CS354: Operating Systems

CS252: Systems Programming
Ninghui Li
Program Interview Questions
String Problem
Input is a string containing a set of words
separated by white space(s). Write a function
that will transform it to a string in which the
words appear in the reverse order. For
example "this is really good" becomes "good
really is this".
• We require that only a constant amount
of extra space, and linear time is used.
String Problem Solution
Solution: two-pass processing
• First pass reverse whole string
• Second pass, find each word and reverse
it locally
Similar Problem
Consider strings over alphabet {a,b,c,d},
write a function that takes an input string, and
(1) removes each “b”, (2) replaces each “a”
by “dd”, and does so in place (assume the
string is in a buffer long enough for the result)
using constant additional storage.
Linked List: Deletion from a
Singly Linked List
Given a singly linked list, you are given one
node and are asked to delete it from the list.
• You are asked to do it in constant time.
• What is the challenge?
• This is possible when the node is not the
last node in the list, and copying the data
field in the node takes constant time.
• How?
Linked List: Checking for
Although a linked list is supposed to end
with NULL, it is possible to have it points to
another element in the list, creating a cycle
Write code to check whether a singly linked
list ends in NULL or contains a cycle
• C1: no modifying the list.
• C2: constant extra space.
• C3: linear time running time
• How to find the first element of the cycle?
Linked List: Checking for
Solution 1: Satisfies C2, C3; violate C1;
• Remember the head, reverse the linked
list as one traverses it, and report cycle if
one reaches the head node again.
Solution 2: Satisfies C2, C3; violates C1;
assumes that there is a bit in the data structure
that is initialized to 0
Linked List: Checking for
Solution 3: Satisfies C1, C3; violate C2;
• Record all visiting nodes in a hash table,
and do a lookup for each newly
encountered node.
Solution 4: Satisfies C1, C2; violates C3;
• For each newly encountered node, check
whether its next will be visited if one
starts from head again and try to traverse
to current node.
Linked List: Checking for
Solution 5: Satisfies C1, C2, C3;
Uses 2 pointers. One fast and one slow. Fast
one advances two steps each time, and show
one advances one step each time, they meet if
and only if there is a cycle.
To figure out where cycle starts,
• First figure out cycle length X, and pinning
fast pointer, and count how many steps it
takes slow pointer to meet fast
• Then set fast pointer at position X, and slow
at beginning, advance both at 1 step at a
time, when they meet, slow is pointing at
beginning of cycle
Additional LinkedList Problem
How to find the middle of a linked list in
one pass of the linked list?
Tower of Hanoi
Move n rings stacked on Peg 1 (rings are
from small to large from top to bottom) to
Peg 2,
• Peg 3 is empty.
Write a program to output the necessary
steps to move n rings on Peg 1 to Peg 2.
Implementing a Queue with two
Queue (FIFO) supports two operation
• enqueue (add an element)
• dequeue (remove the first added element)
Stack (LIFO) supports two operations
• push (add an element)
• pop (remove the last added element)
How to implement a queue using two
Implementing a Queue with two
Enqueue(e) {
if (stack2.isEmpty()) {
while(! stack1.isEmpty()) {
return stack2.pop();
Implement a Reentrant (or
Recursive) Lock
• If a thread currently holding the lock tries
to lock it again, for a non-reentrant or nonrecursive lock, we have a deadlock
• A reentrant (or recursive) lock allows the
thread to proceed; the thread must unlock it
correct number of times for other process
to obtain the lock
Big Data with Limited Memory
• Given an input file with four billion nonnegative integers, provide an algorithm to
generate an integer which is not contained
in the file. Assume you have 1GB of
memory available for this task.
• Use a bit vector, each bit to represent whether
an integer is present; this requires 2^31 bit, or
about 256M bytes.
Taken from Gayle Laakmann McDowell: “Cracking the Coding
Interview”, 5th Edition.
Big Data with Memory Limit
• What if you have only 10MB of memory?
Assume that for input you have no more than one
billion non-negative integers, and they are all
First scan determines # of integers in each thousands,
e.g., between 0 and 1023, 1024 and 2047, etc.
Find a region that is not full, second scan finds a
number in that region
• What if the integers are not guaranteed to be
Taken from Gayle Laakmann McDowell: “Cracking the Coding
Interview”, 5th Edition.
Hash Table
How to design a hash function for words in a
dictionary, e.g., for use in a hash table of
size N?
What makes a good hash function?
• Distribution of output is close of uniform over
• Is sum of all characters a good hash function?
• Is product of all characters a good hash
• Two words are anagrams if they contain
the same set of characters.
• Write a program to partition a dictionary
into sets of words such that all words in
each set are anagrams.
Divide and Conquer
Divide: break down the problem into two or
more sub problems:
Conquer: solve the sub problems, and then
combine their results
Example: Merge sort
Divide and Conquer Question
Compute the number of reversed pairs in an array A of
numbers. A pair (i,j) is reversed if A[i]>A[j].
Solution idea (do merge sort while computing this):
# of inverse in whole =
# of inverse in left half
+ # of inverse in right half
+ # of inverse involving one element in left and one in
right (can be computed while merging left and right)
Code available
Divide and Conquer Question
Diameter of a tree
Consider a tree where each edge has a number
(distance), compute the maximal distance between
two leaf nodes.
Solution Idea:
For each internal node, computes the max distance
as well as the max height
Pseudo Code
max_dist(node *p, &dt, &ht) { // dt:distance, ht:height
int cd, ch, ht2=0, nht; // ht2 stores second highest height
if (p is leaf) return;
for each child c of p {
max_dist(c, &cd, &ch);
if (cd>*dt) *dt=cd;
// current max distance is from subtree at c
nht = ch + e(c);
// e(c) is distance on the edge to child c
if (nht > *ht) { ht2 = *ht; *ht = nht;}
else if (nht > ht2) { ht2 = nht; }
if (ht2>0 && *ht+ht2>*dt)
{ *dt = *ht+ht2; }
// current max distance from leafs in two subtrees

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