Java Program To Find The Minimum Number Of Operations Required To Reach N Steps

Solving the task of reaching a specific number N from a starting value by applying defined operations is a common challenge in programming. This type of problem helps to develop problem-solving skills and improve algorithmic thinking. Combinatorial search involves determining the precise sequence of operations needed to achieve the desired number N within a specified number of steps.

Problem Definition

Let's assume that the initial value is set at one, and the final value is N. There are two operations that can be conducted during the process:

  • Multiplying the current number by 2.
  • Incrementing the current number by 1.

Example

To reach N=10, you can perform the following steps:

  • Start at 1.
  • Multiply 1 by 2 → 2.
  • Multiply 2 by 2 → 4.
  • Multiply 4 by 2 → 8.
  • Add 1 in 8→ 9.
  • Add 1 in 9→ 10.

Minimum number of operations necessary is 6.

Approach to Solve the Problem

In order to determine the maximum number of operations required, we utilize the Breadth-First Search (BFS) technique. BFS proves to be beneficial for scenarios in which a decision-maker must identify the shortest route or the minimum number of operations, as it systematically explores solutions level by level.

Why BFS?

Using Breadth First Search (BFS) ensures the discovery of the shortest path in an unweighted graph. This algorithm thoroughly evaluates all feasible states that can be reached from the current state. It halts as soon as it reaches the target integer N, thus minimizing the number of required operations.

Initialize BFS

To handle the retrieval and updating of two specific data elements - the present count and the count of operations executed so far - a queue can be utilized efficiently.

Let's at least begin with the 1 and 0 operations.

Traverse States

  • At each step, perform the two operations: Double the number add 1 to number.
  • In the case of each new state, compare it to N. If Yes, then it returns Operation count.
  • Double the number
  • add 1 to number.

Avoid Reprocessing

A HashSet is effective for storing a log of visited numbers in order to avoid revisiting the same number.

Terminate Early

To minimize the number of operations, it is advisable to halt the traversal at N, ensuring efficiency in the process.

File Name: MinimumOperations.java

Example

import java.util.*;

public class MinimumOperations {

    // Method to calculate the minimum number of operations to reach N

    public static int minOperations(int N) {

        // Base case

        if (N == 1) {

            return 0;

        }

        // Queue to perform BFS

        Queue<int[]> queue = new LinkedList<>();

        // Set to track visited states

        Set<Integer> visited = new HashSet<>();

        // Add the initial state: number 1 with 0 operations

        queue.add(new int[]{1, 0});

        visited.add(1);

        // BFS loop

        while (!queue.isEmpty()) {

            // Get the current state

            int[] current = queue.poll();

            int number = current[0];

            int operations = current[1];

            // Perform the two operations

            int multiply = number * 2;

            int add = number + 1;

            // Check if either operation reaches N

            if (multiply == N || add == N) {

                return operations + 1;

            }

            // Add new states to the queue if not already visited

            if (!visited.contains(multiply) && multiply <= N) {

                queue.add(new int[]{multiply, operations + 1});

                visited.add(multiply);

            }

            if (!visited.contains(add) && add <= N) {

                queue.add(new int[]{add, operations + 1});

                visited.add(add);

            }

        }

        // Should never reach here if the problem guarantees a solution

        return -1;

    }

    public static void main(String[] args) {

        // Example usage

        int N = 10;

        System.out.println("Minimum operations to reach " + N + ": " + minOperations(N));

    }

}

Output:

Output

Minimum operations to reach 10: 4

Explanation

The code functions by exploring all attainable states starting from state 1 without any operations, utilizing the BFS algorithm. It initializes a queue for storing the upcoming states and a HashSet to prevent duplicate values, thus preventing potential infinite loops or redundant computations.

In the process of BFS traversal, every value goes through two transformations: it is essential to highlight that staff members' bonus upgrades involve doubling the amount and then adding 1. The resultant outcomes are included in the queue solely if they have not been encountered previously.

The algorithm halts as soon as an operation reaches the target number N to reduce the number of operations required. This approach ensures effectiveness as BFS systematically explores all potential paths to reach N level by level.

Complexity Analysis

Time Complexity

Breadth-first search (BFS) traverses every state exactly once, with each state having the potential to produce two fresh states. Consequently, the complexity is O(N), with N representing the desired number.

Space Complexity

Breadth-first search (BFS) utilizes a queue and a set for storing states, leading to a space complexity of O(N).

Conclusion

In scenarios where the minimum number of operations needed to reach a certain point N needs to be calculated, the importance of the BFS algorithm in finding the shortest path becomes evident. A Java implementation that is effective in minimizing operations and preventing precomputation involves utilizing BFS along with a hash set.

By effectively solving these challenges, programmers can enhance their understanding of graph traversal algorithms and how they are applied in practice.

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