The answer that fits the blank is XOR or Exclusive OR. Exclusive OR is a logical operation wherein it demonstrates the logic presented by "either/or". This means that if the output given is true, either of the inputs, but never both, should also be true.
A string parameter and returns new string with all the letters of the alphabet that are not in the argument string. The letters in the returned string should be in alphabetical order. The implementation should uses a histogram from the histogram function
Explanation:
The below code is written in python :
alphabet = "abcdefghijklmnopqrstuvwxyz"
test_dups = ["zzz","dog","bookkeeper","subdermatoglyphic","subdermatoglyphics"]
test_miss = ["zzz","subdermatoglyphic","the quick brown fox jumps over the lazy dog"]
def histogram(s):
d = dict()
for c in s:
if c not in d:
d[c] = 1
else:
d[c] += 1
return d
def has_duplicates(s):
for v in histogram(s).values():
if v > 1:
return True
return False
def test_dups_loop():
for s in test_dups:
print(s + ':', has_duplicates(s))
def missing_letters(s):
r = list('abcdefghijklmnopqrstuvwxyz')
s = s.lower()
for c in s.lower():
if c in r:
r.remove(c) # the first matching instance
return ''.join(r)
def test_miss_loop():
for s in test_miss:
print(s + ':', missing_letters(s))
def main():
test_dups_loop()
test_miss_loop()
if __name__ == '__main__':
main()
The answer is yes I hope this help ya out
Answer:
Explanation:
The following code is written in Java. Both functions traverse the linkedlist, until it reaches the desired index and either returns that value or deletes it. If no value is found the function terminates.
public int GetNth(int index)
{
Node current = head;
int count = 0;
while (current != null)
{
if (count == index)
return current.data;
count++;
current = current.next;
}
assert (false);
return 0;
}
public int removeNth(int index)
{
Node current = head;
int count = 0;
while (current != null)
{
if (count == index)
return current.remove;
count++;
current = current.next;
}
assert (false);
return 0;
}
