Mapping digits to their appearances in a list of numbers

I need to take in a sequence of numbers. The last number in this sequence gives a sequence of digits. For each digit, I need to create a list of all the numbers that contain that digit. If no numbers contain that digit, the list should contain 0 rather than an empty list.

Input:

 96 23 43 113 6 
 315

I need output like this:

[[23,43,113],[113],0]

need to store:
 1st element in k is 3 -> [23,43,113]
 2nd element in k is 1 -> [113]
 3rd element in k is 5 -> 0 (No 5 in any element in LIST(l))

I tried the following:

l=list(map(str,input().split()))
k=input()
ans=[[l[j]] for i in range(len(k)) for j in range(0,len(l)) if k[i] in l[j]]
print(ans)  

but received the following output:

[['23'], ['43'], ['113'], ['113'], ['113']]

>Solution :

Why don’t you just do it not in one line, This way it would be more transparent:

goal = []
for char in list(k):
    temp = [element for element in l if char in element]
    temp = temp if temp else 0
    goal.append(temp)

But if you still want your code to be short in one line:

goal = [[element for element in l if char in element] or 0 for char in k]