Java Program to Count Number of Hops

While solving practical issues, programmers often come across mathematical tasks, which presuppose a definite approach. One of them is a problem of finding out how many steps a subject cover getting to some point with certain conditions of movement.

This problem, besides being an interesting logical challenge, forms a good ground for practicing programming features such as loops, conditions and recursion. In this essay the subject matter is the development of a Java program which involves finding out the number of 'hops' required for the subject to get to a certain destination.

It will be preceded by a description of the problem and its importance and followed by an understanding of the approach taken.

Problem Statement

Let's consider an object, say a frog, with a start position and a required position or a goal or a target position. Our frog can travel one unit, two units, or even three units at any given time. The purpose is to find the number of possible outcomes by which the frog reaches the desired distance.

Example:

If n=3, the frog has the following ways to reach the destination:

Hence, total numbers of ways are 4.

This issue may be used in robotics to find out movement trajectories, in combinatorics, and in game design.

Approaches to the Solution

Recursive Approach

This is a recursive approach to solve this kind of problems. Such a solution of n depends on the solutions of other problems, minor in comparison with it (n−1, n−2, and n−3):

The base cases are:

hops(0)=1: That's where the frog is already at the final location.

hops(n)=0 for n<0: It also means that the frog cannot get into negative positions.

Dynamic Programming

Recursion is neat but can be slow for large value of n because it will do same computation over and over. Dynamic programming optimizes this because the results of subproblems are stored in the data base making the solution hole. However, it can be noted that using an array, therefore, we can calculate hops(n) iteratively.

File Name: FrogHops.java

Output:

 Number of hops (Recursive): 13 Number of hops (DP): 13 

Analysis

Recursive Approach

Advantages: Easy to implement and to integrate into existing protocols. Ideal for small values of n.

Disadvantages: Optimal time complexity of O(3^n) because of the feature of overlapping subproblems. Due to the call stack, the memory is heavily used.

Dynamic Programming Approach

Advantages: Time complexity between O(n) and O(nlogn) and a smallest possible memory usage. This is because it will save the results that have been found out to avoid early repetition throughout the computation process.

Disadvantages: Uses an additional space to store the dp array.

Conclusion

Controlling the number of hops is a good exercise in designing algorithms, checking the orientation of a programmer and his desire to make the code faster. The recursive approach gives the precise and brief solution of small-scale problem while the dynamic programming gives the concern for more important problem of much more large input.

When solved with the right level of optimization, this fundamental problem also illustrates how an application of mathematical logic directly forms a subset of a programming discipline.

Modern computational problems, however, still require such elementary questions to be also at the forefront because their effective solutions are critical for solving actual-world problems.