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# Python program for implementation of Bubble Sortdef bubbleSort(arr):n = len(arr)# Traverse through all array elementsfor 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 placefor 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 elementif arr[j] > arr[j+1] :arr[j], arr[j+1] = arr[j+1], arr[j]# Driver code to test abovearr = [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 Sortimport random# Sorts array a[0..n-1] using Bogo sortdef bogoSort(a):n = len(a)while (is_sorted(a)== False):shuffle(a)# To check if array is sorted or notdef is_sorted(a):n = len(a)for i in range(0, n-1):if (a[i] > a[i+1] ):return Falsereturn True# To generate permuatation of the arraydef 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 abovea = [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 imagedef dfs(r, c):if image[r][c] == color:image[r][c] = newColorif 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.pyfrom itertools import productV='∀'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 breturn cdef 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 breturn cprint("part A:")tt(parta,3)print("part B:")tt(partb,3)
from collections import Counterdef find_parity_outliers(nums):return [x for x in numsif 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 numbersdef prime(n): # define prime numbersif n <= 1:return False# divide n by 2... up to n-1for i in range(2, n):if n % i == 0: # the remainder should'nt be a 0return Falseelse:prime_lists.append(n)return Truefor n in range(30,1000): # calling function and passing starting point =30 coz we need primes >30prime(n)check=0 # a var to limit the output to 10 onlyfor n in prime_lists:for x in prime_lists:val= n *xif (val > 1000 ):check=check +1if (check <10) :print("the num is:", val , "=",n , "* ", x )break