is None:
            self.store[index] = nd
        else:
            if n.key == key:
                n.value = value
            else:
                while n.next:
                    if n.key == key:
                        n.value = value
                        return
                    else:
                        n = n.next
                n.next = nd

hm = HashMap()
hm.put("1", "sachin")
hm.put("2", "sehwag")
hm.put("3", "ganguly")
hm.put("4", "srinath")
hm.put("5", "kumble")
hm.put("6", "dhoni")
hm.put("7", "kohli")
hm.put("8", "pandya")
hm.put("9", "rohit")
hm.put("10", "dhawan")
hm.put("11", "shastri")
hm.put("12", "manjarekar")
hm.put("13", "gupta")
hm.put("14", "agarkar")
hm.put("15", "nehra")
hm.put("16", "gawaskar")
hm.put("17", "vengsarkar")
print(hm.get("1"))
print(hm.get("2"))
print(hm.get("3"))
print(hm.get("4"))
print(hm.get("5"))
print(hm.get("6"))
print(hm.get("7"))
print(hm.get("8"))
print(hm.get("9"))
print(hm.get("10"))
print(hm.get("11"))
print(hm.get("12"))
print(hm.get("13"))
print(hm.get("14"))
print(hm.get("15"))
print(hm.get("16"))
print(hm.get("17"))

Output:

sachin
sehwag
ganguly
srinath
kumble
dhoni
kohli
pandya
rohit
dhawan
shastri
manjarekar
gupta
agarkar
nehra
gawaskar
vengsarkar

class HashMap:
    def __init__(self):
        self.size = 64
        self.map = [None] * self.size

    def _get_hash(self, key):
        hash = 0

        for char in str(key):
            hash += ord(char)
        return hash % self.size

    def add(self, key, value):
        key_hash = self._get_hash(key)
        key_value = [key, value]

        if self.map[key_hash] is None:
            self.map[key_hash] = list([key_value])
            return True
        else:
            for pair in self.map[key_hash]:
                if pair[0] == key:
                    pair[1] = value
                    return True
                else:
                    self.map[key_hash].append(list([key_value]))
                    return True

    def get(self, key):
        key_hash = self._get_hash(key)
        if self.map[key_hash] is not None:
            for pair in self.map[key_hash]: 
                if pair[0] == key:
                    return pair[1]
        return None

    def delete(self, key):
        key_hash = self._get_hash(key)

        if self.map[key_hash] is None :
            return False
        for i in range(0, len(self.map[key_hash])):
            if self.map[key_hash][i][0] == key:
                self.map[key_hash].pop(i)
                return True

    def print(self):

        print('---Phonebook---')
        for item in self.map:
            if item is not None:
                print(str(item))

h = HashMap()

Python Counter is also a good option in this case:

from collections import Counter

counter = Counter(["Sachin Tendulkar", "Sachin Tendulkar", "other things"])

print(counter)

This returns a dict with the count of each element in the list:

Counter({'Sachin Tendulkar': 2, 'other things': 1})

In python you would use a dictionary.

It is a very important type in python and often used.

You can create one easily by

name = {}

Dictionaries have many methods:

# add entries:
>>> name['first'] = 'John'
>>> name['second'] = 'Doe'
>>> name
{'first': 'John', 'second': 'Doe'}

# you can store all objects and datatypes as value in a dictionary
# as key you can use all objects and datatypes that are hashable
>>> name['list'] = ['list', 'inside', 'dict']
>>> name[1] = 1
>>> name
{'first': 'John', 'second': 'Doe', 1: 1, 'list': ['list', 'inside', 'dict']}

You can not influence the order of a dict.

A dictionary in Python is the best way to implement this. We can create the following dictionary using the given <key,value> pairs:

d = {"1": "Sachin Tendulkar", "2": "Dravid", "3": "Sehwag", "4": "Laxman", "5": "Kohli"}

To extract the value of a particular key, we can directly use d[key]:

name = d["1"] # The value of name would be "Sachin Tendulkar" here

That is my solution for LeetCode problem 706:

A hash map class with three methods: get, put and remove

class Item:
    def __init__(self, key, value):
        self.key = key
        self.value = value
        self.next = None


class MyHashMap:

    def __init__(self, size=100):
        self.items = [None] * size
        self.size = size

    def _get_index(self, key):
        return hash(key) & self.size-1

    def put(self, key: int, value: int) -> None:
        index = self._get_index(key)
        item = self.items[index]
        if item is None:
            self.items[index] = Item(key, value)
        else:
            if item.key == key:
                item.value = value

            else:
                while True:
                    if item.key == key:
                        item.value = value
                        return
                    else:
                        if not item.next:
                            item.next = Item(key, value)
                            return
                        item = item.next



    def get(self, key: int) -> int:
        index = self._get_index(key)
            
        if self.items[index] is None:
            return -1
        item = self.items[index]
        while True:
            if item.key == key:
                return item.value
            else:
                if item.next:
                    item = item.next
                else:
                    return -1
        
        

    def remove(self, key: int) -> None:
        value = self.get(key)
        if value > -1:
            index = self._get_index(key)
            item = self.items[index]
            if item.key == key:
                self.items[index] = item.next if item.next else None
                return
                
            while True:
                if item.next and item.next.key == key:
                    item.next = item.next.next
                    return
                else:
                    if item.next:
                        item = item.next
                    else:
                        return