Java Program To Determine Can All Numbers Of An Array Be Made Equal

Working with arrays often involves transforming elements to meet certain conditions. One such challenge is determining if all numbers within an array can be made equal through specific, limited operations. In this article, you will learn how to determine if all numbers in a given array can be made equal by repeatedly dividing them by 2 or 3.

Problem Statement

The problem requires checking if a collection of positive integers can all be reduced to an identical value using only division by 2 or 3. This scenario arises in various computational tasks, such as standardizing data points or optimizing resource distribution where only specific scaling factors are permissible. The core idea is to find if all numbers share the same "base" component after removing all factors of 2 and 3.

Example

Consider the array [6, 12, 18]. Can all numbers in this array be made equal by repeatedly dividing by 2 or 3? Let's apply the operations:

• For 6: No division by 2 or 3 reduces it further to its base (if we only divide). If we remove factors of 2 and 3, it remains 6.
• For 12: Divide by 2 (12 / 2 = 6). Then 6 remains 6.
• For 18: Divide by 3 (18 / 3 = 6). Then 6 remains 6.

Since all numbers can be reduced to 6 using the allowed operations, the answer is true.

Background & Knowledge Prerequisites

To understand the solution, a basic grasp of Java programming concepts is helpful:

• Variables and Data Types: Understanding int for integer numbers.
• Arrays: How to declare, initialize, and iterate over arrays.
• Loops: for and while loops for iteration.
• Conditional Statements: if statements for logic.
• Methods: Defining and calling helper methods.
• Modulo Operator (%): Used to check divisibility.

No complex data structures or advanced algorithms are required beyond these fundamental concepts.

Use Cases or Case Studies

This type of problem, involving specific transformations to achieve equality, has several practical applications:

• Data Normalization: In some data processing pipelines, values might need to be normalized by specific common factors (e.g., powers of two in digital signal processing) to ensure consistency.
• Resource Allocation: Imagine a system where resources can only be scaled by factors of 2 or 3. This problem helps determine if different initial resource pools can be balanced.
• Game Development: In certain game mechanics, item values or character stats might be subject to specific scaling rules. This check could determine if a set of items can be brought to equal power.
• Number Theory Puzzles: It's a classic problem in competitive programming and number theory, testing understanding of prime factorization and common divisors.
• Load Balancing: In distributed systems, if tasks or loads can only be adjusted by specific factors, this logic could help determine if an even distribution is achievable from a given state.

Solution Approaches

The most effective approach for this problem involves normalizing each number by repeatedly dividing out its factors of 2 and 3. If, after these operations, all numbers in the array become identical, then the original array's elements can be made equal.

Approach 1: Reduce by Factors 2 and 3

One-line summary: For each number, repeatedly divide by 2 until it's odd, then repeatedly divide by 3 until it's not divisible by 3. Check if all resulting numbers are equal.

This approach works because if two numbers a and b can be made equal by only dividing by 2s and 3s, it implies that once all factors of 2 and 3 are removed from both a and b, their remaining "core" value must be the same. If these core values are different, no amount of division by 2 or 3 can make them equal.

Code Example:

// Can All Numbers Be Made Equal by Dividing by 2 or 3
import java.util.Scanner; // Not strictly needed for this program, but good practice for interactive inputs

// Main class containing the entry point of the program
public class Main {

    /**
     * Helper method to reduce a number by repeatedly dividing by a specified factor.
     * Continues dividing until the number is no longer divisible by the factor
     * or becomes 0/1.
     *
     * @param n The number to reduce.
     * @param factor The factor to divide by (e.g., 2 or 3).
     * @return The reduced number.
     */
    private static int reduceNumberByFactor(int n, int factor) {
        if (n == 0) return 0; // Handle zero case if necessary, though problem typically assumes positive integers
        while (n > 0 && n % factor == 0) {
            n /= factor;
        }
        return n;
    }

    /**
     * Determines if all numbers in an array can be made equal by repeatedly
     * dividing by 2 or 3.
     *
     * @param arr The input array of integers.
     * @return true if all numbers can be made equal, false otherwise.
     */
    public static boolean canBeMadeEqual(int[] arr) {
        if (arr == null || arr.length == 0) {
            return true; // An empty array or null array can be considered 'equal' trivially
        }

        // Step 1: Process the first number to get its reduced base value.
        // This will be our target for comparison.
        int firstReduced = arr[0];
        firstReduced = reduceNumberByFactor(firstReduced, 2);
        firstReduced = reduceNumberByFactor(firstReduced, 3);

