Algorithms and Data Structures Interview Preparation & Walkthrough — Part 2, Array and String

In my previous post: , we talked about how to do Complexity Time and Space analysis, and also see the common Big-O factors with examples.

In this post, I will start talking about Array in depth, and cover some interview questions and hopefully by the end of the reading, you would have a good glance about Array. Once we are familiar with Array, I will talk about String and how to use Array to solve String problems.

Array is a data structure that contains a group of elements. The most basic implementation of an array is a static array. The reason it’s called static is the size is fixed. The read/write access to a certain position is O(1).

Implementation of Static Array

class StaticArraydef initialize(length)self.store = Array.new(length)end

# O(1)def [](index)self.store[index]end

# O(1)def []=(index, value)self.store[index] = valueend

protectedattr_accessor :storeend

We create a Dynamic Array from Static Array as follow. The read/write access is also O(1). Let’s implement some general methods i.e. pop(), push(), shift() and unshift() for it. The key here is, when we reach the array size, we want to resize it and double its space, in order to push() or unshift() new elements to the array.

Implementation of Dynamic Array

require_relative "static_array"

class DynamicArrayattr_reader :length

def initialize = 0 = 8 = StaticArray.new(8)end

# O(1)def [](index)check_index(index)[index]end

# O(1)def []=(index, value)check_index(index)[index] = valueend

# O(1)def popcheck_index(0) -= 1[length + 1]end

# O(1) amortized; O(n) worst case.def push(val)resize! if == [ + 1] = val += 1end

# O(n): has to shift over all the elements.def shiftcheck_index(0)idx = 0first_el = [0]while idx < - 1[idx] = [idx + 1]idx += 1end -= 1first_elend

# O(n): has to shift over all the elements.def unshift(val)resize! if == idx = while idx > 0[idx] = [idx - 1]idx -= 1end[0] = val += 1end

protectedattr_accessor :capacity, :storeattr_writer :length

def check_index(index)raise "out of bounds" if ( < index + 1 || index < 0)end

# O(n): has to copy over all the elements to the new store.def resize!new_store = StaticArray.new( * 2)idx = 0while idx < new_store[idx] = [idx]idx += 1end = new_store *= 2endend

What is amortization?

If you read carefully enough, you would notice there is a keyword “amortized” in the code snippet. What does that mean? When we want to append (or push) a new element to the Array and it reaches its size limit, we want to double the size. However, resize! method allocates a larger region, moves the whole array, and deletes the previous. This is a O(n) operation. But if we're only doing it every O(1/n) times, then on average it can still come out to O(n * 1/n) = O(1). That’s called amortized cost.

Time Complexity and Space Complexity for Dynamic Array

In average and worst cases,

Access O(1)

Search O(n)

Insertion O(n)(at the end of Array is O(1) amortized, at the beginning or middle of Array is O(n)

Deletion O(n)

Space O(n)

We now know accessing an element in Array is fast (O(1)), whereas searching/adding/removing is relatively slow (O(n)), which sometimes requires looping through the whole array.

Ring Buffer

It is a data structure that uses a Static Array as if it were connected end-to-end.

require_relative "static_array"

class RingBufferattr_reader :length

def initialize = 0 = 8 = 0 = StaticArray.new()end

# O(1)def [](index)check_index(index)ring_index = (index + ) % [ring_index]end

# O(1)def []=(index, val)check_index(index)ring_index = (index + ) % [ring_index] = valend

# O(1)def popcheck_index(0) -= 1;val = [( + ) % ][( + ) % ] = nilvalend

# O(1) amortizeddef push(val)resize! if == [( + ) % ] = val += 1end

# O(1)def shiftcheck_index(0)val = [][] = nil = ( + 1) % -= 1valend

# O(1) amortizeddef unshift(val)resize! if == = ( - 1) % [] = val += 1valend

protectedattr_accessor :capacity, :start_idx, :storeattr_writer :length

def check_index(index)raise "index out of bounds" if (index < 0 || index > - 1)end

def resize!new_store = StaticArray.new( * 2)idx = 0while idx < new_store[idx] = [( + idx) % ]idx += 1end = new_store = 0 *= 2endend

To master Array data structure and questions, we at least need to be very familiar with:

1. Loop operation.
2. Sense of using pointer(s) to record location(s).
3. Swap technique.
4. Basic Math.
5. Common array methods and the time complexity of them, i.e. pop(), push(), shift(), unshift(), forEach(), sort(), slice(), splice(), reverse(), concat(), filter(), map() … etc.

Some advantages of using Array:

● Constant time access and allows random access

● Grouping like items

And some disadvantages…

● Insertion and deletion can be expensive for large arrays

● Dynamic arrays have cost to resize, and limited by the size allocated

Enough said, here are some popular interview questions for your practice:

1. Move Zeros — Given an array of numbers, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements. ()
2. Stocks 101 — When to buy/sell the stock? Given an array contains the daily price for a stock. Try to find the maximum profit. ()
3. Find Duplicates — Given an array of numbers, return true if any number appears more than once in the array, otherwise return false. ()

String

Now we know about Array, let’s talk about String — which is nothing but a Character-based Array. You just need to learn techniques to solve Array questions, and you are a String Master by nature!

Before diving into String related questions, we need to get familiar with:

1. String related methods, i.e. charAt(), includes(), trim(), concat(), slice(), split(), substring(), toUpperCase(), toLowerCase(), toString()…etc.
2. Two pointers.
3. Swap elements within Array.
4. Recursion.

In my , I will talk about Data Structures for Queue, Stack and Linked List.

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