Business Board Problem in Java

The business board Problem is a classic problem in computer science that involves dividing a long wall into several partitions and painting each partition with a different colour. The goal is to minimize the amount of time it takes to paint the wall, while ensuring that no two adjacent partitions have the same colour.

In this section, we will discuss how to solve the Business board Problem using Java. We will start by explaining the problem in more detail and then walk through the steps to implement a solution.

Problem Statement

The business board Problem can be stated as follows: Given a long wall of length L and n painters, each painter can paint a certain amount of wall per unit time. The task is to divide the wall into n partitions, each of arbitrary length, such that the time required to paint the entire wall is minimized. Furthermore, adjacent partitions must be painted by different painters.

To solve this problem, we can use a dynamic programming approach. We will define a 2D array DP[i][j] to represent the minimum amount of time required to paint a wall of length i using j painters. The base cases are DP[0][j] = 0 (it takes zero time to paint a wall of length 0) and DP[i][1] = sum(0 to i) (it takes a single painter the sum of all partition lengths to paint the entire wall).

Approach

The recurrence relation for DP[i][j] can be derived as follows:

DP[i][j] = min(max(DP[k][j-1], sum(k+1 to i))) for all k < i

The above recurrence relation can be interpreted as follows: we divide the wall into a prefix of length k and a suffix of length i-k, and we assume that the prefix is painted by j-1 painters while the suffix is painted by the j-th painter. The minimum amount of time required to paint the wall is the minimum of all such prefix lengths.

Algorithm:

• Read input values from the user: the length of the wall, the number of painters, and the speed of each painter.
• Call the minTime function, passing in the array of painter speeds, the number of painters, and the length of the wall.
• Inside the minTime function, initialize a two-dimensional array dp with n+1 rows and arr.length+1 columns.
• Fill in the base cases of the dp array, which represent the minimum time required to paint the wall when there is only one painter or only one section of the wall to paint.
• For each additional painter, iterate through each possible partition of the wall and find the minimum time required to paint it.
• Return the result, which is the minimum time required to paint the entire wall with n painters.

Implementation

Let's now take a look at how to implement the business board Problem in Java. We will start by defining a function that takes in the length of the wall, the number of painters, and an array of painting speeds for each painter:

Businessboard.java

import java.util.*;
public class Businessboard {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        // Get input values from user
        System.out.print("Enter the length of the wall: ");
        int length = scanner.nextInt();
        System.out.print("Enter the number of painters: ");
        int painters = scanner.nextInt();
        int[] speeds = new int[painters];
        System.out.println("Enter the speed of each painter:");
        for (int i = 0; i < painters; i++) {
            speeds[i] = scanner.nextInt();
        }
        // Call the minTime function and print the result
        int result = minTime(speeds, painters, length);
        System.out.println("The minimum amount of time required to paint the wall is: " + result);
    }
    public static int minTime(int[] arr, int n, int k) {
        int[][] dp = new int[n + 1][arr.length + 1];


        // Fill in base cases
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= arr.length; j++) {
                if (i == 1) {
                    dp[i][j] = dp[i][j - 1] + arr[j - 1];
                } else if (j == 1) {
                    dp[i][j] = arr[j - 1];
                } else {
                    int sum = arr[j - 1];
                    dp[i][j] = Integer.MAX_VALUE;
                    for (int l = j - 1; l >= i - 1; l--) {
                        dp[i][j] = Math.min(dp[i][j], Math.max(dp[i - 1][l], sum));
                        sum += arr[l - 1];
                    }
                }
            }
        }
        // Return the result
        return dp[n][arr.length];
    }
}

Output:

Enter the length of the wall: 10
Enter the number of painters: 2
Enter the speed of each painter:
2 5
The minimum amount of time required to paint the wall is: 25

In the above example, the user inputs a wall length of 10, 2 painters with speeds of 2 and 5 units of length per unit time, respectively. The output is the minimum amount of time required to paint the wall, which is 25 units of time.

Complexity:

The time complexity of the minTime function is O(n^2 * k), where n is the number of painters and k is the length of the wall. This is because the function fills in a two-dimensional array with n+1 rows and arr.length+1 columns, and the innermost loop iterates from j-1 to i-1 at each iteration. Since there are k possible partitions of the wall, the overall time complexity of the algorithm is O(k * n^3).