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 Python.

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 Python. 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 Python

The computation of all the Fibonacci numbers again and again by the recursive algorithm

What is an obvious reason for the lack of speed of the recurring algorithm of Fibonacci sequence generation (given below)?

$\underline{RecFibo(n):} 
\\ \hspace{0.4cm}
 if\space n = 0
\\ \hspace{1cm}
return\space 0 
\\ \hspace{0.4cm}
 else\space if\space n = 1
\\ \hspace{1cm}
return\space 1
\\ \hspace{0.4cm}
 else
\\ \hspace{0.4cm}
return\space RecFibo(n − 1) + RecFibo(n − 2)$



What is an obvious reason for the lack of speed of the recurring algorithm of Fibonacci sequence generation (given below)?
<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><munder accentunder="true"><mrow><mi>R</mi><mi>e</mi><mi>c</mi><mi>F</mi><mi>i</mi><mi>b</mi><mi>o</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>:</mo></mrow><mo stretchy="true">‾</mo></munder><mspace linebreak="newline"></mspace><mspace width="1.1381em"/><mi>i</mi><mi>f</mi><mtext> </mtext><mi>n</mi><mo>=</mo><mn>0</mn><mspace linebreak="newline"></mspace><mspace width="2.8453em"/><mi>r</mi><mi>e</mi><mi>t</mi><mi>u</mi><mi>r</mi><mi>n</mi><mtext> </mtext><mn>0</mn><mspace linebreak="newline"></mspace><mspace width="1.1381em"/><mi>e</mi><mi>l</mi><mi>s</mi><mi>e</mi><mtext> </mtext><mi>i</mi><mi>f</mi><mtext> </mtext><mi>n</mi><mo>=</mo><mn>1</mn><mspace linebreak="newline"></mspace><mspace width="2.8453em"/><mi>r</mi><mi>e</mi><mi>t</mi><mi>u</mi><mi>r</mi><mi>n</mi><mtext> </mtext><mn>1</mn><mspace linebreak="newline"></mspace><mspace width="1.1381em"/><mi>e</mi><mi>l</mi><mi>s</mi><mi>e</mi><mspace linebreak="newline"></mspace><mspace width="1.1381em"/><mi>r</mi><mi>e</mi><mi>t</mi><mi>u</mi><mi>r</mi><mi>n</mi><mtext> </mtext><mi>R</mi><mi>e</mi><mi>c</mi><mi>F</mi><mi>i</mi><mi>b</mi><mi>o</mi><mo stretchy="false">(</mo><mi>n</mi><mo>−</mo><mn>1</mn><mo stretchy="false">)</mo><mo>+</mo><mi>R</mi><mi>e</mi><mi>c</mi><mi>F</mi><mi>i</mi><mi>b</mi><mi>o</mi><mo stretchy="false">(</mo><mi>n</mi><mo>−</mo><mn>2</mn><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">\underline{RecFibo(n):} 
\\ \hspace{0.4cm}
 if\space n = 0
\\ \hspace{1cm}
return\space 0 
\\ \hspace{0.4cm}
 else\space if\space n = 1
\\ \hspace{1cm}
return\space 1
\\ \hspace{0.4cm}
 else
\\ \hspace{0.4cm}
return\space RecFibo(n − 1) + RecFibo(n − 2)</annotation></semantics></math>RecFibo(n):​if n=0return 0else if n=1return 1elsereturn RecFibo(n−1)+RecFibo(n−2)

What does the time complexity of the recursive algorithm drop down to after the “memoization” technique (shown below)?

$\underline{MemFibo(n):} 
\\ \hspace{0.4cm}
 if\space n = 0
\\ \hspace{1cm}
return\space 0 
\\ \hspace{0.4cm}
 else\space if\space n = 1
\\ \hspace{1cm}
return\space 1
\\ \hspace{0.4cm}
 else
\\ \hspace{1cm}
if\space F[n]\space is\space undefined
\\ \hspace{1.7cm}
F [n] ← MemFibo(n − 1) + MemFibo(n − 2)
\\ \hspace{1cm}
return\space F[n]$


