In binary search, for a sorted list, we start from the middle of the list and then check if the middle word is less than, equal to or greater than the search query string.
Based on the result we can generally halve the size of the search, and so recursively solve the problem.
eng2sp = {}
eng2sp["one"] = "uno"
eng2sp["two"] = "dos"
sp2eng = {"uno":"one", "dos":"two"}
x = set(["Internet", "Radio", "TV"])
print(x)
# Prints {'Internet', 'TV', 'Radio'}
y = {"Internet","Radio","Internet"}
print(y)
# Prints {'Internet', 'Radio'}