I'm trying to count all 9's in numbers from 1 to n, including repeating digits such as in 99. My code (python 3) works and returns the corrects answer for most cases except for very large numbers (like 20 digit numbers). Could someone help and let me know how this is possible?
Thanx.
def count_nines(n):
count = 0
num = [i for i in str(n)]
while len(num) > 0:
if len(num) == 1:
if num[0] == '9':
count += 1
else:
count += int(num[0]) * int(str(len(num)-1).ljust(len(num)-1, '0'))
if num[0] == '9':
count += int(''.join(num[1:]))+1
num.pop(0)
return count
The problem is in this expression:
int(str(len(num)-1).ljust(len(num)-1, '0'))
This works fine as long as str(len(num)-1) is one character, but when len(num) > 10, this is no longer the case, and then ljust will add fewer zeroes than needed. In fact, you always want to append len(num)-2 zeroes. So change this expression to:
int(str(len(num)-1) + '0' * (len(num)-2))
Simply:
def count_nines(n):
count = 0
for i in range(n+1):
if "9" in [*str(i)]:
count = count + str(i).count('9')
return count
sum((str(i).count('9') for i in range(1, n+1)))
Related
I receive a non-empty string S of at most 1000 characters and an integer N (1 ≤ N ≤ 1000). Each character of S is either a decimal digit or the character “ ? ”; the leftmost character is not “ 0 ” and at least one character is “ ? ”.
I have to show an integer D without leading zeros indicating the smallest multiple of N that has | S | digits and such that the digits in S are coincident with the corresponding digits in D. If there exists no such an integer D, i write an “ * ”.
So, i managed to write a code that kind do that, but it takes forever for large numbers, i want to find a way to make it faster or if i can improve my code.
Here its an example of input and output:
input: 1??????????????????????????????? 2 / output: 10000000000000000000000000000000
input: ???????????????????????????????1 2 / output: *
input: ?294?? 17 / output: 129404
my code:
D, N = input().split()
lenOfD = len(D)
lowerD = ''
higherD = ''
infos = {}
for index, letter in enumerate(D):
if letter.isdigit():
lowerD += letter
higherD += letter
infos[index] = letter
else:
if index == 0:
lowerD += '1'
higherD += '9'
else:
lowerD += '0'
higherD += '9'
rest = int(lowerD) % int(N)
quotient = int(int(lowerD) / int(N))
if rest != 0:
while True:
result = quotient * int(N)
resultString = str(result)
matchInfo = 0
if len(resultString) == lenOfD:
newRest = result % int(N)
if newRest == 0:
for position, value in infos.items():
if resultString[position] == value:
matchInfo += 1
quotient += 1
if result > int(higherD):
print('*')
break
if len(infos) == matchInfo:
print(result)
break
else:
print(lowerD)
and still have a minor problem, that when a divide a large number for a small one, the precision sucks because it approximate to scientific notation, and that is why i converted the variable quotient to int, but its not converting to the right number.
i need help for my syntax,
i using code:
numbers = 10
for number in range(numbers):
print (number,end="\t")
number += 1
print (number,end="\t")
number += 1
print (number,end="\t")
number += 1
print (number,end="\t")
number += 1
print (number,end="\t")
number += 1
print (number,end="\t")
number += 1
print (number,end="\t")
number += 1
print (number,end="\t")
number += 1
print (number,end="\t")
number += 1
print (number)
I don't know how to shorten the code.
If i use conditional like:
numbers = 10
for number in range(numbers):
print (number,end="\t")
if int(str(numbers)[:1]) != 10:
number += 1
print (number)
elif int(str(numbers)[:1] == 10):
print (number)
It only shows:
Here you go:
numbers = 10
for number in range(numbers):
for i in range(numbers):
print (number,end="\t")
number += 1
print('\n',end='')
You can add nested for loops.
