I really do not know how to figure it out!!!
remove_middle_character("apple")
'aple'
remove_middle_character("banana")
'bana'
remove_middle_character("discount")
‘disunt‘
”“”
Since it's probably classwork, I'm not going to give you the code (which you would be unwise using anyway since your educational institution probably knows about Stack Overflow and would catch you out with plagiarism). Instead, I'll provide guidance on what you need to know.
If the string or of size two or less (we handle this early so that we don't have to worry about edge cases in the following steps), just return an empty string.
Otherwise, if the length of the string is odd, get the length x, and return the first x // 2 (the // operator is integer division) characters concatenated with the character starting at x // 2 + 1.
Otherwise, return the first x // 2 - 1 characters concatenated with the characters starting at x // 2 + 1.
In terms of turning that into code:
you can get the length of a string x with len(x);
you can get the first n characters with x[:n];
you can get the characters starting at m with x[m:];
you can concatenate strings with +; and
you can test if a value p is even by using if (p % 2) == 0 (and, of course, odd numbers will cause (p % 2) == 1 to evaluate as true).
That should be all you need to write the code yourself.
Using all the excellent suggestions above, I suggest a refactored approach where the edge conditions mentioned by paxdiablo and enhanced by mhawke are catered for but remove the need for using an if-else statement.
As you can see from paxdiablo if the string length is odd (the modulo function returns 1) the fist part of the string is:
x // 2 (subtract 0).
If the string length is even (the modulo function returns 0) the fist part of the string is:
x // 2 - 1 (subtract 1).
In both cases (odd and even length), the second part of the string is:
x // 2 + 1
So we need to subtract the reverse of what the modulo 2 of the string length would give us. We can do this by making an even length odd and an odd length even. One possible way is: mod = (len(s) + 1) % 2. So a possible solution might look like:
def remove_middle_character(s):
h = len(s)//2
mod = (len(s) + 1) % 2
return s[:h - mod] + s[h + 1:]
if __name__ == "__main__":
print(remove_middle_character(""))
print(remove_middle_character("x"))
print(remove_middle_character("xy"))
print(remove_middle_character("apple"))
print(remove_middle_character("banana"))
print(remove_middle_character("discount"))
Related
I have not really used maps before in my programming experience, so I am having trouble understanding the more complex versions of maps. So let's say that the problem is you are given an integer in minutes, in this case n = 808. What you are to do with this number is convert it to 24 hour time, so hh:mm. This would give you 13:28. Once this is done, add up the digits of that time to get the answer. So, the answer would be 14. I saw a really nice one-liner to this solution and I am trying to understand it because my solution took about 5 more lines of code.
This is the solution:
sum(map(int, str(n // 60 * 100 + n % 60)))
So I understand that maps apply the same function over an iteration of numbers, but what throws me off is the int,str(...) part. I am not sure what is going on behind the scenes.
There are two mathematical operators used here:
// represents floor division, i.e. extract the integer portion of the result after division.
% represents modulus, i.e. the remainder after division.
Therefore, for n = 808, the algorithm returns:
str(808 // 60 * 100 + 808 % 60) = str(13 * 100 + 28) = '1328'
map(int, '1328') then takes each character in the string '1328' and converts it into an integer, itself returning an iterable. map requires an iterable as its second (and subsequent) arguments. Strings can be iterated to extract each character one at a time.
Finally, sum takes each of the integers returned from map and adds them together.
An equivalent formulation of the logic is possible via sum with a generator expression:
sum(int(i) for i in str(n // 60 * 100 + n % 60))
map ,as you stated, applies a function over a iterable.
So, when you do
map(int, str(n // 60 * 100 + n % 60))
You are using the function int over the iterable str(n // 60 * 100 + n % 60). As you probably know, strings are iterables (because, of course, you can iterate over them) - that can be easily checked
for char in "abcd":
print(char)
a
b
c
d
The return from str(n // 60 * 100 + n % 60) is '1328'. When you apply int to each char, you transform each to an integer. You can easily see this by instead of taking the sum right away, getting a lsit
list(map(int, str(n // 60 * 100 + n % 60)))
[1, 3, 2, 8]
I guess now it is easy to see that the sum will get the sum of these numbers, which is what you wanted from the beginning :)
Both int() and str() are functions. In this particular example, when n=808 the argument to the str() function is calculated as 1328, which when converted to the string becomes '1328'. A string is iterable, so the map function is simply applying int to each character of the string, producing the sequence [1,3,2,8].
just wondering if a better solution exists for this sort of problem.
