How do I handle cases with numbers or the other cases?

Question

I am doing a code coach, and I have passed half the cases, but there seem to be cases with digits, as outlined in the brief.

Here's my attempt.

str_input = input().split(" ")

def mutate(x):
start = 0
end = 1

while start < len(x):
i = 0
while i < len(x):
yield x[start:end]
end += 1
i += 1
start += 1

mutations = []

for i in str_input:
for j in mutate(i):
mutations.append(j)

commons = []

for i in mutations:
common = set([x for x in str_input if(i in x)])
if len(common) == len(str_input) and len(i) > 2:
commons.append(i)

res = [x for x in commons if commons.count(x) > 1]

res = list(sorted(res, key=str.lower))

res = set(res)

res = list(res)

if len(res) > 1:
res = list(sorted(res, key=len))
res = list(sorted(res, key=str.lower))
print(res[0])
else:
print(res[0])

Answer 1

the code coach is: *longest common substring*
^^^^^

Answer 2

some thoughts from my side:

> using permutations for this task does create too many unnecessary samples, since it creates all possible *combinations*.

> what we need to use are only samples that have partially the same sequence in all strings.

> using this strings ['lean', 'isle', 'clean'] to find the longest common substring is giving the result: 'le'.
^^ ^^ ^^ ^^
> a recommended way to get this done is using 2 nested for loops with a *sliding and sizeable window* that can cut the chunks for the comparison.
each of these samples has to be taken and checked if it is in the supplied strings.

here a sample code that should work:

https://code.sololearn.com/cPB4r3J6SiMC/?ref=app

Answer 3

Which task is it you're trying to solve?

Answer 4

I think you are complicating the simple task even for large inputs..!

Some observation!
do you don't need to reset end value in inner while loop like start ?

for i in str_input : why to all list of string to mutate?
str_input[0] is not enough, first string??
Hope it helps, 🤔not confuse instead...

Answer 5

What's your first for loop intention?
And why you are applying set on for 'commons'...? I think it's leading to wrong output.. if am understood correctly :
'aba' should be compared totally, applying set makes compare list of 'ab' only.. .
Taking all individual character from entire input is no need.

Simple approach:
compare substring from word1 in other words just and take max length matched string.

Answer 6

sry, iam not able to understand your approach, why permutations are needed there in the task..!

Ok. I hope this sample may helps you find mistake by your own. I tested sample as
12 123 1234
It should output common substring as 12 but return empty 'res'. So res[0] is an error.

Answer 7

Once try :
if len(common) == len(str_input) :
Instead of
if len(common) == len(str_input) and len(i) > 2:

Answer 8

William Mabotja
Last one i guess, that you are sorting by len so it will be ascending order. Last one will be longest one, not the first.

Answer 9

So then try another : Don't convert to lower , output should in original input. .

Answer 10

if common substrings are multiple like "xyz", "abcd" then smallest one in alphabetical order is "abcd"..

Save code and share link, that easies debug...

Answer 11

Longest Common Substring

Answer 12

I have to use set on commons because the resulting list includes duplicates. The intention is to create all possible permutations for a string, and then ensure that the permutation occurs in each item in the list.

if it occurs in all items, it is therefore common. Then I proceed to check the list's length, if it is above one. If that's true, I get the smallest item in alphabetical order, and print that out.

I understand what you're saying on the suggested approach, but I'm making room for large inputs, and inverse matches.

Answer 13

It turns out that it's nit numeric data that's the issue, made some changes and got 4/6 cases right. By Moore's law, I could be close.

# Grab input and split to list

str_input = input().split(" ")

def mutate(x):
"""
Creates all possible permutations for a
given string.
"""
start = 0
end = 1

while start < len(x):
i = 0
while i < len(x):
yield x[start:end]
end += 1
i += 1
start += 1

# List to hold all possible permutations

mutations = []

# Run generator function on all list elememts and append to mutations list

for i in str_input:
for j in mutate(i):
mutations.append(j)

# List to hold common petmutations

commons = []

# Find words that occur in each list item

for i in mutations:
common = set([x for x in str_input if(i in x)])
if len(common) == len(str_input) and len(i) > 2:
commons.append(i)

# Remove words that are less than 1 in length

Answer 14

Thanks for the tip, all I had to change was len > 2 to len > 1, and worked for that case, and the other cases still worked. But the solution still fails. I even tried another case from the forum ["abcekde", "acekt", "zaqacekyi"]

And managed to fix that bug too, but the solution still failed. I'll keep checking. Thanks.

Answer 15

I'll try that

Answer 16

It breaks my pseudo code. It's a different code altogether. I've seen comments in the forum about complexity, and pushing the compiler. That could be the issue.

Answer 17

If I don't get it working I'll rewrite that way

Answer 18

https://code.sololearn.com/cosNSDqq4qHa/?ref=app

Answer 19

Thanks Jayakrishna 🇮🇳 , got 5/6 cases correct, left with test case #4

Answer 20

Thanks, but I've already used reversed=True on the sort to make the first item the longest