What does the time complexity of the recursive algorithm drop down to after the “memoization” technique (shown below)?
<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><munder accentunder="true"><mrow><mi>M</mi><mi>e</mi><mi>m</mi><mi>F</mi><mi>i</mi><mi>b</mi><mi>o</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>:</mo></mrow><mo stretchy="true">‾</mo></munder><mspace linebreak="newline"></mspace><mspace width="1.1381em"/><mi>i</mi><mi>f</mi><mtext> </mtext><mi>n</mi><mo>=</mo><mn>0</mn><mspace linebreak="newline"></mspace><mspace width="2.8453em"/><mi>r</mi><mi>e</mi><mi>t</mi><mi>u</mi><mi>r</mi><mi>n</mi><mtext> </mtext><mn>0</mn><mspace linebreak="newline"></mspace><mspace width="1.1381em"/><mi>e</mi><mi>l</mi><mi>s</mi><mi>e</mi><mtext> </mtext><mi>i</mi><mi>f</mi><mtext> </mtext><mi>n</mi><mo>=</mo><mn>1</mn><mspace linebreak="newline"></mspace><mspace width="2.8453em"/><mi>r</mi><mi>e</mi><mi>t</mi><mi>u</mi><mi>r</mi><mi>n</mi><mtext> </mtext><mn>1</mn><mspace linebreak="newline"></mspace><mspace width="1.1381em"/><mi>e</mi><mi>l</mi><mi>s</mi><mi>e</mi><mspace linebreak="newline"></mspace><mspace width="2.8453em"/><mi>i</mi><mi>f</mi><mtext> </mtext><mi>F</mi><mo stretchy="false">[</mo><mi>n</mi><mo stretchy="false">]</mo><mtext> </mtext><mi>i</mi><mi>s</mi><mtext> </mtext><mi>u</mi><mi>n</mi><mi>d</mi><mi>e</mi><mi>f</mi><mi>i</mi><mi>n</mi><mi>e</mi><mi>d</mi><mspace linebreak="newline"></mspace><mspace width="4.837em"/><mi>F</mi><mo stretchy="false">[</mo><mi>n</mi><mo stretchy="false">]</mo><mo>←</mo><mi>M</mi><mi>e</mi><mi>m</mi><mi>F</mi><mi>i</mi><mi>b</mi><mi>o</mi><mo stretchy="false">(</mo><mi>n</mi><mo>−</mo><mn>1</mn><mo stretchy="false">)</mo><mo>+</mo><mi>M</mi><mi>e</mi><mi>m</mi><mi>F</mi><mi>i</mi><mi>b</mi><mi>o</mi><mo stretchy="false">(</mo><mi>n</mi><mo>−</mo><mn>2</mn><mo stretchy="false">)</mo><mspace linebreak="newline"></mspace><mspace width="2.8453em"/><mi>r</mi><mi>e</mi><mi>t</mi><mi>u</mi><mi>r</mi><mi>n</mi><mtext> </mtext><mi>F</mi><mo stretchy="false">[</mo><mi>n</mi><mo stretchy="false">]</mo></mrow><annotation encoding="application/x-tex">\underline{MemFibo(n):} 
\\ \hspace{0.4cm}
 if\space n = 0
\\ \hspace{1cm}
return\space 0 
\\ \hspace{0.4cm}
 else\space if\space n = 1
\\ \hspace{1cm}
return\space 1
\\ \hspace{0.4cm}
 else
\\ \hspace{1cm}
if\space F[n]\space is\space undefined
\\ \hspace{1.7cm}
F [n] ← MemFibo(n − 1) + MemFibo(n − 2)
\\ \hspace{1cm}
return\space F[n]</annotation></semantics></math>MemFibo(n):​if n=0return 0else if n=1return 1elseif F[n] is undefinedF[n]←MemFibo(n−1)+MemFibo(n−2)return F[n]

The time taken by the parent algorithm was very high.

Why was there a need for a technique like “memoization”?

Why was there a need for a technique like “memoization”?

Row-major order (from top down, each row from left to right)

What is the evaluation order for filling the two-dimensional array in the dynamic programming algorithm for edit distance?

What is the evaluation order for filling the two-dimensional array in the dynamic programming algorithm for edit distance?

We can say that “dynamic programming is recursion without ________”?

We can say that “dynamic programming is recursion without ________”?

Test your knowledge of dynamic programming.

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

Quiz: Dynamic Programming