We know that for a X/Y percentage split of an even number we can get an exact split of the data - for example for data size 10:
10 * .6 = 6
10 * .4 = 4
10
Splitting data this way is easy, and we can guarantee we have all of the data and nothing is lost. However where I am struggling is on less friendly numbers - take 11
11 * .6 = 6.6
11 * .4 = 4.4
11
However we can't index into an array at i = 6.6 for example. So we have to decide how to to do this. If we take JUST the integer portion we lose 1 data point -
First set = 0..6
Second set = 6..10
This would be the same case if we floored the numbers.
However, if we take the ceiling of the numbers:
First set = 0..7
Second set = 7..12
And we've read past the end of our array.
This gets even worse when we throw in a 3rd or 4th split (30,30,20,20 for example).
Is there a standard splitting procedure for these kinds of problems? Is data loss accepted? It seems like data loss would be unacceptable for dependent data, such as time series.
Thanks!
EDIT: The values .6 and .4 are chosen by me. They could be any two numbers that sum to 1.
First of all, notice that your problem is not limited to odd-sized arrays as you claim, but any-sized arrays. How would you make the 56%-44% split of a 10 element array? Or a 60%-40% split of a 4 element array?
There is no standard procedure. In many cases, programmers do not care that much about an exact split and they either do it by flooring or rounding one quantity (the size of the first set), while taking the complementary (array length - rounded size) for the other (the size of the second).
This might be ok in most cases when this is an one-off calculation and accuracy is not required. You have to ask yourself what your requirements are. For example: are you taking thousands of 10-sized arrays and each time you are splitting them 56%-44% doing some calculations and returning a result? You have to ask yourself what accuracy do you want. Do you care if your result ends up being
the 60%-40% split or the 50%-50% split?
As another example imagine that you are doing a 4-way equal split of 25%-25%-25%-25%. If you have 10 elements and you apply the rounding technique you end up with 3,3,3,1 elements. Surely this will mess up your results.
If you do care about all these inaccuracies then the first step is consider whether you can to adjust either the array size and/or the split ratio(s).
If these are set in stone then the only way to have an accurate split of any ratios of any sized array is to make it probabilistic. You have to split multiple arrays for this to work (meaning you have to apply the same split ratio to same-sized arrays multiple times). The more arrays the better (or you can use the same array multiple times).
So imagine that you have to make a 56%-44% split of a 10 sized array. This means that you need to split it in 5.6 elements and 4.4 elements on the average.
There are many ways you can achieve a 5.6 element average. The easiest one (and the one with the smallest variance in the sequence of tries) is to have 60% of the time a set with 6 elements and 40% of the time a set that has 5 elements.
0.6*6 + 0.4*5 = 5.6
In terms of code this is what you can do to decide on the size of the set each time:
import random
array_size = 10
first_split = 0.56
avg_split_size = array_size * first_split
floored_split_size = int(avg_split_size)
if avg_split_size > floored_split_size:
if random.uniform(0,1) > avg_split_size - floored_split_size:
this_split_size = floored_split_size
else:
this_split_size = floored_split_size + 1
else:
this_split_size = avg_split_size
You could make the code more compact, I just made an outline here so you get the idea. I hope this helps.
Instead of using ciel() or floor() use round() instead. For example:
>>> round(6.6)
7.0
The value returned will be of float type. For getting the integer value, type-cast it to int as:
>>> int(round(6.6))
7
This will be the value of your first split. For getting the second split, calculate it using len(data) - split1_val. This will be applicable in case of 2 split problem.
In case of 3 split, take round value of two split and take the value of 3rd split as the value of len(my_list) - val_split_1 - val_split2
In a Generic way, For N split:
Take the round() value of N-1 split. And for the last value, do len(data) - "value of N round() values".
where len() gives the length of the list.
Let's first consider just splitting the set into two pieces.
Let n be the number of elements we are splitting, and p and q be the proportions, so that
p+q == 1
I assert that the parts after the decimal point will always sum to either 1 or 0, so we should use floor on one and ceil on the other, and we will always be right.
