GO Classes DPP | GATE DA | Python & DSA | Mutating Iteration

Answer 1

Initially:

L1 = [1, 2, 3, 4]
L2 = [1, 2, 5, 6]

The loop starts with $\texttt{e = 1}$. Since $1$ is present in $\texttt{L2}$, it is removed from $\texttt{L1}$.

L1 = [2, 3, 4]

Now the important point is that the list is being changed while the loop is running. Python's loop moves to the next index, so it skips $2$.

Therefore, $2$ is not removed.

Final output:

[2, 3, 4]

Answer 2

Edit: The metaphor loop in the second code is better written as 

i = 0
while i < len(L1):
    e = L1[i]
    i += 1

    if e in L2:
        L1.remove(e)

Answer: B

Answer 3

This can be a reason of skipping the element :-

In python everthing is a object and every variable refernce some object

so now,

let say we have a list L

if we remove the list in which we are traversing so because of the referancing we unkowingly vhange the size of the list so it feels like we skip the element but internally it happens becasuse we changed the no of element in the list which cause the skipping

Ex-  

L=[1,2,2,3,4]

we are traversing on the list and let say we want to remove the element 2

index = 0

it compares the list L[0] = 1

nothing happens

list is L=[1,2,2,3,4] [becsue we have the referance of the list]

goes to next index

 

index = 1

it compares list L[1] = 2

removes the element

now the list is L=[1,2,3,4]

goes to next index

 

index = 2

L=[1,2,3,4]

in this list L[2] = 3

goes to next index

 

index = 3

L=[1,2,3,4]

index 3 L[3] = 4

nothing happens

goes to next index

 

condition false "terminate the loop"

 

Correct me if i am wrong in here

thank you 😄