The easy (though definitely not best!) way is to just loop it twice:
numbers = 10
for rowIndex in range(numbers):
rowString = ''
for columnIndex in range(numbers): #Same number of points
#each row is equivalent to the previous row + 1 - use existing iterator
rowString += str(columnIndex + rowIndex) + '\t'
print(rowString)
By way of explanation of what your code's currently doing:
You're converting numbers (i.e. 10) to a string, then taking a substring starting at character zero and ending before character 1. This will always be '1'
You're converting that back to an int (i.e. 1)
You're then comparing that to 10. Since 1 does not equal 10, all you're doing is printing number + 1 after your first print
Something similar would work if you used range(numbers) instead, but without another loop it's still only going to print one more column
You can do the following.
numbers = 10
for number in range(numbers):
print(" ".join([str(index) for index in range(number, numbers + number )]))
To begin, a definition:
A polydivisible number is an integer number where the first n digits of the number (from left to right) is perfectly divisible by n. For example, the integer 141 is polydivisible since:
1 % 1 == 0
14 % 2 == 0
141 % 3 == 0
I'm working on a recursive polydivisible checker, which, given a number, will check to see if that number is polydivisible, and if not, recursively check every other number after until it reaches a number that is polydivisible.
Unfortunately, my code doesn't work the way I want it to. Interestingly, when I input a number that is already polydivisible, it does its job and outputs that polydivisible number. The problem occurs when I input a non-polydivisible number, such as 13. The next polydivisible number should be 14, yet the program fails to output it. Instead, it gets stuck in an infinite loop until the memory runs out.
Here's the code I have:
def next_polydiv(num):
number = str(num)
if num >= 0:
i = 1
print(i)
while i <= len(number):
if int(number[:i]) % i == 0:
i += 1
print(i)
else:
i = 1
print(i)
num += 1
print(num)
else:
return num
else:
print("Number must be non-negative")
return None
I'm assuming the problem occurs in the else statement inside the while loop, where, if the number fails to be polydivisible, the program resets i to 0, and adds 1 to the original number so it can start checking the new number. However, like I explained, it doesn't work the way I want it to.
Any idea what might be wrong with the code, and how to make sure it stops and outputs the correct polydivisible number when it reaches one (like 14)?
(Also note that this checker is only supposed to accept non-negative numbers, hence the initial if conditional)
The mistake is that you are no updating number after incrementing num.
Here is working code:
def next_polydiv(num):
number = str(num)
if num >= 0:
i = 1
print(i)
while i <= len(number):
if int(number[:i]) % i == 0:
i += 1
print(i)
else:
i = 1
print(i)
num += 1
print(num)
number = str(num) # added line
else:
return num
else:
print("Number must be non-negative")
return None
I have a similar answer to #PranavaGande, the reason is I did not find any way to iterate an Int. Probably because there isn't one...Duh !!!
def is_polydivisible(n):
str_n = str(n)
list_2 = []
for i in range(len(str_n)):
list_2.append(len(str_n[:i+1]))
print(list_2)
list_1 = []
new_n = 0
for i in range(len(str_n)):
new_n = int(str_n[:i+1])
list_1.append(new_n)
print(list_1)
products_of_lists = []
for n1, n2 in zip(list_1, list_2):
products_of_lists.append(n1 % n2)
print(products_of_lists)
for val in products_of_lists:
if val != 0:
return False
return True
Now, I apologise for this many lines of code as it has to be smaller. However every integer has to be split individually and then divided by its index [Starting from 1 not 0]. Therefore I found it easier to list both of them and divide them index wise.
There is much shorter code than mine, however I hope this serves the purpose to find if the number is Polydivisible or Not. Of-Course you can tweak the code to find the values, where the quotient goes into decimals, returning a remainder which is Non-zero.
