For this project, you will be given five technical interviewing questions on a variety of topics discussed in the technical interviewing course. You should write up a clean and efficient answer in Python, as well as a text explanation of the efficiency of your code and your design choices. A qualified reviewer will look over your answer and give you feedback on anything that might be awesome or lacking—is your solution the most efficient one possible? Are you doing a good job of explaining your thoughts? Is your code elegant and easy to read?

# Question 1¶

Given two strings s and t, determine whether some anagram of t is a substring of s. For example: if s = "udacity" and t = "ad", then the function returns True. Your function definition should look like: question1(s, t) and return a boolean True or False.

In [29]:
# issubset() is used to check whether one set is a subset of the other set
def question1(s,t):
return set(list(t)).issubset(list(s)) # Complexity Class O(len(t))

s1 = 'udacity'
s2 = 'udacity'
t2 = 'da'
s3 = 'udacity'
t3 = 'xyz'

print question1(s1,t1), question1(s2,t2), question1(s3,t3)

True True False


Case1 : t1 is a subset of s1

Case2 : t2 is a subset of s2

Case3 : t3 is a subset of s3

Case4 : t4 is not a subset of s4

# Question 2¶

Given a string a, find the longest palindromic substring contained in a. Your function definition should look like question2(a), and return a string.

In [27]:
"""
Question 2
Given a string a, find the longest palindromic substring contained in a.
Your function definition should look like question2(a), and return a string.
"""
# @param {string} s input string
# @return {bool} if string is palindrome or not
def isPalindrome(s):
if not s:
return False
# reverse compare
return s == s[::-1]

# @param {string} s input string
# @return {string} the longest palindromic substring
def question2(s):
if not s:
return ""

n = len(s)
longest, low, high = 0, 0, 0
for i in xrange(0, n):
for j in xrange(i + 1, n + 1):
substr = s[i:j]
if isPalindrome(substr) and len(substr) > longest:
longest = len(substr)
low, high = i, j
# construct longest substr
result = s[low:high]
return result

print question2("aaabc")
print question2("forgeeksskeegfor")
print question2(None)

aaa
geeksskeeg



# Question 3¶

Given an undirected graph G, find the minimum spanning tree within G. A minimum spanning tree connects all vertices in a graph with the smallest possible total weight of edges. Your function should take in and return an adjacency list structured like this:

{'A': [('B', 2)],
'B': [('A', 2), ('C', 5)],
'C': [('B', 5)]}

Vertices are represented as unique strings. The function definition should be question3(G)

In [44]:
"""
Gained intuitions from these videos:
"""
parent = {}
rank = {}

# initialize disjoint sets. each set contains one vertice. rank is used to keep the
# tree MST flat as much as possible for faster search.
def make_set(vertice):
parent[vertice] = vertice
rank[vertice] = 0

# find the set to which this vertice belongs
def find(vertice):
if parent[vertice] != vertice:
parent[vertice] = find(parent[vertice])
return parent[vertice]

# merge the sets represented by these two given root nodes
def union(vertice1, vertice2):
root1 = find(vertice1)
root2 = find(vertice2)
if root1 != root2:
if rank[root1] > rank[root2]:
parent[root2] = root1
else:
parent[root1] = root2
if rank[root1] == rank[root2]: rank[root2] += 1

# perform kruskal algorithm to find mst
def kruskal(vertices, edges):
minimum_spanning_tree = set()
for vertice in vertices:
make_set(vertice)

# sort edges by increasing weights
edges = sorted(edges, key=lambda x : x[2])

for edge in edges:
vertice1, vertice2, wt = edge
if find(vertice1) != find(vertice2):
union(vertice1, vertice2)

return minimum_spanning_tree

# main
def question3(G):

graph = G
vertices = []
edges = []

# pre process given input graph and extract all vertices and edges
for vertice in graph.keys():
# collect vertices
vertices.append(vertice)
# build edge tuples
verticeEdges = graph[vertice]
for verticeEdge in verticeEdges:
fromNode = vertice
toNode, weight = verticeEdge
edges.append((fromNode, toNode, weight))

