So, here is the code
Module - 1 (Populating Note)
n = len(l)
note = [None]*(n**2)
for inner in range(len(l)):
if type(l[inner])==list:
for j in range(len(l[inner])):
if type(l[inner][j])==str:
print(f"In list {inner } Index of str element {l[inner][j]} is {j}")
if note[inner]==None:
note[inner] = []
note[inner].append(j)
else:
note[inner].append(j)
Module - 2 (Removing String elements)
for i in range(len(note)):
if note[i]!=None:
count=0
while count < len(note[i]):
j = note[i][count]
#If pop has occured once update pop index to reflect resizing
if count>0:
j-=count
l[i].pop(j)
count+=1
Yup, that's it.
To understand the code in a better fashion , try to break the code in chunks and see the output also use intermediate print statements , that helps.
Please let me know in case I have missed or misplaced any syntax.
NOTE - Take care of the indentation as it has been provided manually only to clear the purpose of what each Line of Code does and where it belongs.
Time Complexity - This is a O(N^2) solution where we have assumed that each nested list will have at most N element where N is the len of the outer list.
Scope of Improvement - Obviously turning the solution into O(n) but I don't think it will be possible as traversing of nested list is concerned.
Hope it helps, please let me know your suggestions.
Till then,
Namaste
