Gain insights into essential algorithms for effective problem-solving in Java. Delve into recursion, dynamic programming, and graph algorithms to enhance your coding and tackle complex problems confidently.

As a developer, mastering the concepts of algorithms and being proficient in implementing them is essential to improving problem-solving skills. This course aims to equip you with an in-depth understanding of algorithms and how they can be utilized for problem solving in Java.

Starting with the basics, you'll gain a foundational understanding of what algorithms are, with topics ranging from simple multiplication algorithms to analyzing algorithms. Then, you’ll delve into more advanced topics like recursion, backtracking, dynamic programming, and greedy algorithms. You'll also learn about basic graph algorithms, depth-first search, shortest paths, minimum spanning trees, and all-pairs shortest paths. 

By the end of this course, you'll have acquired a wide range of skills that will significantly enhance your ability to solve problems efficiently in Java. Through this course, not only will you improve your coding skills, but you will also gain confidence in tackling complex problems.

Mastering Algorithms for Problem Solving in Java

# Optimizing the text segmentation problem
For our next dynamic programming algorithm, let’s consider the text segmentation problem from the previous chapter. We are given a string $A[1 .. n]$ and a subroutine $IsWord$ that determines whether a given string is a word (whatever that means), and we want to know whether $A$ can be partitioned into a sequence of words.

We solved this problem by defining a function $Splittable(i)$ that returns `True` if and only if the $suffix\space A[i .. n]$ can be partitioned into a sequence of words. We need to compute $Splittable(1)$. This function satisfies the recurrence

$$Splittable(i)=\begin{cases}
 & \text{ True \hspace{5.8cm}}  if\space i > n \\
 &   \overset{n}{\underset{j=1}{\vee}} \left ( IsWord(i,j) \wedge Splittable(j+1) \right )  \hspace{1cm} otherwise 
\end{cases}$$


where $IsWord(i, j)$ is shorthand for $IsWord(A[i .. j])$. This recurrence translates directly into a recursive backtracking algorithm that calls the $IsWord$ subroutine $O(2^n)$ times in the worst case.

But for any fixed string $A[1..n]$, there are only $n$ different ways to call the recursive function $Splittable(i)$—one for each value of $i$ between 1 and $n + 1$—and only $O(n^2)$ different ways to call $IsWord(i, j)$—one for each pair $(i, j)$ such that $1 ≤ i ≤ j ≤ n$.

# Memoization
Why are we spending exponential time computing only a polynomial amount of stuff?

Each recursive subproblem is specified by an integer between $1$ and $n + 1$, so we can memoize the function $Splittable$ into an array $SplitTable[1 .. n + 1]$. Each subproblem $Splittable(i)$ depends only on the results of subproblems $Splittable(j)$ where $j > i$, so the memoized recursive algorithm fills the array in decreasing index order. If we fill the array in this order deliberately, we obtain the dynamic programming algorithm shown in the pseudocode below. The algorithm makes $O(n^2)$ calls to $IsWord$, an exponential improvement over our earlier backtracking algorithm.

Explore the text segmentation problem to take it to the next level.

Getting Started

Introduction to Algorithm

Recursion

Backtracking

Dynamic Programming

Greedy Algorithms

Prove Your Skills: A Five-Chapter Assessment

Basic Graph Algorithms

Depth-First Search

Minimum Spanning Trees

Shortest Paths

All-Pairs Shortest Paths

Pushing Your Limits: A Comprehensive Assessment

Wrapping up

Text Segmentation

Optimizing the text segmentation problem