Here is a function that does that, along with a test. I left the print statements in but they are commented out.
def simpleSplitN(n, p, q):
"split n into proportions p and q and return indices"
np = math.ceil(n*p)
nq = math.floor(n*q)
#print n, sum([np, nq]) #np and nq are the proportions
return [0, np] #these are the indices we would use
#test for simpleSplitN
for i in range(1, 10):
p = i/10.0;
q = 1-p
simpleSplitN(37, p, q);
For the mathematically inclined, here is the proof that the decimal proportions will sum to 1
-----------------------
We can express p*n as n/(1/p), and so by the division algorithm we get integers k and r
n == k*(1/p) + r with 0 <= r < (1/p)
Thus r/(1/p) == p*r < 1
We can do exactly the same for q, getting
q*r < 1 (this is a different r)
It is important to note that q*r and p*r are the part after the decimal when we divide our n.
Now we can add them together (we've added subscripts now)
0 <= p*(r_1) < 1
0 <= q*(r_2) < 1
=> 0 < p*r + q*r == p*n + q*n + k_1 + k_2 == n + k_1 + k_2 < 2
But by closure of the integers, n + k_1 + k_2 is an integer and so
0 < n + k_1 + k_2 < 2
means that p*r + q*r must be either 0 or 1. It will only be 0 in the case that our n is divided evenly.
Otherwise we can now see that our fractional parts will always sum to 1.
-----------------------
We can do a very similar (but slightly more complicated) proof for splitting n into an arbitrary number (say N) parts, but instead of them summing to 1, they will sum to an integer less than N.
Here is the general function, it has uncommented print statements for verification purposes.
import math
import random
def splitN(n, c):
"""Compute indices that can be used to split
a dataset of n items into a list of proportions c
by first dividing them naively and then distributing
the decimal parts of said division randomly
"""
nc = [n*i for i in c];
nr = [n*i - int(n*i) for i in c] #the decimal parts
N = int(round(sum(nr))) #sum of all decimal parts
print N, nc
for i in range(0, len(nc)):
nc[i] = math.floor(nc[i])
for i in range(N): #randomly distribute leftovers
nc[random.randint(1, len(nc)) - 1] += 1
print n,sum(nc); #nc now contains the proportions
out = [0] #compute a cumulative sum
for i in range(0, len(nc) - 1):
out.append(out[-1] + nc[i])
print out
return out
#test for splitN with various proportions
c = [.1,.2,.3,.4]
c = [.2,.2,.2,.2,.2]
c = [.3, .2, .2, .3]
for n in range( 10, 40 ):
print splitN(n, c)
If we have leftovers, we will never get an even split, so we distribute them randomly, like #Thanassis said. If you don't like the dependency on random, then you could just add them all at the beginning or at even intervals.
Both of my functions output indices but they compute proportions and thus could be slightly modified to output those instead per user preference.
I came across this problem Unlucky number 13! recently but could not think of efficient solution this.
Problem statement :
N is taken as input.
N can be very large 0<= N <= 1000000009
Find total number of such strings that are made of exactly N characters which don't include "13". The strings may contain any integer from 0-9, repeated any number of times.
# Example:
# N = 2 :
# output : 99 (0-99 without 13 number)
# N =1 :
# output : 10 (0-9 without 13 number)
My solution:
N = int(raw_input())
if N < 2:
print 10
else:
without_13 = 10
for i in range(10, int('9' * N)+1):
string = str(i)
if string.count("13") >= 1:
continue
without_13 += 1
print without_13
Output
The output file should contain answer to each query in a new line modulo 1000000009.
Any other efficient way to solve this ? My solution gives time limit exceeded on coding site.
I think this can be solved via recursion:
ans(n) = { ans([n/2])^2 - ans([n/2]-1)^2 }, if n is even
ans(n) = { ans([n/2]+1)*ans([n/2]) - ans([n/2])*ans([n/2]-1) }, if n is odd
Base Cases:
ans(0) = 1
ans(1) = 10
It's implementation is running quite fast even for larger inputs like 10^9 ( which is expected as its complexity is O(log[n]) instead of O(n) like the other answers ):
cache = {}
mod = 1000000009
def ans(n):
if cache.has_key(n):
return cache[n]
if n == 0:
cache[n] = 1
return cache[n]
if n == 1:
cache[n] = 10
return cache[n]
temp1 = ans(n/2)
temp2 = ans(n/2-1)
if (n & 1) == 0:
cache[n] = (temp1*temp1 - temp2*temp2) % mod
else:
temp3 = ans(n/2 + 1)
cache[n] = (temp1 * (temp3 - temp2)) % mod
return cache[n]
print ans(1000000000)
Online Demo
Explanation:
Let a string s have even number of digits 'n'.
