Answer:
# Code in Python
dictionary={'A':1,'B':2,'C':3,'D':4}
other_dictionary={}
for keys in dictionary:
if dictionary[keys]&1==1:
temp=dictionary[keys]*dictionary[keys]-10*10
other_dictionary[keys]=temp
else:
other_dictionary[keys]=dictionary[keys]
print(other_dictionary)
assert other_dictionary
Explanation:
- Initialize a sample example dictionary and other_dictionary.
- Do a binary comparision for checking odd number
.
- Update the the value stored in the dictionary to store the squared difference of the original value and '10'.
- For even: store the original value (from dictionary).
Answer:
c) it will always return with an answer of 7 or fewer comparisons of data.
Explanation:
As maximum number of comparison in worst case for sorted binary search is log₂n, here n = 100, so maximum search is at least 6.6 which is approximately 7.
