chore(CPlusPlus): add 0/1 Knapsack problem to Dynamic Programming (#404)

algorithms/CPlusPlus/Dynamic-Programming/01-knapsack.cpp

@@ -0,0 +1,77 @@
+// 0/1 knapsack (Bottom Up or memorization) Dynamic Programming
+#include <bits/stdc++.h>
+using namespace std;
+
+int knapsack(int wait[], int price[], int n, int capacity, vector<vector<int>> dp)
+{
+    // base condition - when the bag capacity is 0 and wait[] and price[] size is 0.
+    if (n == 0 || capacity == 0)
+        return 0;
+
+    // for each call first check the table if value are present(not -1) then directly return
+    if (dp[n][capacity] != -1)
+        return dp[n][capacity];
+
+    /*
+    * two case here:
+    * case 1: include (if wait of the element is less than the bag capacity).
+    * case 2: else not include
+    */
+    if (wait[n - 1] <= capacity)
+    {
+        dp[n][capacity] = max(price[n - 1] + knapsack(wait, price, n - 1, capacity - wait[n - 1], dp),
+                              knapsack(wait, price, n - 1, capacity, dp));
+    }
+    else
+    {
+        dp[n][capacity] = knapsack(wait, price, n - 1, capacity, dp);
+    }
+
+    return dp[n][capacity];
+}
+
+int main()
+{
+    // dp table initialized with -1
+    vector<vector<int>> dp(1001, vector<int>(1001, -1));
+
+    int n;
+    cout << "Enter the no. of items" << endl;
+    cin >> n;
+
+    int wait[n];
+    cout << "Enter the Wait of every items" << endl;
+    for (int i = 0; i < n; i++)
+        cin >> wait[i];
+
+    int price[n];
+    cout << "Enter the Price of every items" << endl;
+    for (int i = 0; i < n; i++)
+        cin >> price[i];
+
+    cout << "Enter the Capacity of Knapsack" << endl;
+    int capacity;
+    cin >> capacity;
+
+    // knapsack function return the maximum profit
+    int max_profit = knapsack(wait, price, n, capacity, dp);
+    cout << "Maximum Profit = " << max_profit << endl;
+
+    return 0;
+}
+
+/*
+Complexity Analysis:
+    Time:  O(n*capacity)
+    Space: O(n*capacity)
+The use of 2D vector data structure for storing results of intermediate states
+
+Test case 1:
+    input:
+        3           -> No. of items
+        10 20 30    -> wait[]
+        60 100 120  -> price[]
+        50          -> capacity of knapsack
+    output:
+        220         -> maximum profit
+*/

algorithms/CPlusPlus/README.md

@@ -15,6 +15,7 @@
 ## Dynamic-Programming
 
 1. [Longest Common Subsequence](Dynamic-Programming/longest-common-subsequence.cpp)
+2. [0/1-knapsack](Dynamic-Programming/01-knapsack-bottom-up.cpp)
 
 
 ## Graphs