Added multiplication matrix problem and solution

Author: Murillo2380

src/medium-apple-multiplication-table/README.md

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+# Description
+
+Good morning! Here's your coding interview problem for today.
+
+This problem was asked by **Apple**.
+
+Suppose you have a multiplication table that is N by N. That is, a 2D array where the value at the i-th row and j-th column is `(i + 1) * (j + 1)` (if 0-indexed) or `i * j` (if 1-indexed).
+
+Given integers N and X, write a function that returns the number of times X appears as a value in an N by N multiplication table.
+
+For example, given N = 6 and X = 12, you should return 4, since the multiplication table looks like this:
+
+```
+| 1 | 2 | 3 | 4 | 5 | 6 |
+
+| 2 | 4 | 6 | 8 | 10 | 12 |
+
+| 3 | 6 | 9 | 12 | 15 | 18 |
+
+| 4 | 8 | 12 | 16 | 20 | 24 |
+
+| 5 | 10 | 15 | 20 | 25 | 30 |
+
+| 6 | 12 | 18 | 24 | 30 | 36 |
+```
+
+And there are 4 12's in the table.
+
+# Source
+
+Received by email from the [Daily Coding Problem](https://www.dailycodingproblem.com)

src/medium-apple-multiplication-table/java/Idea.md

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+# Idea behind the solution
+
+We can simply check the divisors of the given X and count how many of those divisors multiplied by each other gets X.
+
+For the `i-th` divisor we check all the previous `j-th` divisor, for `0 <= j <= i`. When `divisor_i * divisor_j == X` is `false` for every `j` in the on going iteration, we can return the current count since `divisor_i` will be smaller in the next iteration and every upcoming check of `divisor_i * divisor_j == X` will be `false`.
+
+The running time in the worst case is `O(n^2)` for n being the length of the matrix.

src/medium-apple-multiplication-table/java/Solution.java

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+import java.util.*;
+
+class Solution {
+
+    private static List<Integer> divisors(int n, int maxDivisor) {
+
+        final int min = Math.min(n / 2, maxDivisor);
+        List<Integer> divisors = new ArrayList<>(min);
+
+        for (int i = 2; i <= min; i++)
+            if (n % i == 0)
+                divisors.add((i));
+
+        return divisors;
+    }
+
+    private static int solve(int n, int x) {
+
+        int count = 0;
+        List<Integer> divisors = divisors(x, n);
+        System.out.println(divisors);
+        boolean foundAny = true;
+        for (int i = divisors.size() - 1; foundAny && i >= 0; i--) {
+            foundAny = false;
+            for (int j = i; j >= 0; j--) {
+                if (divisors.get(i) * divisors.get(j) != x) {
+                    continue;
+                }
+
+                foundAny = true;
+
+                if (i == j)
+                    count++;
+                else
+                    count += 2;
+            }
+        }
+
+        return count;
+    }
+
+    public static void main(String[] args) {
+        System.out.println(solve(6, 12)); // 4
+        System.out.println(solve(6, 9)); // 1
+        System.out.println(solve(6, 36)); // 1
+        System.out.println(solve(6, 20)); // 2
+        System.out.println(solve(6, 30)); // 2
+    }
+}