package com.thealgorithms.dynamicprogramming;
/**
* 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() {
}
/**
* 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) {
if (n == 0 || w == 0) {
return 0;
}
if (wt[n - 1] > w) {
return 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));
}
}
}