Algorithm Techniques

An algorithm is a well-defined computational procedure that takes in a set of values as inputs and produces a set of values as outputs. An algorithm is thus a sequence of computational steps that transform input into output.

Please see Algorithm Problems for problem explanations.

Nondecreasing order = an ordering where the next value is greater than or equal to the previous. Ex.
For values $x(n)$ and $x(n+1),\ x(n+1)>=x(n)$

Instance = an instance is a set of inputs to an algorithm that satisfy the input the input requirements.

Correct = an algorithm is said to be correct if, for every input instance, the computation halts when the correct output has been calculated.

Convex Hull = the smallest convex polygon that contains all points in a set of numbers that lie on a plane.

NP-complete problem = a problem for which there is currently no efficient algorithm that solves it, neither is there a proof that proves that the algorithm exists.

To prove an algorithm works, one way is by induction.

Start with the loop invariant, which is the subset of the input that is acted upon by the algorithm.

Then check for correctness in each of the following parts of the algorithm:

Initialization - Is it true prior to the first iteration of the loop?
Maintenance - Is it true at the start of the loop and also true at the end of the loop?
Termination - Is it true at the end of the loop?

Analyzing an algorithm is predicting how much resources it will use. You have to model the implemented technology.

Input size - A measure of the input must be selected that has the most significant impact on the resources required for an operation. For instance in a multiplication algorithm the input size should be number of bits used in the input numbers.

Running time - This should be a machine independent number. Number of primitive steps executed.

An algorithm's running time is dependent on the input data and there will be a best case, average case and worst case running time. The average case is often roughly as bad as the worst case. For instance with insertion sort.

Order of growth if the rate of growth of the running time. We consider the leading, most significant term of the formula of the worst case running time, called theta notation.

For an algorithm with a running time of $an^2 + bn + c$ the theta notation would be $Θ(n^2)$

Here's a breakdown of different sorting algorithms. There's sorting algorithms that sort in place, rely on element comparisons, others that don't.

For an unsorted list A of n length, loop through a growing subset of A by comparing the last element of the subset with each previous element and compare. If greater/smaller swap elements and iterate through again through a subset that is 1 larger than the previous.

1. for j = 1 to n
2.     key = A[j]
3.     i = j-1;
4.     while i > 0 and key < A[i]
5.        A[i+1] = A[i]
6.        i--
7.     A[i+1] = key

Find the smaller number and place it in A[1]. Then find smallest number in A[2..n] and place it in A[2]. Repeat

1. for i = 1 to A.length-1
2. 	//find the smallest number in A(i to n)
3. 	cur_smallest_index = i
4.	for j = i+1 to A.length
5.		if A[j] < A[cur_smallest_index]
6.			cur_smallest_index = j
7.	A[i] = temp
8.	A[i] = A[cur_smallest)index]
9.	A[cur_smallest_index] = temp;

Merge sort is an O(n lg n) sorting algorithm that uses the dive and conquer paradigm.

Heap sort is an O(n lg n) sorting algorithm that sorts in place with a constant number of array elements stored outside the array at any time.

Heapsort uses a data structure called a “heap” to manage information.

A heap is a binary-tree that consists of a root, and parent nodes that have up to two child nodes. The math to retreive the the parent, left element or right element of a node is really simple.

Parent - $return\lfloor i/2 \rfloor$
Left - $return 2*i$
Right - $return 2*i+1$

There are two kinds of binary heaps, max-heaps and min-heaps. In both kinds the values in the nodes satisfy a heap property. In max-heap the max-heap property is that for every node $i$ other than the root,

$A[Parent(i)]\geq A[i]$

For min-heap the min-heap property is that for every node $i$ other than the root,

$A[Parent(i)]\leq A[i]$

There are a set of operation algorithms that are used in conjuction with heaps. They are as follows:

• Max-Heapify
• Build-Max-Heap
• Heapsort
• Max-Heap-Insert, Heap-Extract-Max, Heap-Increase-Key, Heap-Maximum

Runs in O(lg n), whose inputs are A and index i into the array. When it is called, Max-Heapify assumes that the binary trees rooted at Left(i) and Right(i) are max-heaps, but that A[i] might be smaller than its children. Max-Heapify lets the A[i] value “float” down so that the subtree rooted at index i obeys the max-heap property.

1. Is index left smaller than A.heap-size and A[l] > A[i]?
   1. If yes, largest = l
2. Else, largest = i
3. Is index right smaller than A.heap-size and A[r] > A[i]?
   1. If yes, largest = r
4. Else, largest = i
5. If largest != i
   1. exchange A[i] with A[largest]
   2. Max-Heapify(A, largest)

Runs in $O(lg\ n)$ time.

