Computer Science
Predict the output :
def h_t(NLst):
from_back = NLst.pop()
from_front = NLst.pop(0)
NLst.append(from_front)
NLst.insert(0, from_back)
NLst1 = [[21, 12], 31]
NLst3 = NLst1.copy()
NLst2 = NLst1
NLst2[-1] = 5
NLst2.insert(1, 6)
h_t(NLst1)
print(NLst1[0], NLst1[-1], len(NLst1))
print(NLst2[0], NLst2[-1], len(NLst2))
print(NLst3[0], NLst3[-1], len(NLst3))
Python
Linear Lists
1 Like
Answer
5 [21, 12] 3
5 [21, 12] 3
[21, 12] 31 2
Working
def h_t(NLst):
— This line defines a function namedh_t
that takes a single argumentNLst
.from_back = NLst.pop()
— This line removes and returns the last element from the listNLst
, storing it in the variablefrom_back
.from_front = NLst.pop(0)
— This line removes and returns the first element from the listNLst
, storing it in the variablefrom_front
.NLst.append(from_front)
— This line appends the value stored infrom_front
to the end of the listNLst
.NLst.insert(0, from_back)
— This line inserts the value stored infrom_back
at the beginning of the listNLst
.NLst1 = [[21, 12], 31]
— This line initializesNLst1
as a nested list.NLst3 = NLst1.copy()
— This line creates a shallow copy ofNLst1
and assigns it toNLst3
.NLst2 = NLst1
— This line assigns the reference ofNLst1
toNLst2
, meaning bothNLst1
andNLst2
point to the same list object.NLst2[-1] = 5
— This line modifies the last element ofNLst2
to 5. SinceNLst1
andNLst2
reference the same list object, this change also affectsNLst1
.NLst2.insert(1, 6)
— This line inserts 6 at index 1 inNLst2
. Again, sinceNLst1
andNLst2
reference the same list object, this change also affectsNLst1
.h_t(NLst1)
— This line calls the functionh_t
withNLst1
as an argument.print(NLst1[0], NLst1[-1], len(NLst1))
— This line prints the first element, last element, and length ofNLst1
.print(NLst2[0], NLst2[-1], len(NLst2))
— This line prints the first element, last element, and length ofNLst2
.print(NLst3[0], NLst3[-1], len(NLst3))
— This line prints the first element, last element, and length ofNLst3
.
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