Let ans(n) be the answer for the input n, i.e. the number of strings without the substring 13 in them.
Therefore, the answer for string s having length n can be written as the multiplication of the answer for the first half of the string (ans([n/2])) and the answer for the second half of the string (ans([n/2])), minus the number of cases where the string 13 appears in the middle of the number n, i.e. when the last digit of the first half is 1 and the first digit of the second half is 3.
This can expressed mathematically as:
ans(n) = ans([n/2])^2 - ans([n/2]-1)*2
Similarly for the cases where the input number n is odd, we can derive the following equation:
ans(n) = ans([n/2]+1)*ans([n/2]) - ans([n/2])*ans([n/2]-1)
I get the feeling that this question is designed with the expectation that you would initially instinctively do it the way you have. However, I believe there's a slightly different approach that would be faster.
You can produce all the numbers that contain the number 13 yourself, without having to loop through all the numbers in between. For example:
2 digits:
13
3 digits position 1:
113
213
313 etc.
3 digits position 2: 131
132
133 etc.
Therefore, you don't have to check all the number from 0 to n*9. You simply count all the numbers with 13 in them until the length is larger than N.
This may not be the fastest solution (in fact I'd be surprised if this couldn't be solved efficiently by using some mathematics trickery) but I believe it will be more efficient than the approach you have currently taken.
This a P&C problem. I'm going to assume 0 is valid string and so is 00, 000 and so on, each being treated distinct from the other.
The total number of strings not containing 13, of length N, is unsurprisingly given by:
(Total Number of strings of length N) - (Total number of strings of length N that have 13 in them)
Now, the Total number of strings of length N is easy, you have 10 digits and N slots to put them in: 10^N.
The number of strings of length N with 13 in them is a little trickier.
You'd think you can do something like this:
=> (N-1)C1 * 10^(N-2)
=> (N-1) * 10^(N-2)
But you'd be wrong, or more accurately, you'd be over counting certain strings. For example, you'd be over counting the set of string that have two or more 13s in them.
What you really need to do is apply the inclusion-exclusion principle to count the number of strings with 13 in them, so that they're all included once.
If you look at this problem as a set counting problem, you have quite a few sets:
S(0,N): Set of all strings of Length N.
S(1,N): Set of all strings of Length N, with at least one '13' in it.
S(2,N): Set of all strings of Length N, with at least two '13's in it.
...
S(N/2,N): Set of all strings of Length N, with at least floor(N/2) '13's in it.
You want the set of all strings with 13 in them, but counted at most once. You can use the inclusion-exclusion principle for computing that set.
Let f(n) be the number of sequences of length n that have no "13" in them, and g(n) be the number of sequences of length n that have "13" in them.
Then f(n) = 10^n - g(n) (in mathematical notation), because it's the number of possible sequences (10^n) minus the ones that contain "13".
Base cases:
f(0) = 1
g(0) = 0
f(1) = 10
g(1) = 0
When looking for the sequences with "13", a sequence can have a "13" at the beginning. That will account for 10^(n-2) possible sequences with "13" in them. It could also have a "13" in the second position, again accounting for 10^(n-2) possible sequences. But if it has a "13" in the third position, and we'd assume there would also be 10^(n-2) possible sequences, we could those twice that already had a "13" in the first position. So we have to substract them. Instead, we count 10^(n-4) times f(2) (because those are exactly the combinations in the first two positions that don't have "13" in them).
E.g. for g(5):
g(5) = 10^(n-2) + 10^(n-2) + f(2)*10^(n-4) + f(3)*10^(n-5)
We can rewrite that to look the same everywhere:
g(5) = f(0)*10^(n-2) + f(1)*10^(n-3) + f(2)*10^(n-4) + f(3)*10^(n-5)
Or simply the sum of f(i)*10^(n-(i+2)) with i ranging from 0 to n-2.
