# Python program for implementation of Bubble Sort

def bubbleSort(arr):
	n = len(arr)

	# Traverse through all array elements
	for i in range(n-1):
	# range(n) also work but outer loop will repeat one time more than needed.

		# Last i elements are already in place
		for j in range(0, n-i-1):

			# traverse the array from 0 to n-i-1
			# Swap if the element found is greater
			# than the next element
			if arr[j] > arr[j+1] :
				arr[j], arr[j+1] = arr[j+1], arr[j]

# Driver code to test above
arr = [64, 34, 25, 12, 22, 11, 90]

bubbleSort(arr)

print ("Sorted array is:")
for i in range(len(arr)):
	print ("%d" %arr[i]),

# Python program for implementation of Bogo Sort
import random

# Sorts array a[0..n-1] using Bogo sort
def bogoSort(a):
	n = len(a)
	while (is_sorted(a)== False):
		shuffle(a)

# To check if array is sorted or not
def is_sorted(a):
	n = len(a)
	for i in range(0, n-1):
		if (a[i] > a[i+1] ):
			return False
	return True

# To generate permuatation of the array
def shuffle(a):
	n = len(a)
	for i in range (0,n):
		r = random.randint(0,n-1)
		a[i], a[r] = a[r], a[i]

# Driver code to test above
a = [3, 2, 4, 1, 0, 5]
bogoSort(a)
print("Sorted array :")
for i in range(len(a)):
	print ("%d" %a[i]),

class Solution(object):
    def floodFill(self, image, sr, sc, newColor):
        R, C = len(image), len(image[0])
        color = image[sr][sc]
        if color == newColor: return image
        def dfs(r, c):
            if image[r][c] == color:
                image[r][c] = newColor
                if r >= 1: dfs(r-1, c)
                if r+1 < R: dfs(r+1, c)
                if c >= 1: dfs(r, c-1)
                if c+1 < C: dfs(r, c+1)

        dfs(sr, sc)
        return image

# question3.py
from itertools import product
V='∀'
E='∃'

def tt(f,n) :
  xss=product((0,1),repeat=n)
  print('function:',f.__name__)
  for xs in xss : print(*xs,':',int(f(*xs)))
  print('')

# this is the logic for part A (p\/q\/r) /\ (p\/q\/~r) /\ (p\/~q\/r) /\ (p\/~q\/~r) /\ (~p\/q\/r) /\ (~p\/q\/~r) /\ (~p\/~q\/r) /\ (~p\/~q\/~r)

def parta(p,q,r) :
  
  a=(p or q or r) and (p or q or not r) and (p or not q or r)and (p or not q or  not r)
  b=(not p or q or r ) and (not p or q or not r) and (not p or not q or r) and (not p or not q or not r)
  c= a and b
  return c

def partb(p,q,r) :
  a=(p or q and r) and (p or not q or not r) and (p or not q or not  r)and (p or q or  not r)
  b=(not p or q or r ) and (not p or q or not r) and (not p or not q or r) and (not p or not q or not r)
  c= a and b
  return c

print("part A:")
tt(parta,3)

print("part B:")
tt(partb,3)

from collections import Counter

def find_parity_outliers(nums):
  return [
    x for x in nums
    if x % 2 != Counter([n % 2 for n in nums]).most_common()[0][0]
  ]

find_parity_outliers([1, 2, 3, 4, 6]) # [1, 3]

prime_lists=[] # a list to store the prime numbers

def prime(n): # define prime numbers
    if n <= 1:
        return False
    # divide n by 2... up to n-1
    for i in range(2, n):
        if n % i == 0:  # the remainder should'nt be a 0
            return False
    else:
        prime_lists.append(n)
        return True



for n in range(30,1000):  # calling function and passing starting point =30 coz we need primes >30
    prime(n)

check=0 # a var to limit the output  to 10 only
for n in prime_lists:
    for x in prime_lists:
        val=  n *x
        if (val > 1000 ):
            check=check +1
        if (check <10) :
            print("the num is:", val , "=",n , "* ", x )
        break