Write a statement to create an instance of a class Dictionary that implements DictionaryInterface. Your dictionary contains Word objects that are lists of synonyms for words. The key value is a String representing the word to look up. Call your dictionary thesaurus. DictionaryInterface thesaurus = new DictionaryInterface<>(); Word> thesaurus = new Dictionary<>(); DictionaryInterface thesaurus = new DictionaryInterface<>(); DictionaryInterface thesaurus = new Dictionary<>();
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- Assume that, you have a Linked List class "LL" with a function named "fun1". You can assume that "value" is the value of the nodes and "Next" is the next pointer of the node. You created an instance of the LL class, named "L1", where "head" is the head node. And you inserted the following numbers: 31 -> 32 -> 43 -> 17 -> 34 -> 57 ->91. What will be the output of the linked list L1 after the "fun1" operation to it?* 31 -> 32 -> 43 -> 17 -> 34 -> 57 ->91 31 -> 32 -> 17 -> 34 -> 57 ->91 31 -> 32 -> 43 -> 17 -> 57 ->91 31 -> 32 -> 43 -> 17 -> 34 -> 57 31 -> 32 -> 43 -> 57 ->91 31 -> 32 -> 43 -> 34 -> 57 ->91A map is a container that stores a collection of ordered pairs, each pair consists of a key and a value, <key, value>. Keys must be unique. Values need not be unique, so several keys can map to the same values. The pairs in the map are sorted based on keys. Group of answer choices True FalseConsider the following class that inherits from the built-in list class: class BoundedList (list): def _init_(self, bound): list._init_(self) self.bound = bound Suppose that we want this class to act like a list, except that when append is called, if there are self.bound items in the list already, the first item in the list gets removed. Which of the following methods will properly overload append to work this way? def append(self, item): if len(self) >= self.bound: self.pop(0) list.append(self, item) def append (self, item): if len(self) >= self.bound: self.remove (0) list. append (self, item) def append(self, item): if len(self) >= self.bound: self.pop(0) self.append (item) def append(self, item): list. append (self,item) if len(self) > self.bound: self.pop(0)
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