370 midterm notes
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cc=javac
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cc=javac
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fxlib=--module-path /home/shockrah/Downloads/javafx-sdk-11.0.2/lib
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ctrl=--add-modules javafx.controls
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env=java
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env=java
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cfile="adsf"
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#fxlib=--module-path /home/shockrah/Downloads/javafx-sdk-11.0.2/lib
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#ctrl=--add-modules javafx.controls
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jfile="Main.java"
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# Target class file to run(no extension)
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cfile="Main"
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default: %.java
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default: %.java
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# takes a java file as entry to build
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# takes a java file as entry to build
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@ -21,7 +21,12 @@ Time: O(VlogV + E)
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## Dijkstra's
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## Dijkstra's
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O(V)
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O(V^2 + E)
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## A\*
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## A\*
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O(V^2 = E)
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Worst case is the same as Dijkstra's time
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O(V^2 + E)
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@ -7,7 +7,7 @@ _time complexity will be in terms of height_
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* impl: _root will typically have the largest height value_
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* impl: _root will typically have the largest height value_
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> Balance:
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> Balance:
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| left_height - right_height |
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left_height - right_height
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we can discard the || if we want negative balance vals but it really shouldn't matter
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we can discard the || if we want negative balance vals but it really shouldn't matter
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basically: we want the balance for each node to be 1 or 0.
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basically: we want the balance for each node to be 1 or 0.
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17
370/notes/single-source.md
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370/notes/single-source.md
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# Single Source Shortest Path
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# Bellman-ford Algorithm
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Time: O(VE)
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# Floyd-Warshall
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Space: O(V^2)
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That space is because we're using a matrix to store the paths, which of course is going to take up two dimensions
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Main idea: Shortest path between any two nodes in a graph w/ V nodes _will_ go through at most V nodes
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Iterative idea: Path's can visit i intermediary node. Does that many any path shorter?
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Advantages: we can find negative cycles(Cycle where there is a negative edge)
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53
370/past-midterms/2.md
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53
370/past-midterms/2.md
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# Time/space complexity
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For basically everthing, just list it out somewhere on sheet
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# AVL Tree's
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Get some functions for them written out
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Insert/Deletion/Lookup Functions
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# No pathfinding code
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_mainly cuz we never did any for homework/class_
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Time/space complexities should be straightforward memorization
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# Tries
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# Huffman
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Decoding:
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: Time - O(n) {n = number of input bits}
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Encoding
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: Time O(nlog(n))
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: Where n is the number of unique bits in the string
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# Pathfinding - Conceptual things
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Admissability: allowed to underestimate but not overestimate.
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> Heuristic?
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* A-star
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* Best-first
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> Multiple sources
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* Floyd's Algorithm
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> Multiple destination
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* Floyd's
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* Bellman-Ford
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> Negative edges
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* Floyd
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* Bellman-Ford
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> Negative Cycles
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None
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96
370/past-midterms/2.py
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370/past-midterms/2.py
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class HuffmanNode:
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# $ will represent a dummy node
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# anything else is a regular node
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def __init__(self, data, freq):
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self.data = data
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self.freq = freq
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class HuffmanTree:
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def __init__(self, node):
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self.root = node
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def frequencyMap(self, string):
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ret = {}
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for i in string:
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if i not in ret:
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ret[i] = Node(i, 1)
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else:
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ret[i].freq += 1
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return ret
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def encode(self, string):
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f = self.frequencyMap(string)
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class TrieNode:
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def __init__(self):
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# 1 ref for each character in the alphabet
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self.isleaf = False
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self.edges = [0] * 26
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class TrieTree:
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def __init__(self, root):
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self.root = root
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# just not going to bother with capitals
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def _charIndex(self, char):
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return ord(char) - ord('a')
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def insert(self, string):
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curr = self.root
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for i in string:
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alphabetIndex = _charIndex(i)
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if curr[alphabetIndex] is None:
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curr.edges[alphabetIndex] = Node()
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# go to the nexxt thing
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curr = curr.edges[alphabetIndex]
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else:
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# Finally mark the last node as a leaf
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curr.isleaf = True
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def lookup(self, string):
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curr = self.root
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# First we should go straight to the end of the string
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# That's why we don't do any checks
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for i in string:
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alphaIndex = _charIndex(i)
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curr = curr.edges[alphaIndex]
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return curr.isleaf
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def remove(self, string):
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pass
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def traverse(self, node, preFix):
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if node:
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if n.isleaf:
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print(prefix)
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# recursively go through the tree
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for i in prefix:
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traverse(node.edges[_i], prefix+_charIndex(i))
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class AVLNode:
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def __init__(self, data):
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self.data = data
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self.height = 0
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self.left = None
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self.Right = None
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class AVLTree:
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def __init__(self,node):
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self.root = node
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def rotateLeft(self, node):
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tmp = node.right
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node.left = tmp.right
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tmp.right = node
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# Set the heights on each node
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node.height = 1 + max(node.left.height, node.right.height)
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tmp.height = 1 + max(tmp.left.height, tmp.right.height)
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return tmp
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def rotateRight(self, node):
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pass
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BIN
370/past-midterms/midterm2.pdf
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BIN
370/past-midterms/midterm2.pdf
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370/past-midterms/tmp.py
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class Trie:
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def __init__(self, *words):
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self.root = {}
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for word in words:
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self.insert(word)
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def insert(self, word):
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curr = self.root
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for c in word:
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if c not in curr:
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curr[c] = {}
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else:
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curr = curr[c]
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# add a marker to show that we are a leaf
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curr['.'] = '.'
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self.root = curr
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def remove(self, word):
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pass
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def printWord(self, word):
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pass
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def all(self):
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pass
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59
370/past-midterms/trie.py
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370/past-midterms/trie.py
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class Node:
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def __init__(self, leaf=False):
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# refs list of next valid characters
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self.letters = [None] * 26
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self.isLeaf = leaf
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class Trie:
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def __init__(self):
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self.root = Node()
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def _charIndex(self, char):
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return ord(char) - ord('a')
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def insert(self, string):
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curr = self.root
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for i in string:
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refIndex = self._charIndex()
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# create nodes if we have to
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if curr.letters[refIndex] is None:
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curr.letters[refIndex] = Node()
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curr = curr.letters[refIndex]
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# Set the last letter to be a leaf
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curr.isLeaf = True
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def remove(self, string):
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# Lazy removal
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curr = self.root
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for i in string:
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refIndex = self._charIndex(i)
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if curr.letters[refIndex] is None:
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break
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curr = curr.letters[refIndex]
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curr.isLeaf = False
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# We're also going to be printing things as we go along the thing
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def lookup(self, string):
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curr = self.root
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for i in string:
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refIndex = self._charIndex(i)
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# Make sure we don't walk into nothing
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if curr.letters[refIndex] is None:
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return False
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curr = curr.letters[refIndex]
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print(i, end='')
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return curr.isLeaf
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def all(self, node, word):
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if node is None:
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return
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if node.isLeaf:
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# Print out the word we have so far
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self.lookup(word)
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