Add tests, remove main method, improve docs in BruteForceKnapsack (#5641)

src/main/java/com/thealgorithms/dynamicprogramming/BruteForceKnapsack.java

@@ -1,39 +1,67 @@
 package com.thealgorithms.dynamicprogramming;
 
-/* A Naive recursive implementation
-of 0-1 Knapsack problem */
+/**
+ * A naive recursive implementation of the 0-1 Knapsack problem.
+ *
+ * <p>The 0-1 Knapsack problem is a classic optimization problem where you are
+ * given a set of items, each with a weight and a value, and a knapsack with a
+ * fixed capacity. The goal is to determine the maximum value that can be
+ * obtained by selecting a subset of the items such that the total weight does
+ * not exceed the knapsack's capacity. Each item can either be included (1) or
+ * excluded (0), hence the name "0-1" Knapsack.</p>
+ *
+ * <p>This class provides a brute-force recursive approach to solving the
+ * problem. It evaluates all possible combinations of items to find the optimal
+ * solution, but this approach has exponential time complexity and is not
+ * suitable for large input sizes.</p>
+ *
+ * <p><b>Time Complexity:</b> O(2^n), where n is the number of items.</p>
+ *
+ * <p><b>Space Complexity:</b> O(n), due to the recursive function call stack.</p>
+ */
 public final class BruteForceKnapsack {
     private BruteForceKnapsack() {
     }
-    // Returns the maximum value that
-    // can be put in a knapsack of
-    // capacity W
+
+    /**
+     * Solves the 0-1 Knapsack problem using a recursive brute-force approach.
+     *
+     * @param w   the total capacity of the knapsack
+     * @param wt  an array where wt[i] represents the weight of the i-th item
+     * @param val an array where val[i] represents the value of the i-th item
+     * @param n   the number of items available for selection
+     * @return    the maximum value that can be obtained with the given capacity
+     *
+     * <p>The function uses recursion to explore all possible subsets of items.
+     * For each item, it has two choices: either include it in the knapsack
+     * (if it fits) or exclude it. It returns the maximum value obtainable
+     * through these two choices.</p>
+     *
+     * <p><b>Base Cases:</b>
+     * <ul>
+     * <li>If no items are left (n == 0), the maximum value is 0.</li>
+     * <li>If the knapsack's remaining capacity is 0 (w == 0), no more items can
+     * be included, and the value is 0.</li>
+     * </ul></p>
+     *
+     * <p><b>Recursive Steps:</b>
+     * <ul>
+     * <li>If the weight of the n-th item exceeds the current capacity, it is
+     * excluded from the solution, and the function proceeds with the remaining
+     * items.</li>
+     * <li>Otherwise, the function considers two possibilities: include the n-th
+     * item or exclude it, and returns the maximum value of these two scenarios.</li>
+     * </ul></p>
+     */
     static int knapSack(int w, int[] wt, int[] val, int n) {
-        // Base Case
         if (n == 0 || w == 0) {
             return 0;
         }
 
-        // If weight of the nth item is
-        // more than Knapsack capacity W,
-        // then this item cannot be included
-        // in the optimal solution
         if (wt[n - 1] > w) {
             return knapSack(w, wt, val, n - 1);
-        } // Return the maximum of two cases:
-        // (1) nth item included
-        // (2) not included
-        else {
-            return Math.max(val[n - 1] + knapSack(w - wt[n - 1], wt, val, n - 1), knapSack(w, wt, val, n - 1));
+        } else {
+            return Math.max(knapSack(w, wt, val, n - 1), val[n - 1] + knapSack(w - wt[n - 1], wt, val, n - 1));
         }
     }
-
-    // Driver code
-    public static void main(String[] args) {
-        int[] val = new int[] {60, 100, 120};
-        int[] wt = new int[] {10, 20, 30};
-        int w = 50;
-        int n = val.length;
-        System.out.println(knapSack(w, wt, val, n));
-    }
 }