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arr = [3,4,5,7,8,9] # This is a List
print(arr) # output will be [3,4,5,7,8,9]

# for array in python you'll need array module or nympy package
# let's import that

from array import *
arr = array('i', [3,4,5,7,8,9]) # array() is a function defined in array module

# to print array 

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arr = [3,4,5,7,8,9] # list in python
# throught this we will refer it as a array

# 1. append(value)
arr.append(10) # now the array will be [3,4,5,7,8,9,10]
# 2. insert(i, value) ------ i refer index
arr.insert(0,2) # now the array will be [2,3,4,5,7,8,9,10]
# 3. extend(arr) ------ arr refer another array 
arr.extend([11,12]) # now the array will be [2,3,4,5,7,8,9,10,11,12]
# 4. remove(value)

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# define string in python
s = "hack your code"
# now let's see some of the operations that can be perform on the string
# accessing characters in string is same as accessing elements in array
s[0] # == "h"
s[2] # == "c"
s[-1] # == "e" last character of string
# you can not assign or update a single character in string
s[3] = "u" # this is not valid this will give you error

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# our task is to reverse an array 
# Their are lots of method to reverse an array in Python

# 1. Reverse an array using Slicing method.
arr = [10,20,30,40,50] # the original array
res = arr[::-1] # reversing given array using slicing
print(res) 
# output wil be [50,40,30,20,10]

# 2. Reverse an array using reverse method.

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# reverse an array in place
# We can do this using two pointer approach with O(n) time and O(1) space complexity.
# here we are going to define function 
def reverse_in_place(arr):
    i = 0               # i hold starting index which is 0
    j = len(arr) - 1    # j hold last index which is 5 (0-based indexing)
   
    while(j > i):
        arr[i], arr[j] = arr[j], arr[i] # swapping an elements
        i += 1

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# remove duplicate from array in place
# below function returns modified array
# time complexity of this program is O(nlogn) + O(n) that is nothing but O(nlogn).
def remove_duplicate(arr):
    arr.sort()         # sort array that will take O(nlogn) time
    l = len(arr)
    if l == 0 or l == 1: # check for a base condition
        return arr
      
    x = 0 # extra pointer

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# Method 1 ----- first travese the an array then keep record of first occurring element
arr = [10,20,10,20,30,40,30]
dummy = [] ------------ # O(n) space
for i in arr: --------- # O(n)
  if i not in dummy: -- # O(n^2) this is a time complexity of this method
    dummy.append(i)
print(dummy) # this will give you desired output
# the output will be [10,20,30,40]

# Methos 2 using set() function which is in-build in python

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# we can use dictionary to store key-value pair
# by using this we can slove this problem in O(n) time complexity
# to do that we need to import Counter from collections module 

from collections import Counter

def find_duplicate(arr):
    dc = Counter(arr) # accoring to python's docs this will take O(n)
    
    for val, count in dc.items(): # this will take O(n)

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# this problem can be solve using nested for loops
# this solution is not quite efficient

def find_duplicate(arr):
  l = len(arr)
  res = []
  for i in range(l):
    for j in range(i,l):
      if arr[i] == arr[j] and not in res:
        res.append(arr[i])     # you can also print values

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# implimentation of quicksort algorithm 
# it can be implimented both recursively and iteratively.
def quicksort(arr):
    l = len(arr)
    
    if l < 2:
        return arr
    
    pivot = 0      # pivot element