Max-Heap Insert. Input: heap array A, with data from 1 to length, element to insert p
   // stuff the element at the end of the heap array
1. A[length+1] = p
   // iterate up through the heap
2. int i = length + 1;
3. while i >= 0
   // check if the parent node of our inserted element is less than it. 
4.    if (i/2 < p)
        // swap the inserted element with its parent
5.      A[i] = A[i/2]
6.      A[i/2] = p
7.      i = i/2
8.    else
9.      break

Build-Max-Heap uses Max-Heapify to a bottom-up manner to convert an array A[1..n], where n = A.length, in a max-heap. This procedure goes through the non-leaf nodes of the tree ($A[\lfloor n/2\rfloor..1]$) and runs Max-Heapify on each one.

Build-Max-Heap(A)
1. A.heap-size = A.length
2. for i = [A.length/2] down to 1
3.  Max-Heapify(A,i)

Heapsort works by building a max heap and then deconstructing it in order which in turn constructs a sorted array! Super neat and works in O(n lg n) time.

The algorithm works as follows:

Heapsort(A)

1. Build-Max-Heap(A)
2. for i = A.length down to 2 
3.  Exchange A[1] with A[i]
4.  A.heap-length -- 
5.  Max-Heapify(A)

Quick sort selects a random location within the array called a pivot.

Then it creates two lists, one containing all elements smaller than the pivot, and one containing all elements greater than the pivot.

Then it recursively calls its with each of those two new lists. When the lists end up being only one element long, it all gets merged.

Apparently heapsort is great and all, but quicksort beats it. What people do use is the heap data structure, with its most popular application being a priority queue. A priority queue is a data structure that contains a set of keys sorted by their priority. We use the following procedures when utilizing priority queues.

• Insert
• Find-Max

There are others that you can implement, but will make it difficult to address a element within the queue¹⁾.

• Delete-Max
• Change-Key

https://leetcode.com/explore/learn/card/recursion-ii/470/divide-and-conquer/2897/

Many algorithms are recursive, they call themselves recursively to deal with closely related sub-problems.

Divide - divide the problem into smaller subsets of the big set

Conquer - conquer the subproblems

Combine - combine the solutions to come up with the main solution of the bigger problem

A recurrence is an equation or inequality that describes a function in terms of its value of smaller inputs.

Methods for solving recurrences:

• substitution method guess a bound and then use mathematical induction to prove our guess correct
• recursion-tree method convert the recurrance into a tree whose nodes represent the costs incurred at various levels of the recursion.
• master method provides bounds for recurrences of the form $T(n) = aT(n/b) + f(n)$ where $a \geq 1,\ b\ > \ 1$ and $f(n)$ is a given function.

Validating a binary search tree can be solved with a divide and conquer approach. We check each node against an upper and lower bound. We then recurs and check both children.

Binary trees. Height is log(n).

Bottom level of a perfect binary tree has roughly n/2 elements.

When traversing recursively, always thing of the following:

1. What is my base case that I am checking for?
2. What nodes will I call again recursively?

A greedy algorithm is used to find optimal soltions - minimum or maximums. Problems are solved in stages, with each stage choosing the optimal solutions. If a solution is found, add it to your solutions.

n = 5

a = { a₁, a₂, a₃, a₄, a₅ }

1. Algorithm Greedy(a,n)
2. for i=1 to n do:
3.    x = Select(a)
4.    if Feasible(x) Then:
5.        Solution = Solution + x

An approach for scanning through data is to use a sliding window technique.

For example, take this problem: find the length of the longest substring without repeating characters. Ex. Input: “abcabcbb” output = 3 (abc) and for “vadva” output = 3 (vadv)

To do this we have to scan through the input string while keeping track of the characters used. One such answer is as the following:

${\rm L{\small ONGEST}-S{\small UBSTRING}} (string)$

   // our sliding window will be bound by i and j
1. int i = 0; j = 0; max = 0
   // this is our hash table that contains a flag for every character
2. bool characterArray[128] = {false}
   // iterate up starting from i = 0
3. for int j from 0 to < string.length
       // if we've never seen this character before
       // and tally our max and update our character hash key  
4.     if characterArray(string[j]) = false 
5.         max = (max > (j+1-i)) ? max : (j+1-i)
6.         characterArray(string[j]] =  true)
       // if we have seen it before, reset it, and slide i forward one
       // and try again
7.     else
8.         characterArray(string[i]] = false)
9.         i++, j--
10. return max

Asymptotic efficiency is when we use an input size large enough to make only the growth of the running time relevant.

