DP Solution to Knapsack problem

Algorithms/Dynamic Programming/Knapsack_Problem.md

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+# Knapsack Problem
+
+Knapsack problem states that you are given a bag with a finite capacity, and a set of items which carry a certain weight and a profit value.
+The goal is to maximize the profit with the constraint that the weight of the bag must not exceed the capacity of the bag. \
+There are two kinds of Knapsack problems, one that can hold fractional values, and the more popular one is 0/1 Knapsack Problem.
+
+Example of Fractional Knapsack problem:- A bag that needs to contain W kg of varied fruits such that the profit is maximised.
+Example of 0/1 Knapsack problem:- A bag that needs to contain electronic gadgets whose total weight cannot exceed W kg.
+
+We'll tackle here the more popular one, which is **0/1 Knapsack Problem**.
+
+So,
+Let's say we are given the following data. \
+$N$ - Number of elements.
+$P$ - Array containing the profit value of each item.
+$W$ - Weight of each element.
+$K$ - Capacity of the bag.
+
+Let's denote a binary variable $x$ which denotes whether an item is included.
+
+**Objective Function**: $ max \sum_{i=1}^n P_i  x_i $
+Subject to **Constraint**: $ \sum_{i=1}^n W_i x_i \le K $
+
+Here I propose a solution using *Dynamic Programming*.
+
+```
+#include <bits/stdc++.h>
+using namespace std;
+
+// A utility function that returns
+// maximum of two integers
+int max(int a, int b)
+{
+    return (a > b) ? a : b;
+}
+
+// Returns the maximum value that
+// can be put in a knapsack of capacity W
+int knapSack(int W, int wt[], int val[], int n)
+{
+    int i, w;
+      vector<vector<int>> K(n + 1, vector<int>(W + 1));
+
+    // Build table K[][] in bottom up manner
+    for(i = 0; i <= n; i++)
+    {
+        for(w = 0; w <= W; w++)
+        {
+            if (i == 0 || w == 0)
+                K[i][w] = 0;
+            else if (wt[i - 1] <= w)
+                K[i][w] = max(val[i - 1] +
+                                K[i - 1][w - wt[i - 1]],
+                                K[i - 1][w]);
+            else
+                K[i][w] = K[i - 1][w];
+        }
+    }
+    return K[n][W];
+}
+
+// Driver Code
+int main()
+{
+    int val[] = { 60, 100, 120 };
+    int wt[] = { 10, 20, 30 };
+    int W = 50;
+    int n = sizeof(val) / sizeof(val[0]);
+
+    cout << knapSack(W, wt, val, n);
+
+    return 0;
+}
+```