In Python:
from functools import lru_cache
#lru_cache(maxsize=1024)
def f(n):
return 10**n - g(n)
#lru_cache(maxsize=1024)
def g(n):
return sum(f(i)*10**(n-(i+2)) for i in range(n-1)) # range is exclusive
The lru_cache is optional, but often a good idea when working with recursion.
>>> [f(n) for n in range(10)]
[1, 10, 99, 980, 9701, 96030, 950599, 9409960, 93149001, 922080050]
The results are instant and it works for very large numbers.
In fact this question is more about math than about python.
For N figures there is 10^N possible unique strings. To get the answer to the problem we need to subtract the number of string containing "13".
If string starts from "13" we have 10^(N-2) possible unique strings. If we have 13 at the second possition (e.i. a string like x13...), we again have 10^(N-2) possibilities. But we can't continue this logic further as this will lead us to double calculation of string which have 13 at different possitions. For example for N=4 there will be a string "1313" which we will calculate twice. To avoid this we should calculate only those strings which we haven't calculated before. So for "13" on possition p (counting from 0) we should find the number of unique string which don't have "13" on the left side from p, that is for each p
number_of_strings_for_13_at_p = total_number_of_strings_without_13(N=p-1) * 10^(N-p-2)
So we recursevily define the total_number_of_strings_without_13 function.
Here is the idea in the code:
def number_of_strings_without_13(N):
sum_numbers_with_13 = 0
for p in range(N-1):
if p < 2:
sum_numbers_with_13 += 10**(N-2)
else:
sum_numbers_with_13 += number_of_strings_without_13(p) * 10**(N-p-2)
return 10**N - sum_numbers_with_13
I should say that 10**N means 10 in the power of N. All the other is described above. The functions also has a surprisingly pleasent ability to give correct answers for N=1 and N=2.
To test this works correct I've rewritten your code into function and refactored a little bit:
def number_of_strings_without_13_bruteforce(N):
without_13 = 0
for i in range(10**N):
if str(i).count("13"):
continue
without_13 += 1
return without_13
for N in range(1, 7):
print(number_of_strings_without_13(N),
number_of_strings_without_13_bruteforce(N))
They gave the same answers. With bigger N bruteforce is very slow. But for very large N recursive function also gets mush slower. There is a well known solution for that: as we use the value of number_of_strings_without_13 with parameters smaller than N multiple times, we should remember the answers and not recalculate them each time. It's quite simple to do like this:
def number_of_strings_without_13(N, answers=dict()):
if N in answers:
return answers[N]
sum_numbers_with_13 = 0
for p in range(N-1):
if p < 2:
sum_numbers_with_13 += 10**(N-2)
else:
sum_numbers_with_13 += number_of_strings_without_13(p) * 10**(N-p-2)
result = 10**N - sum_numbers_with_13
answers[N] = result
return result
Thanks to L3viathan's comment now it is clear. The logic is beautiful.
Let's assume a(n) is a number of strings of n digits without "13" in it. If we know all the good strings for n-1, we can add one more digit to the left of each string and calculate a(n). As we can combine previous digits with any of 10 new, we will get 10*a(n-1) different strings. But we must subtract the number of strings, which now starts with "13" which we wrongly summed like OK at the previous step. There is a(n-2) of such wrongly added strings. So a(n) = 10*a(n-1) - a(n-2). That is it. Such simple.
What is even more interesting is that this sequence can be calculated without iterations with a formula https://oeis.org/A004189 But practically that doesn't helps much, as the formula requires floating point calculations which will lead to rounding and would not work for big n (will give answer with some mistake).
