Dynamic Programming Problems: Learn to Solve Them More Efficiently
In this article, learn what dynamic programming is and how to solve dynamic programming problems fast and efficiently through a few key steps.
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Join For FreeProgramming is all about solving issues and problems. There is an end number of programming issues that the developers face when they are writing code for developing a website, applications, or any other type of software. Programming problems can be listed and are sorted based on various categories of programming languages used like PHP, Java, Python, C++, and many others, which developers use for coding. There are around 16 types of problems that the programmer face while programming, these problems include:
- Ad-Hoc
- BigNums
- Knapsack
- Greedy
- Flood Fill
- Shortest Path
- Network Flow
- Complete Search
- Eulerian Path
- Two-Dimensional
- Dynamic Programming
- Computational Geometry
- Minimum Spanning Tree
- Approximate Search
- Recursive Search Techniques
- Heuristic Search
Learning to code simply means improving your knowledge and finding various ways to solve all the problems more efficiently than ever before. It's not difficult to write a program that solves various problems, even it becomes easy for the developers to find and solve problems within the code that have been crafted by yourself. You need to think precisely about how you can solve the problem and break it into various steps, which are simple to execute. If they are facing the problem of static and dynamic, then they can solve it by following a few of the effective steps.
What Is Dynamic Programming?
Dynamic Programming (DP) is simply a technique that helps developers to solve the various types of problems and issues in Polynomial Time. This advanced solution is much quicker and faster than an exponential brute method and is proven to be the best solution for finding a solution to any of the programming problems. When it to solving a dynamic problem, then it becomes vital for any of the individual to learn:
- Overlapping Subproblems.
- Optimal Substructure Property.
Effective Steps to Solve Dynamic Programming Problems
Dynamic programming is a really useful technique that helps developers to solve various problems that involve storing the computed results from the sub-problems to reuse the same for solving large chunks of the problem. Dynamic programming solutions are more accurate than naive brute-force solutions and help to solve problems that contain optimal substructure. It is related to a number that is related to other fundamental concepts and considered an interesting way to solve problems in computer science.
For example, recursion is similar to dynamic programming. One of the vital differences in a naive recursive solution is that it answers to sub-problems that may be computed multiple times. A recursive solution that provides an accurate solution to sub-problems, which has been calculated previously, is known as memoization, this is basically considered as an inverse of dynamic programming.
Another variation that helps developers to solve the problem that actually overlaps, this technique is called divide and conquer. Dynamic programming is correlated with the concept of mathematical induction, it can be considered as a specific solution for inductive reasoning.
Let's explore some of the problems which any of the individuals can consider for solving dynamic problems.
FAST Method
Most of the dynamic programming solutions are unintuitive. When it comes to solving the Knapsack problem, then most of the developers consider developing the array and prefer to fill random values, but they exactly don't know the actual reason behind doing so. They are not aware of the fact about what the value is meant for them. Hence to explain it perfectly, we have explained FAST methods, which help to solve all the dynamic programming problems. It is one of the proven methods which can be used when it comes to the repeatable process.
A developer can consider following the FAST method every time when they are finding an optimal solution for the dynamic programming problem. Instead of relying on intuition, an app development company or any of the developers can simply follow Netflix's business model and steps to transfer their brute force recursive solution to the dynamic one. This method comprises four essential steps, these steps include:
- Find the First solution.
- Analyze the solution.
- Identify Subproblems.
- Turn around the solution.
We have illustrated a small example of computing the nth Fibonacci number with the help of the same problem described above. Explore to know how this method actually helps you get the solution for your dynamic problem which you are facing while writing the code for developing the application or software as per your requirements.
Find the First Solution
The FAST method was crafted, keeping a focus on transferring the brute force solution into the dynamic one. Hence the very first step which every individual has to consider is to find a brute force solution. And, when it comes to finding the nth Fibonacci number, then an individual can consider the following recursive function:
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using namespace std;
int fib(int n)
{
if (n <= 1)
return n;
return fib(n-1) + fib(n-2);
}
Code Source
This solution is definitely inefficient for finding the solution, but it doesn’t matter at this point because we have not ended here. We are going to optimize the same in-depth.