        // Step 2: Iterate through the rest of the array and reduce each number.
        // Compare each reduced number with the firstReduced value.
        for (int i = 1; i < arr.length; i++) {
            int currentReduced = arr[i];
            currentReduced = reduceNumberByFactor(currentReduced, 2);
            currentReduced = reduceNumberByFactor(currentReduced, 3);

            // If any reduced number doesn't match the firstReduced, they cannot all be made equal.
            if (currentReduced != firstReduced) {
                return false;
            }
        }

        // If the loop completes, all numbers reduced to the same value.
        return true;
    }

    public static void main(String[] args) {
        // Test Cases
        int[] arr1 = {6, 12, 18};
        System.out.println("Can [6, 12, 18] be made equal? " + canBeMadeEqual(arr1)); // Expected: true

        int[] arr2 = {2, 3, 5};
        System.out.println("Can [2, 3, 5] be made equal? " + canBeMadeEqual(arr2)); // Expected: false (reduces to 1, 1, 5)

        int[] arr3 = {1, 1, 1};
        System.out.println("Can [1, 1, 1] be made equal? " + canBeMadeEqual(arr3)); // Expected: true

        int[] arr4 = {7, 14, 21};
        // 7 -> 7
        // 14 -> 7 (14/2)
        // 21 -> 7 (21/3)
        System.out.println("Can [7, 14, 21] be made equal? " + canBeMadeEqual(arr4)); // Expected: true

        int[] arr5 = {10, 20, 30};
        // 10 -> 5 (10/2)
        // 20 -> 5 (20/2/2)
        // 30 -> 5 (30/2/3)
        System.out.println("Can [10, 20, 30] be made equal? " + canBeMadeEqual(arr5)); // Expected: true

        int[] arr6 = {4, 6, 8, 9, 12};
        // 4 -> 1 (4/2/2)
        // 6 -> 1 (6/2/3)
        // 8 -> 1 (8/2/2/2)
        // 9 -> 1 (9/3/3)
        // 12 -> 1 (12/2/2/3)
        System.out.println("Can [4, 6, 8, 9, 12] be made equal? " + canBeMadeEqual(arr6)); // Expected: true
    }
}

Run

Sample Output:

Can [6, 12, 18] be made equal? true
Can [2, 3, 5] be made equal? false
Can [1, 1, 1] be made equal? true
Can [7, 14, 21] be made equal? true
Can [10, 20, 30] be made equal? true
Can [4, 6, 8, 9, 12] be made equal? true

Stepwise Explanation for Clarity:

1. Handle Edge Cases: Check if the array is null or empty. In such cases, it trivially satisfies the condition, so true is returned.
2. Define reduceNumberByFactor Helper: Create a private helper method reduceNumberByFactor(int n, int factor). This method takes an integer n and a factor (e.g., 2 or 3). It repeatedly divides n by factor as long as n is greater than 0 and divisible by factor. This ensures all instances of that factor are removed.
3. Process First Element: Take the first element of the array (arr[0]). Apply reduceNumberByFactor first with 2 and then with 3. The result (firstReduced) is the target base value that all other numbers must match.
4. Iterate and Compare: Loop through the rest of the array (starting from the second element, arr[1]).

• For each element arr[i], apply reduceNumberByFactor with 2 and then with 3, just like the first element.
• Compare this currentReduced value with firstReduced.
• If currentReduced is ever different from firstReduced, it means that specific number cannot be made equal to the others, so the function immediately returns false.

1. Return True: If the loop completes without finding any mismatch, it means all numbers can be reduced to the same base value. Therefore, return true.

Conclusion

Determining if all numbers in an array can be made equal by specific division operations involves understanding their underlying prime factors. By reducing each number to its "core" value—after removing all allowed factors (in this case, 2 and 3)—we can efficiently check for equality. This method provides a clear and robust solution to a common algorithmic problem, demonstrating the power of normalization through factorization.

Summary

• Problem: Check if array numbers can be made equal by dividing by 2 or 3.
• Core Idea: Remove all factors of 2 and 3 from each number.
• Method:
• Create a helper to repeatedly divide a number by a given factor.
• Reduce the first number to its "base" value (without factors 2 or 3).
• Reduce every subsequent number similarly.
• If any reduced number does not match the first reduced number, return false.
• If all match, return true.
• Relevance: Useful in data normalization, resource allocation, and number theory.