# perform Kruskal algo
ms_tree = kruskal(vertices, edges)

# post process results into the required output format
output = {}
for node in ms_tree:
fromNode, toNode, weight = node

if toNode < fromNode:
fromNode = node[1]
toNode = node[0]

if fromNode in output:
output[fromNode].append((toNode, weight))
else:
output[fromNode] = [(toNode, weight)]

return output

print question3({})
print question3({'A': [('B', 2)],
'B': [('A', 2), ('C', 5)],
'C': [('B', 5)]})
print question3({'A': [('B', 2), ('C', 7)],
'B': [('A', 1), ('C', 5), ('D', 3), ('E', 4)],
'C': [('C', 7), ('D', 7), ('E', 6)],
'D': [('B', 3), ('C', 4), ('E', 2)],
'E': [('B', 4), ('D', 5)],
})

{}
{'A': [('B', 2)], 'B': [('C', 5)]}
{'A': [('B', 1)], 'C': [('D', 4)], 'B': [('D', 3)], 'D': [('E', 2)]}


# Question 4¶

Find the least common ancestor between two nodes on a binary search tree. The least common ancestor is the farthest node from the root that is an ancestor of both nodes. For example, the root is a common ancestor of all nodes on the tree, but if both nodes are descendents of the root's left child, then that left child might be the lowest common ancestor. You can assume that both nodes are in the tree, and the tree itself adheres to all BST properties. The function definition should look like question4(T, r, n1, n2), where T is the tree represented as a matrix, where the index of the list is equal to the integer stored in that node and a 1 represents a child node, r is a non-negative integer representing the root, and n1 and n2 are non-negative integers representing the two nodes in no particular order. For example, one test case might be

question4([[0, 1, 0, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 0],
[1, 0, 0, 0, 1],
[0, 0, 0, 0, 0]],
3,
1,
4)

and the answer would be 3.

In [53]:
head = None

class Node(object):
def __init__(self, data):
self.data = data
self.right = None
self.left = None

# Function to insert a new node at the beginning
def push_right(node, new_data):
new_node = Node(new_data)
node.right = new_node
return new_node

# Function to insert a new node at the beginning
def push_left(node, new_data):
new_node = Node(new_data)
node.left = new_node
return new_node

# Function to find LCA of n1 and n2. The function assumes
# that both n1 and n2 are present in BST
# Base Case
return None

# If both n1 and n2 are smaller than root, then LCA
# lies in left

# If both n1 and n2 are greater than root, then LCA
# lies in right

def question4(mat, root, n1, n2):
# Make BST
node_value = 0
tmp_right, tmp_left = None, None
node_list = []
for elem in mat[root]:
if elem:
if(node_value>root):
else:
node_value += 1

tmp_node = node_list.pop(0)
while tmp_node != None:
node_value = 0
for elem in mat[tmp_node.data]:
if elem:
if(node_value>tmp_node.data):
node_list.append(push_right(tmp_node, node_value))
else:
node_list.append(push_left(tmp_node, node_value))
node_value += 1
if node_list == []:
break
else:
tmp_node = node_list.pop(0)

# Main program
print question4([[0, 1, 0, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 0],
[1, 0, 0, 0, 1],
[0, 0, 0, 0, 0]],
3,
1,
4)

3


# Question 5¶

Find the element in a singly linked list that's m elements from the end. For example, if a linked list has 5 elements, the 3rd element from the end is the 3rd element. The function definition should look like question5(ll, m), where ll is the first node of a linked list and m is the "mth number from the end". You should copy/paste the Node class below to use as a representation of a node in the linked list. Return the value of the node at that position.

class Node(object):
def __init__(self, data):
self.data = data
self.next = None
In [22]:
head = None

class Node(object):
def __init__(self, data):
self.data = data
self.next = None

# Function to insert a new node at the beginning
def push(new_data):
new_node = Node(new_data)

count  = 0

while(count < n ):
if(ref_ptr is None):
print "%d is greater than the no. of nodes in list" %(n)
return

ref_ptr = ref_ptr.next
count += 1

while(ref_ptr is not None):
main_ptr = main_ptr.next
ref_ptr = ref_ptr.next

return main_ptr.data

# Main program
def main():

20