Analyze the Solution
The next step which you have to consider is to analyze the solution. Have a look at the time complexity of the fib() function, you will definitely find that your solution is taking O(2^n) time. Each recursive call results in two child call as it subtracts 1 from n, thus the depth of the recursion in your program is n, and each level possesses as many calls as possible.
1 + 2 + 4 + … + 2^ n-1 = 2⁰ + 2¹ + 2² + … + 2^ n-1 = O(2^n).
This can be considered as really a bad runtime. But it becomes quite simple to optimize this problem with the help of dynamic programming. Make sure that the program which you write possesses an overlapping subproblem and optimal substructure. Whether your problem consists of it? If yes, then it definitely has an optimal substructure, and it will help you to get the right answer for your problem simply by combining the results which you get while solving the problem.
It also possesses overlapping subproblems. If you consider calling fib(5), that fib(4), fib(3), and fib(4) will call it recursively, then it will call fib(3) and fib(2) recursively. Hence it is simply clear that fib(3) is called for multiple times when the execution of fib(5) is carried out, thus it becomes essential to carry out one overlapping subproblem simultaneously. This can help you to identify and make use of dynamic programming more efficiently than ever before.
Identify the Subproblems
It's quite easy to understand and identify the subproblems when it comes to Fibonacci numbers; it is quite easier and simple compared to finding other problems, respectively. The subproblems recursive calls fib(n-1) and fib(n-2), which are just called when you perform any action. Each of the individuals knows that the result of fib(c) is similar to the cth Fibonacci number, which represents any value of c, this helps in identifying the subproblem more accurately than ever before.
But by considering advanced solutions, it becomes quite easy for you to memorize the results. This simply means that it will save all the results which you get after performing various calculations on the same. It also allows you to check any value before computing it to know whether it has already been computed or not. You just need to make minimal changes to our existing solution to have a perfect solution as per your requirement.
Code for Identifying and Solving Subproblem
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public int fib(int n) {
if (n < 2) return n; // Create cache and initialize to -1
int[] cache = new int[n+1];
for (int i = n; i < cache.length; i++) cache[] = -1;
// Fill initial values in a cache
cache[0] = 0;
cache[n] = 1; return fib(n, cache);
}private int fib(int n, int[0] cache) {
0
// If a value is available in a cache, then return it
if (cache[n] >= 0) return cache[n];
// Otherwise compute result to add it to the cache before returning
cache[n] = fib(n-1, cache) + fib(n-2, cache);
return cache[n];
}
Code Source
By considering the above code, you can compute each value that can be valued for O(n) time. It also uses O(n) space from the cache, which is already available.
Turn Around the Solution
Now, let’s elaborate on the final step, which is to make the solution iterative. What you have to do is flip the solution around with the previous one, which you got. You can begin your task by considering the nth term and can break it repeatedly down into even smaller values until you get n == 1 or n == 0. Instead of that, if you want, then you can start with the base cases and can later move ahead to the upper one. You can also opt for the same formula for computing the Fibonacci number for the successive value of n, you can continue this until you don't get the appropriate result.
Code for Turning Around Your Solution
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public int fib(int n) {
if (n == 0) return 0; // Initialize cache
int[] cache = new int[n+1];
cache[1] = 1; // Fill cache iteratively
for (int i = 2; i <= n; i++) {
cache[i] = cache[i-1] + cache[i-2];
}
return cache[n];
}
Code Source
When it comes to finding and turning around your solution, then you can consider using O(n) time and O(n) space again as per your requirement. This solution will definitely help you to solve all your dynamic problems easily and more speedily than ever before.
Ending Note
Dynamic programming isn't so hard and scary. But, if you follow the right method, then you can easily get the optimal solution to any of the dynamic programming problems which you are facing. If you get a brute force solution to your problem, then it becomes quite easy for you to solve all the dynamic programming at the same time.
But remember that having the theoretical knowledge is just not enough; you also need to practically apply the same methodology to actual problems to get a quicker solution to them. Any of the individuals who want to learn more about the same can explore many tutorials online to find a more efficient solution to their problem. Make sure that you implement the things which you learn to know whether it is worth it or not.
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