Nevertheless the original sequence is quite easy to calculate and it doesn't need to store all the previous values, just the last two. So here is the code
def number_of_strings(n):
result = 0
result1 = 99
result2 = 10
if n == 1:
return result2
if n == 2:
return result1
for i in range(3, n+1):
result = 10*result1 - result2
result2 = result1
result1 = result
return result
This one is several orders faster than my previous suggestion. And memory consumption is now just O(n)
P.S. If you run this with Python2, you'd better change range to xrange
This python3 solution meets time and memory requirement of HackerEarth
from functools import lru_cache
mod = 1000000009
#lru_cache(1024)
def ans(n):
if n == 0:
return 1
if n == 1:
return 10
temp1 = ans(n//2)
temp2 = ans(n//2-1)
if (n & 1) == 0:
return (temp1*temp1 - temp2*temp2) % mod
else:
temp3 = ans(n//2 + 1)
return (temp1 * (temp3 - temp2)) % mod
for t in range(int(input())):
n = int(input())
print(ans(n))
I came across this problem on
https://www.hackerearth.com/problem/algorithm/the-unlucky-13-d7aea1ff/
I haven't been able to get the judge to accept my solution(s) in Python but (2) in ANSI C worked just fine.
Straightforward recursive counting of a(n) = 10*a(n-1) - a(n-2) is pretty slow when getting to large numbers but there are several options (one which is not mentioned here yet):
1.) using generating functions:
https://www.wolframalpha.com/input/?i=g%28n%2B1%29%3D10g%28n%29+-g%28n-1%29%2C+g%280%29%3D1%2C+g%281%29%3D10
the powers should be counted using squaring and modulo needs to be inserted cleverly into that and the numbers must be rounded but Python solution was slow for the judge anyway (it took 7s on my laptop and judge needs this to be counted under 1.5s)
2.) using matrices:
the idea is that we can get vector [a(n), a(n-1)] by multiplying vector [a(n-1), a(n-2)] by specific matrix constructed from equation a(n) = 10*a(n-1) - a(n-2)
| a(n) | = | 10 -1 | * | a(n-1) |
| a(n-1) | | 1 0 | | a(n-2) |
and by induction:
| a(n) | = | 10 -1 |^(n-1) * | a(1) |
| a(n-1) | | 1 0 | | a(0) |
the matrix multiplication in 2D should be done via squaring using modulo. It should be hardcoded rather counted via for cycles as it is much faster.
Again this was slow for Python (8s on my laptop) but fast for ANSI C (0.3s)
3.) the solution proposed by Anmol Singh Jaggi above which is the fastest in Python (3s) but the memory consumption for cache is big enough to break memory limits of the judge. Removing cache or limiting it makes the computation very slow.
You are given a string S of length N. The string S consists of digits from 1-9, Consider the string indexing to be 1-based.
You need to divide the string into blocks such that the i block contains the elements from the index((i 1) • X +1) to min(N, (i + X)) (both inclusive). A number is valid if it is formed by choosing exactly one digit from each block and placing the digits in the order of their block
number
This question already has answers here:
Why does integer division yield a float instead of another integer?
(4 answers)
Closed 5 months ago.
I am a complete python beginner and I am trying to solve this problem :
A number is called triangular if it is the sum of the first n positive
integers for some n For example, 10 is triangular because 10 = 1+2+3+4
and 21 is triangular because 21 = 1+2+3+4+5+6. Write a Python program
to find the smallest 6-digit triangular number. Enter it as your
answer below.
I have written this program:
n = 0
trinum = 0
while len(str(trinum)) < 6:
trinum = n*(n+1)/2
n += 1
print(trinum)
And it only works in the python I have installed on my computer if I say while len(str(trinum)) < 8: but it is supposed to be while len(str(trinum)) < 6:. So I went to http://www.skulpt.org/ and ran my code there and it gave me the right answer with while len(str(trinum)) < 6: like it's supposed to. But it doesn't work with 6 with the python i have installed on my computer. Does anyone have any idea what's going on?
Short Answer
In Python 3, division is always floating point division. So on the first pass you get something like str(trinum) == '0.5'. Which isn't what you want.
You're looking for integer division. The operator for that is //.
Long Answer
The division operator changed in Python 2.x to 3.x. Previously, the type of the result was dependent on the arguments. So 1/2 does integer division, but 1./2 does floating point division.
To clean this up, a new operator was introduced: //. This operator will always do integer division.
So in Python 3.x, this expression (4 * 5)/2 is equal to 10.0. Note that this number is less than 100, but it has 4 characters in it.
If instead, we did (4*5)//2, we would get the integer 10 back. Which would allow your condition to hold true.
In Python 2, the / operator performs integer division when possible: "x divided by y is a remainder b," throwing away the "b" (use the % operator to find "b"). In Python 3, the / operator always performs float division: "x divided by y is a.fgh." Get integer division in Python 3 with the // operator.