Worst case running time of an algorithm.

$$Θ(g(n)) = \{f(n)\ :\ there\ exist\ positive\ constants\ c_1,\ c_2,\ and\ n_0\ such\ that\\\ 0 \le c_1*g(n)\le f(n) \le c_2*g(n)\ for\ all\ n\ \ge\ n_0\}$$

O(g(n)) is defined as the set of functions which grow no faster than g(n).

$$0(g(n))\ =\ \{ f(n)\ :\ there\ exist\ positive\ constants\ c\ and\ n_0\ such\ that\\\ 0\leq f(n) \leq c*g(n)\ for\ all\ n\geq n_0\}.$$

or

$O(g(n))={f(n)\ :\exists\ c>0,\ n_0>0 \ni 0\leq f(n) \leq\ c*g(n) \forall n\geq n_0}$

Example²⁾:

Using this example of $f(n)=5n$ and $g(n)=n$ we can prove that $O(g(n))=n$.

The definition of $O(g(n))$ is:

$f(n)\in O(g(n))\iff \exists c>0,n_0\ni \forall n \geq n_0,f(n)\leq c*g(n) $

Using this definition we can prove that $f(n)$ is a member of $O(g(n))$:

$n\in O(g(n)) \iff \exists c>0, 5n \leq c*n $

By substituting constant c = 6:

$n\in O(g(n)) \iff \exists 6>0, 5n \leq 6*n$

Ω notation provides an asymptotic lower bound. For a given function $g(n)$, we denote by $Ω(g(n)$ (pronounced “big-omega of g of n” or “omega of g of n”) the set of functions

$$Ω(g(n))=\{f(n)\ :\ there\ exist\ positive\ constants\ c\ and\ n_0\ such\ that\\\ 0\leq c*g(n)\leq f(n)\ for\ all\ n\geq n_0\}$$.

We use o-notation to denote an upper bound that is not asymptotically tight. We define $o(g(n))$ as the set:

$$o(g(n)) = \{f(n)\ :\ for\ any\ positive\ constant\ c>0,\ there\ exists\ a\ constant\\\ n_0>0\ such\ that\ 0\leq f(n)\ < c*g(n)\ for\ all\ n \geq n_0\}.$$

https://stackoverflow.com/questions/55876342/difference-between-ologn-and-onlogn/55876397

Think of it as O(n*log(n)), i.e. “doing log(n) work n times”. For example, searching for an element in a sorted list of length n is O(log(n)). Searching for the element in n different sorted lists, each of length n is O(n*log(n)).

Remember that O(n) is defined relative to some real quantity n. This might be the size of a list, or the number of different elements in a collection. Therefore, every variable that appears inside O(…) represents something interacting to increase the runtime. O(n*m) could be written O(n_1 + n_2 + … + n_m) and represent the same thing: “doing n, m times”.

Let's take a concrete example of this, mergesort. For n input elements: On the very last iteration of our sort, we have two halves of the input, each half size n/2, and each half is sorted. All we have to do is merge them together, which takes n operations. On the next-to-last iteration, we have twice as many pieces (4) each of size n/4. For each of our two pairs of size n/4, we merge the pair together, which takes n/2 operations for a pair (one for each element in the pair, just like before), i.e. n operations for the two pairs.

From here, we can extrapolate that every level of our mergesort takes n operations to merge. The big-O complexity is therefore n times the number of levels. On the last level, the size of the chunks we're merging is n/2. Before that, it's n/4, before that n/8, etc. all the way to size 1. How many times must you divide n by 2 to get 1? log(n). So we have log(n) levels. Therefore, our total runtime is O(n (work per level) * log(n) (number of levels)), n work.

Backtracking is an algorithm that looks at all possible solutions for a problem and evaluates each one.

Decent writeup here.

To check if intervals overlap simply do this!

            min
            end
  |---------->
         |------------>
        max
        strt

    |------------>
        |---->
      max   min
      strt  end

             |------------>
 |---->     max
     min    strt
     end

if(MININUM_END_TIME > MAXIMUM_START_TIME) 
  WE OVERLAP!

Undirected Graphs: Graphs don't have edges to them. You can go from A to B or B to A. No problems!

Assumptions to keep in mind. We only ever care about nodes reachable by our given start node.

You can traverse a Graph in two ways, Depth First Traversal and Breath First Traversal.

DFT is accomplished with a Stack. Algorithm:

1. Push first vertex onto the stack.
2. Mark first vertex as visited.
3. Repeat while Stack is not empty
4.    Visit next vertex adjacent to the one on top of the stack
5.    Push this vertex on the stack
6.    Mark this vertex as visited
7.    If there isn't a vertex to visit
8.      pop vertex off top of the stack 

BFT is accomplished with a Queue.

1. Push first vertex into queue
2. Mark first vertex as seen
3. While queue not empty
4.    Pop front of the queue.
4.    Visit all neighboring verticles 
      If they are not seen
5.      Push them onto the queue.

Theres like… all these different kinds of permutations and combinations with repetitions, with unique values, etc.

Youtube video on this

1. Sort input candidates.
2. Make a boolean flag array to track used elements.
3. Recurse over the canditations, choosing and building a solution, and then unchoosing.
4. If the next option is the same, skip it.