diff --git a/knapsack/knapsack_memoization.py b/knapsack/knapsack_memoization.py index 49c22c5f1..6643025ca 100644 --- a/knapsack/knapsack_memoization.py +++ b/knapsack/knapsack_memoization.py @@ -15,21 +15,25 @@ 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 +This problem is implemented here with MEMOIZATION method using the concept of +Dynamic Programming """ """ for more information visit https://en.wikipedia.org/wiki/Memoization """ -def knapsack(values:list, weights:list, num_of_items:int, max_weight:int, dp:list) -> int: +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 + :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 problem using memoization dynamic programming :return: Maximum expected gain @@ -60,26 +64,34 @@ def knapsack(values:list, weights:list, num_of_items:int, max_weight:int, dp:lis >>> 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 + #no profit gain if any of these two become zero + if max_weight == 0 or num_of_items == 0: 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 + #if this case is previously encountered => maximum gain for this case is already + elif dp[num_of_items][max_weight] != -1: #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 + #if the item can be included in the bag + elif weights[num_of_items-1] <= max_weight: + # ans1 stores the maximum profit if the item at + # index num_of_items -1 is included in the bag + incl = knapsack(values,weights,num_of_items-1, + max_weight-weights[num_of_items-1],dp) + ans1 = values[num_of_items - 1] + incl + # 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 + # 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) + dp[num_of_items][max_weight] = knapsack(values, weights, + num_of_items-1, max_weight, dp) return dp[num_of_items][max_weight]