Peer Instruction: Strings Multiple Choice Questions¶

11-9-1: What is the value of s after the following code runs?

s = "abc"
s = "d" * 3 + s
s = s + "" * 3
s = s + "q"

"abcddd q"
Incorrect! The value of s changes in the following order from the first line- s="abc", s="dddabc", s="dddabc" and s="dddabcq".
"abcddd" "" "" "q"
Incorrect! The value of s changes in the following order from the first line- s="abc", s="dddabc", s="dddabc" and s="dddabcq".
"qdddabc"
Incorrect! The value of s changes in the following order from the first line- s="abc", s="dddabc", s="dddabc" and s="dddabcq".
"dddabcq"
Correct! The value of s changes in the following order from the first line- s="abc", s="dddabc", s="dddabc" and s="dddabcq".
I don't know
Incorrect! The value of s changes in the following order from the first line- s="abc", s="dddabc", s="dddabc" and s="dddabcq".

11-9-2: What is the value of y after the following code runs?

s = "Cats"
y = s[len(s)-1]

C
Incorrect! Here, len(s) = 4 and s[len(s)-1] = s[3]. So, the value of y will be the fourth character in the string i.e. "s".
t
Incorrect! Here, len(s) = 4 and s[len(s)-1] = s[3]. So, the value of y will be the fourth character in the string i.e. "s".
s
Correct! Here, len(s) = 4 and s[len(s)-1] = s[3]. So, the value of y will be the fourth character in the string i.e. "s".
Nothing, this will cause an error
Incorrect! Here, len(s) = 4 and s[len(s)-1] = s[3]. So, the value of y will be the fourth character in the string i.e. "s".
I don't know
Incorrect! Here, len(s) = 4 and s[len(s)-1] = s[3]. So, the value of y will be the fourth character in the string i.e. "s".

11-9-3: What will the following code print?

s = "bats"
print(s[0:len(s)])

11-9-4: What will the following code print?

s = "Vampires"
print(s[3:-1]

mpire
Incorrect! Negative index- "-1" has been used in the slice. So, the slice will end at the second last character.
pires
Incorrect! Negative index- "-1" has been used in the slice. So, the slice will end at the second last character.
pire
Correct! Negative index- "-1" has been used in the slice. So, the slice will end at the second last character.
pir
Incorrect! Negative index- "-1" has been used in the slice. So, the slice will end at the second last character.
I don't know
Incorrect! Negative index- "-1" has been used in the slice. So, the slice will end at the second last character.

11-9-5: What does the following code return?

def mystery(s):
   new_s = ""
   for c in s:
      new_s = c + new_s
   return new_s

Returns a copy of s
Incorrect! For example, take s = "xyz". With each step in the loop, the value of new_s will change in the order- new_s = x, new_s = yx and new_s = zyx.
Returns the reverse of s
Correct! For example, take s = "xyz". With each step in the loop, the value of new_s will change in the order- new_s = x, new_s = yx and new_s = zyx.
Returns a string with only the last character of s
Incorrect! For example, take s = "xyz". With each step in the loop, the value of new_s will change in the order- new_s = x, new_s = yx and new_s = zyx.
Returns a string with only the first character of s
Incorrect! For example, take s = "xyz". With each step in the loop, the value of new_s will change in the order- new_s = x, new_s = yx and new_s = zyx.
I don't know
Incorrect! For example, take s = "xyz". With each step in the loop, the value of new_s will change in the order- new_s = x, new_s = yx and new_s = zyx.

11-9-6: What is the value of s after the following code runs?

s = 'abc'
s = 'd' * 3 + s
s = s + ' ' * 3
s = s + 'q'

"abcddd q"
Incorrect! Here, 'd' * 3 + s = 'ddd' + 'abc' = 'dddabc'. Then ' ' adds three spaces at the end of the string followed by a 'q' at the end.
"abcddd q"
Incorrect! Here, 'd' * 3 + s = 'ddd' + 'abc' = 'dddabc'. Then ' ' adds three spaces at the end of the string followed by a 'q' at the end.
"abcdddq"
Incorrect! Here, 'd' * 3 + s = 'ddd' + 'abc' = 'dddabc'. Then ' ' adds three spaces at the end of the string followed by a 'q' at the end.
"qdddabc"
Incorrect! Here, 'd' * 3 + s = 'ddd' + 'abc' = 'dddabc'. Then ' ' adds three spaces at the end of the string followed by a 'q' at the end.
"dddabc q"
Correct! Here, 'd' * 3 + s = 'ddd' + 'abc' = 'dddabc'. Then ' ' adds three spaces at the end of the string followed by a 'q' at the end.

11-9-7: What is a good description of what the following function does?

def mystery(s):
   new_s = ''
   for c in s:
      new_s = c + new_s
   return new_s

Returns a copy of s
Incorrect! Consider s = 'abc'. In the first iteration, new_s = c + new_s = 'a' + '' = 'a'. In the second iteration, new_s = 'b' + 'a' = 'ba' and so on. So, it will return the reverse of s.
Returns the reverse of s
Correct! Consider s = 'abc'. In the first iteration, new_s = c + new_s = 'a' + '' = 'a'. In the second iteration, new_s = 'b' + 'a' = 'ba' and so on. So, it will return the reverse of s.
Returns a string consisting of only the final character of s
Incorrect! Consider s = 'abc'. In the first iteration, new_s = c + new_s = 'a' + '' = 'a'. In the second iteration, new_s = 'b' + 'a' = 'ba' and so on. So, it will return the reverse of s.
Returns a string consisting of only the first character of s
Incrrect! Consider s = 'abc'. In the first iteration, new_s = c + new_s = 'a' + '' = 'a'. In the second iteration, new_s = 'b' + 'a' = 'ba' and so on. So, it will return the reverse of s.

11-9-8: What is the value of val after this code executes?

val = 0
for i in 'ab':
   for j in 'cd':
      val += 1

1
Incorrect! For each 'a' and 'b', the nested for loop will run twice. Thus, val gets incremented by 1 four times resulting in val = 4.
2
Incorrect! For each 'a' and 'b', the nested for loop will run twice. Thus, val gets incremented by 1 four times resulting in val = 4.
4
Correct! For each 'a' and 'b', the nested for loop will run twice. Thus, val gets incremented by 1 four times resulting in val = 4.
8
Incorrect! For each 'a' and 'b', the nested for loop will run twice. Thus, val gets incremented by 1 four times resulting in val = 4.
16
Incorrect! For each 'a' and 'b', the nested for loop will run twice. Thus, val gets incremented by 1 four times resulting in val = 4.

11-9-9: What is the value of val after this code executes?

val = 0
for i in 'abc':
   for j in 'def':
   val += 1

1
Incorrect! For each 'a', 'b' and 'c', the nested for loop will run thrice. Thus, val gets incremented by 1 nine times resulting in val = 9.
3
Incorrect! For each 'a', 'b' and 'c', the nested for loop will run thrice. Thus, val gets incremented by 1 nine times resulting in val = 9.
6
Incorrect! For each 'a', 'b' and 'c', the nested for loop will run thrice. Thus, val gets incremented by 1 nine times resulting in val = 9.
9
Correct! For each 'a', 'b' and 'c', the nested for loop will run thrice. Thus, val gets incremented by 1 nine times resulting in val = 9.
27
Incorrect! For each 'a', 'b' and 'c', the nested for loop will run thrice. Thus, val gets incremented by 1 nine times resulting in val = 9.

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