[FIXED] Maximize the number of Elements in the Array divisible by M

Issue

I'm working on the following task.

Given an array of n integers and two integer numbers m and k.

You can add any positive integer to any element of the array such that the total value does not exceed k.

The task is to maximize the multiples of m in the resultant array.

Consider the following example.

Input:

n = 5, m = 2, k = 2, arr[] = [1, 2, 3, 4, 5]

Let's add 1 to the element arr[0] and 1 to arr[2] then the final array would be:

[2, 2, 4, 4, 5]

Now there are four (4) elements which are multiples of m (2).

I am not getting correct output.

My code:

public class Main {
    public static void main(String[] args) {
        int n = 5;
        int m = 4;
        int k = 3;
        int count = 0;
        
        int[] arr = {17, 8, 9, 1, 4};
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= k; j++) {
                // check initial
                if (arr[i] % m == 0) {
                    break;
                }
                // add
                arr[i] = arr[i] + j;
                // check again
                if (arr[i] % m == 0) {
                    count++;
                    break;
                }
            }
        }
        
        System.out.println("Final Array : " + Arrays.toString(arr));
        System.out.println("Count : " + count);
    }
}

Solution

This task boils down to a well-known Dynamic programming algorithm called Knapsack problem after a couple of simple manipulations with the given array.

This approach doesn't require sorting and would be especially advantages when k is much smaller n.

We can address the problem in the following steps:

• Iterate over the given array and count all the numbers that are already divisible by m (this number is stored in the variable count in the code below).

• While iterating, for every element of the array calculate the difference between m and remainder from the division of this element by m. Which would be equal to m - currentElement % m. If the difference is smaller or equal to k (it can cave this difference) it should be added to the list (differences in the code below) and also accumulated in a variable which is meant to store the total difference (totalDiff). All the elements which produce difference that exceeds k would be omitted.

• If the total difference is less than or equal to k - we are done, the return value would be equal to the number of elements divisible by m plus the size of the list of differences.

• Otherwise, we need to apply the logic of the Knapsack problem to the list of differences.

The idea behind the method getBestCount() (which is an implementation Knapsack problem) boils down to generating the "2D" array (a nested array of length equal to the size of the list of differences +1, in which every inner array having the length of k+1) and populating it with maximum values that could be achieved for various states of the Knapsack.

Each element of this array would represent the maximum total number of elements which can be adjusted to make them divisible by m for the various sizes of the Knapsack, i.e. number of items available from the list of differences, and different number of k (in the range from 0 to k inclusive).

The best way to understand how the algorithm works is to draw a table on a piece of paper and fill it with numbers manually (follow the comments in the code, some intermediate variables were introduced only for the purpose of making it easier to grasp, and also see the Wiki article linked above).

For instance, if the given array is [1, 8, 3, 9, 5], k=3 and m=3. We can see 2 elements divisible by m - 3 and 9. Numbers 1, 8, 5 would give the following list of differences [2, 1, 1]. Applying the logic of the Knapsack algorithm, we should get the following table:

[0, 0, 0, 0]
[0, 0, 1, 1]
[0, 1, 2, 3]
[0, 1, 2, 3]

We are interested in the value right most column of the last row, which is 3 plus 2 (number of elements divisible by 3) would give us 5.

^{Note: that code provided below can dial only with positive numbers. I don't want to shift the focus from the algorithm to such minor details. If OP or reader of the post are interested in making the code capable to work with negative number as well, I'm living the task of adjusting the code for them as an exercise. Hint: only a small change in the countMultiplesOfM() required for that purpose.}

That how it might be implemented:

public static int countMultiplesOfM(int[] arr, int k, int m) {
    
    List<Integer> differences = new ArrayList<>();
    int count = 0;
    long totalDiff = 0; // counter for the early kill - case when `k >= totalDiff`

    for (int next : arr) {

        if (next % m == 0) count++; // number is already divisible by `m` we can increment the count and from that moment we are no longer interested in it
        
        else if (m - next % m <= k) {
            differences.add(m - next % m);
            totalDiff += m - next % m;
        }
    }
    
    if (totalDiff <= k) { // early kill - `k` is large enough to adjust all numbers in the `differences` list
        return count + differences.size();
    }

    return count + getBestCount(differences, k); // fire the main logic
}
// Knapsack Algorithm implementation
public static int getBestCount(List<Integer> differences, int knapsackSize) {

    int[][] tab = new int[differences.size() + 1][knapsackSize + 1];

    for (int numItemAvailable = 1; numItemAvailable < tab.length; numItemAvailable++) {
        int next = differences.get(numItemAvailable - 1);  // next available item which we're trying to place to knapsack to Maximize the current total
        
        for (int size = 1; size < tab[numItemAvailable].length; size++) {
            int prevColMax = tab[numItemAvailable][size - 1]; // maximum result for the current size - 1 in the current row of the table (with the same number of items available)
            int prevRowMax = tab[numItemAvailable - 1][size]; // previous maximum result for the current knapsack's size
            
            if (next <= size) { // if it's possible to fit the next item in the knapsack
                tab[numItemAvailable][size] = Math.max(prevColMax, tab[numItemAvailable][size - next] + 1); // either the max achieved previous with the Current Size (i.e. value in the previous row) or a value from the previous column of the current row (for size - 1) plus 1 (i.e. with a new item)
            } else {
                tab[numItemAvailable][size] = Math.max(prevRowMax, prevColMax); // either a value in the previous row or a value in the previous column of the current row
            }
        }
    }

    return tab[differences.size()][knapsackSize];
}

main()

public static void main(String[] args) {
    System.out.println(countMultiplesOfM(new int[]{17, 8, 9, 1, 4}, 3, 4));
    System.out.println(countMultiplesOfM(new int[]{1, 2, 3, 4, 5}, 2, 2));
    System.out.println(countMultiplesOfM(new int[]{1, 8, 3, 9, 5}, 3, 3));
}

Output:

3   // input array [17, 8, 9, 1, 4], m = 4, k = 3
4   // input array [1, 2, 3, 4, 5],  m = 2, k = 2
5   // input array [1, 8, 3, 9, 5],  m = 3, k = 3

Answered By - Alexander Ivanchenko
Answer Checked By - Willingham (JavaFixing Volunteer)