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Compute all the Permutation of a String

May 31, 2023CodeCatch
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Bubble sort

Nov 19, 2022CodeCatch

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# 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]),

Bogo Sort

Nov 19, 2022CodeCatch

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# 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]),

LeetCode Flood Fill

Oct 15, 2022CodeCatch

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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

CSCE 2100 Question 3

Nov 18, 2022AustinLeath

0 likes • 11 views

# 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)

find parity outliers

Nov 19, 2022CodeCatch

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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]

primes numbers finder

Mar 12, 2021mo_ak

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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