You have two problems here, that combine to give you the wrong answer.
The first problem is that you're using /, which means integer division in Python 2 (and the almost-Python language that Skulpt implements), but float division in Python 3. So, when you run it on your local machine with Python 3, you're going to get floating point numbers.
The second problem is that you're not checking for "under 6 digits" you're checking for "under 6 characters long". For positive integers, those are the same thing, but for floats, say, 1035.5 is only 4 digits, but it's 6 characters. So you exit early.
If you solve either problem, it will work, at least most of the time. But you really should solve both.
So:
n = 0
trinum = 0
while trinum < 10**6: # note comparing numbers, not string length
trinum = n*(n+1)//2 # note // instead of /
n += 1
print(trinum)
The first problem is fixed by using //, which always means integer division, instead of /, which means different things in different Python versions.
The second problem is fixed by comparing the number as a number to 10**6 (that is, 10 to the 6th power, which means 1 with 6 zeros, or 1000000) instead of comparing its length as a string to 6.
Taking Malik Brahimi's answer further:
from itertools import *
print(next(dropwhile(lambda n: n <= 99999, accumulate(count(1))))
count(1) is all the numbers from 1 to infinity.
accumulate(count(1)) is all the running totals of those numbers.
dropwhile(…) is skipping the initial running totals until we reach 100000, then all the rest of them.
next(…) is the next one after the ones we skipped.
Of course you could argue that a 1-liner that takes 4 lines to describe to a novice isn't as good as a 4-liner that doesn't need any explanation. :)
(Also, the dropwhile is a bit ugly. Most uses of it in Python are. In a language like Haskell, where you can write that predicate with operator sectioning instead of a lambda, like (<= 99999), it's a different story.)
The division method in Py2.x and 3.x is different - so that is probably why you had issues.
Just another suggestion - which doesn't deal with divisions and lengths - so less buggy in general. Plus addition is addition anywhere.
trinum = 0
idx =0
while trinum < 99999: #largest 5 digit number
idx += 1
trinum += idx
print trinum
import itertools # to get the count function
n, c = 0, itertools.count(1) # start at zero
while n <= 99999:
n = n + next(c)
Ive been give a task, it is as follows:
write a function called decToBin that takes in an integer and converts it to an 8-bit binary number represented as a string
As I am new to this im very lost! Having no introduction to my task as thrown me off a little and I really need some help!
I have tried the following code:
#function
def decTobin(integer)
return bin
#main program
decToBin(3)
decToBin(4)
decToBin(5)
However I had no sucess, could someone point me in the right direction, it would be much appreciated, thank you!
Please try to keep your questions tidy. Also, judging from your other questions, you should look at some basic python tutorials. Happy coding!
Try to learn about base conversions. Here is a great place to find a step by step walkthrough for doing it manually.
You will need to use the modulo (%) operator. The modulo operator is a binary operator, meaning it has two inputs. You use it like so:
a % b
It returns the remainder when a is divided by b:
10 % 7 = 3
The following code will do what you need:
def decToBin(x):
if x == 0:
return "00000000"
bits = []
while x:
bits.append(str(x % 2))
x >>= 1
return "".join(bits).zfill(8)
I will explain line by line.
def decToBin(x):
This declares the function.
if x == 0:
return "00000000"
This returns a string of eight zeros if the input is zero. We need this because the while loop only operates when x is not equal to zero.
bits = []
This initializes the array of bits. During the while loop, we will add to this with the append function.
while x:
This begins a while loop, which runs until x is false (or zero).
bits.append(str(x % 2))
This adds to the bits array the remainder when x is divided by 2. str() converts it to a string.
x >>= 1
>>= 1 Shifts the bits in x to the right one time like so:
Before: 1 1 0 1 0 1
After: 0 1 1 0 1 0
It is the same as dividing by 2, without keeping the remainder.
return "".join(bits).zfill(8)
Breakdown:
"abc".join(l)
Joins all the strings in list l, separating it with abc.
"2345".zfill(i)
adds zeros to the beginning of a string until there are i numbers. So
return "".join(bits).zfill(8)
returns the array of bits as one string, and pads the beginning until there are eight characters.