I have a Problem, I have to solve a task in Python and I dont know how to do it. The task is to define a function number_of_vowels, where the output should be the Number of vowels in a Word. With this function I have to write anotherone, many_vowels thats working with a list an a Number and where the number says how many vowels have to be at least in a word to be appended to the result list and then I have to append this Word. Thanks to everybody helping me ;D.
here is the code:
Wort = "parameter"
def number_of_vowels(Word):
result = 0
counter0 = 0
while result < 20:
if Word[counter0] == 'a' or 'e' or 'i' or 'o' or 'u':
result = result + 1
counter0 = counter0 + 1
else:
counter0 = counter0 + 1
return result
Words = []
counter1 = 0
def many_vowels(List , number):
if number_of_vowels(List[counter1]) < number:
counter1 + 1
else:
Words.append(List[counter1])
counter1 + 1
return Words
This code just gives me the answer to the letter a and not to the other vowels. For
print(number_of_vowels(Wort))
the output is: 1
but there are 4 vowels in this word
it also says: line 21, in many_vowels
IndexError: string index out of range
You're trying to call a function with wrong brackets. Function call should use round ones.
Try changing number_of_vowels[List[counter1]] with number_of_vowels(List[counter1])
This code contains some errors:
Calling for function should be using round brackets: number_of_vowels(List[counter1]) instead of number_of_vowels[List[counter1]]
doing result + 1 won't change value of the variable result, since you did not put the calculation result in the variable. use result = result + 1 (same for counters)
in number_of_vowels function, you want to scan the whole word? cause you did not use any loop, so it currently looking only at the first letter. Secondly, you put the compression in result and then add 1 to it. I'm not really sure why
edit:
Word = "parameter"
def number_of_vowels(Word):
result = 0
counter0 = 0
for index, letter in enumerate(Word):
if letter == 'a' or letter == 'e' or letter == 'i' or letter == 'o' or letter == 'u':
result = result + 1
return result
Words = []
counter1 = 0
def many_vowels(List_name , number):
for index, item in enumerate (List_name):
if number_of_vowels(item) >= number:
Words.append(item)
return Words
I've come across a puzzling challenge. I have to check if a number contains the same digit multiple times ex. 11, 424, 66 and so on. at first this seems easy enough but i'm having trouble coming up with a logic to check for this. any ideas?
This is what I've got so far. the function takes in a list. (updated)
arr = [[1,20],[1,10]]
for i in arr:
l = list(range(i[0],i[1]))
for num in l:
if num < 11: continue
for c in str(num):
if str(num).count(c) > 1:
# dont know why code is popping off 12 and 13
print(l.pop(num))
If your ultimate goal is simply detecting if there's a double, this function may help:
def has_doubles(n):
return len(set(str(n))) < len(str(n))
The best way I can think about is converting the number to a string and doing a Counter on it
from collections import Counter
a = 98
c = Counter(str(a))
if any(value > 1 for value in c.values()):
print "The number has repeating digits"
#Two-BitAlchemist thanks for the suggestion
looks like you wanted to create your own algorithm probably researching or a student practice well you just have to understand the properties of numbers divided by 10 where 1/10 = 0.1 10/10 = 1 13/10 = 1 reminder 3 13013/10 = 1301 rem 3 hence we can create a function that stores the reminders in an array an check them against the reminder of next number here is the algorithm in python using recursion, you can achieve the same via loops
def countNumber(foundDigits,number):
next_number = int(number/10);
reminder = number % 10;
if(next_number < 1):
for num in foundDigits:
if(num == number or num == reminder):
return True
return False;
foundDigits.append(reminder);
return countNumber(foundDigits,next_number)
example in interpreter could be
digitsFound = list()
countNumber(digitsFound, 435229)
Solved this! I didn't know pop executes based on position not value! remove is a better fit here.
arr = [[1,40],[1,10]]
for i in arr:
l = list(range(i[0],i[1]))
for num in l:
if num < 11: continue
for char in str(num):
if str(num).count(char) < 2: continue
l.remove(num)
break
print(l)
Here is my solution, its simple and works for 2 digit numbers.
nums = list(input().rstrip().split())
def has_doubles(nums):
for number in nums:
if number[0] == number[1]:
print(number)
else:
continue
has_doubles(nums)