Skip to main content

Python - Find the first recurring character in a list - 3 Solutions - Google Question

Google Question: Find the first recurring character in the list
Given an array = [2,5,1,2,3,5,1,2,4]:
It should return 2
Given an array = [2,1,1,2,3,5,1,2,4]:
It should return 1
Given an array = [2,3,4,5]:
It should return undefined

By using a new blank list

class Recurring:

def __init__(self, list1):
self.list1 = list1

def first_recurring(self):
print(self.list1)
list2 = []
for n in self.list1:
if n in list2:
return n
else:
list2.append(n)
return None


obj = Recurring([1,2,3,4,1,5])
print(obj.first_recurring())
OUTPUT
[1, 2, 3, 4,1]
1
BIG O
O(n^2) since we are using another list, it iterates through the list to find a value

By expanding existing list

class Recurring:

def __init__(self, list1):
self.list1 = list1

def first_recurring(self):
print(self.list1)
list1_len=len(self.list1)+1
for n in self.list1[0:list1_len]:
if n in self.list1[list1_len:]:
return n
else:
self.list1.append(n)
return None


obj = Recurring([1,2,3,4])
print(obj.first_recurring())
OUTPUT
[1, 2, 3, 4]
None
BIG O
O(n^2) since we are using another list, it iterates through the list to find a value

By checking each element with every other element in sequential order (worst method)

class Recurring:

def __init__(self, list1):
self.list1 = list1

def first_recurring(self):
print(self.list1)
for n in self.list1:
for m in self.list1[n:]:
if n == m:
return n
return None

obj = Recurring([1,2,2,3,4])
print(obj.first_recurring())
OUTPUT
[1, 2, 2, 3, 4]
2
BIG O
O(n^2)

By using Dictionary (Hash Table)

class Recurring:

def __init__(self, list1):
self.list1 = list1

def first_recurring(self):
print(self.list1)
dict1 = {}
i = 0
for n in self.list1:
if n in dict1.values():
return n
else:
i += 1
dict1[i] = n
return None

obj = Recurring([1,2,3,4,4])
print(obj.first_recurring())
OUTPUT
[1, 2, 2, 3, 1, 4, 4]
2
BIG O
O(n) since dictionaries are hash tables and the program doesn't iterate through it but finds the value directly using hash values. This is the most efficient way. BUT, this will give wrong answer in some cases. Like in the case used in the example. 1 is repeated too, but later, so it returns 2 which is repeated first.
By using Dictionary (Hash Table) in such a way that program returns first character that would be repeated
class Recurring:

def __init__(self, list1):
self.list1 = list1

def first_recurring(self):
print(self.list1)
dict1 = {}
i = 0
for n in self.list1:
i += 1
dict1[i] = n

for j in self.list1:
if j in dict1.values():
return j
return None

obj = Recurring([1,2,2,3,1,4,4])
print(obj.first_recurring())
OUTPUT
[1, 2, 2, 3, 1, 4, 4]
1
BIG O
O(n)

Comments

Post a Comment

Popular posts from this blog

Python - List - Append, Count, Extend, Index, Insert, Pop, Remove, Reverse, Sort

🐍 Advance List List is widely used and it's functionalities are heavily useful. Append Adds one element at the end of the list. Syntax list1.append(value) Input l1 = [1, 2, 3] l1.append(4) l1 Output [1, 2, 3, 4] append can be used to add any datatype in a list. It can even add list inside list. Caution: Append does not return anything. It just appends the list. Count .count(value) counts the number of occurrences of an element in the list. Syntax list1.count(value) Input l1 = [1, 2, 3, 4, 3] l1.count(3) Output 2 It returns 0 if the value is not found in the list. Extend .count(value) counts the number of occurrences of an element in the list. Syntax list1.extend(list) Input l1 = [1, 2, 3] l1.extend([4, 5]) Output [1, 2, 3, 4, 5] If we use append, entire list will be added to the first list like one element. Extend, i nstead of considering a list as one element, it joins the two lists one after other. Append works in the following way. Input l1 = [1, 2, 3] l1.append([4, 5]) Output...
Post a Comment
Read more

Difference between .exec() and .execPopulate() in Mongoose?

Here I answer what is the difference between .exec() and .execPopulate() in Mongoose? .exec() is used with a query while .execPopulate() is used with a document Syntax for .exec() is as follows: Model.query() . populate ( 'field' ) . exec () // returns promise . then ( function ( document ) { console . log ( document ); }); Syntax for .execPopulate() is as follows: fetchedDocument . populate ( 'field' ) . execPopulate () // returns promise . then ( function ( document ) { console . log ( document ); }); When working with individual document use .execPopulate(), for model query use .exec(). Both returns a promise. One can do without .exec() or .execPopulate() but then has to pass a callback in populate.
1 comment
Read more

Python Class to Calculate Distance and Slope of a Line with Coordinates as Input

🐍  Can be run on Jupyter Notebook #CLASS DESIGNED TO CREATE OBJECTS THAT TAKES COORDINATES AND CALCULATES DISTANCE AND SLOPE class Line:     def __init__(self,coor1,coor2):         self.coor1=coor1         self.coor2=coor2 #FUNCTION CALCULATES DISTANCE     def distance(self):         return ((self.coor2[0]-self.coor1[0])**2+(self.coor2[1]-self.coor1[1])**2)**0.5 #FUNCTION CALCULATES SLOPE         def slope(self):         return (self.coor2[1]-self.coor1[1])/(self.coor2[0]-self.coor1[0]) #DEFINING COORDINATES coordinate1 = (3,2) coordinate2 = (8,10) #CREATING OBJECT OF LINE CLASS li = Line(coordinate1,coordinate2) #CALLING DISTANCE FUNCTION li.distance() #CALLING SLOPE FUNCTION li.slope()
Post a Comment
Read more