From 99a9f1e09385ab3a5343848bbf104acdac32bb39 Mon Sep 17 00:00:00 2001
From: Srinjan Dutta <srinjandutta13@gmail.com>
Date: Thu, 5 Oct 2023 21:26:28 +0530
Subject: [PATCH] knapsack problem with memoization

---
 knapsack/knapsack_memoization.py | 87 ++++++++++++++++++++++++++++++++
 1 file changed, 87 insertions(+)
 create mode 100644 knapsack/knapsack_memoization.py

diff --git a/knapsack/knapsack_memoization.py b/knapsack/knapsack_memoization.py
new file mode 100644
index 000000000000..e6be03a103c9
--- /dev/null
+++ b/knapsack/knapsack_memoization.py
@@ -0,0 +1,87 @@
+"""
+A shopkeeper has bags of wheat that each have different weights and different profits.
+eg.
+no_of_items : 5
+profit [15, 14,10,45,30]
+weight [2,5,1,3,4]
+max_weight that can be carried : 7
+
+Constraints:
+    max_weight > 0
+    profit[i] >= 0
+    weight[i] >= 0
+
+Calculate:
+     The maximum profit that the shopkeeper can make given maxmum weight that can
+be carried.
+
+This problem is implemented here with MEMOIZATION method using the concept of Dynamic Programming
+"""
+
+
+def knapsack(values:list, weights:list, num_of_items:int, max_weight:int, dp:list) -> int:
+    """
+    Function description is as follows-
+    :param weights: Take a list of weights
+    :param values: Take a list of profits corresponding to the weights
+    :param number_of_items: number of items available to pick from
+    :param max_weight: Maximum weight that could be carried
+    :param dp: it is a list of list, i.e, a table whose (i,j) cell represents the maximum profit earned
+                for i items and j as the maximum weight allowed, it is an essential part for implementing this problems
+                using memoization dynamic programming
+    :return: Maximum expected gain
+
+    Testcase 1:
+    >>> values = [1, 2, 4, 5]
+    >>> wt = [5, 4, 8, 6]
+    >>> n = len(values)
+    >>> w = 5
+    >>> dp = [[-1 for x in range(w+1)] for y in range(n+1)]
+    >>> knapsack(values,wt,n,w,dp)
+    2
+
+    Testcase 2:
+    >>> values = [3 ,4 , 5]
+    >>> wt = [10, 9 , 8]
+    >>> n = len(values)
+    >>> w = 25
+    >>> dp = [[-1 for x in range(w+1)] for y in range(n+1)]
+    >>> knapsack(values,wt,n,w,dp)
+    9
+
+    Testcase 3:
+    >>> values = [15, 14,10,45,30]
+    >>> wt = [2,5,1,3,4]
+    >>> n = len(values)
+    >>> w = 7
+    >>> dp = [[-1 for x in range(w+1)] for y in range(n+1)]
+    >>> knapsack(values,wt,n,w,dp)
+    75
+    """
+    if max_weight == 0 or num_of_items == 0: #no profit gain if any of these two become zero
+        dp[num_of_items][max_weight] = 0
+        return 0
+    
+    elif dp[num_of_items][max_weight] != -1: #if this case is previously encountered => maximum gain for this case is already
+        #in dp table
+        return dp[num_of_items][max_weight]
+    
+    elif weights[num_of_items-1] <= max_weight: #if the item can be included in the bag
+        # ans1 stores the maximum profit if the item at index num_of_items -1 is included in the bag
+        ans1 = values[num_of_items - 1] + knapsack(values, weights, num_of_items-1, max_weight-weights[num_of_items-1], dp)
+        # ans2 stores the maximum profit if the item at index num_of_items -1 is not included in the bag
+        ans2 = knapsack(values, weights, num_of_items-1, max_weight, dp)
+        # the final answer is the maximum profit gained from any of ans1 or ans2
+        dp[num_of_items][max_weight] = max(ans1, ans2)
+        return dp[num_of_items][max_weight]
+    
+    # if the item's weight exceeds the max_weight of the bag => it cannot be included in the bag
+    else: 
+        dp[num_of_items][max_weight] = knapsack(values, weights, num_of_items-1, max_weight, dp)
+        return dp[num_of_items][max_weight]
+    
+
+
+if __name__ == '__main__':
+    import doctest
+    doctest.testmod(name='knapsack', verbose=True)
\ No newline at end of file