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Mains Answer Writing Latest Questions

Valina Mendonca
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Begginer
Asked: 2024-07-16T10:42:42+05:30 In:

Dynamic Programming – Knapsack Problem

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Solve the 0/1 Knapsack problem using dynamic programming to maximize the total value of items without exceeding the weight capacity.

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2 Answers

  1. Ganesh M
    Begginer

    The 0/1 Knapsack problem is a classic optimization challenge where the goal is to maximize the value of items placed in a knapsack without exceeding its weight capacity. Using dynamic programming, we can efficiently solve this problem.

    Here’s how it works:

    1. Create a DP Table: We use a table dp where dp[i][w] represents the maximum value achievable with the first i items and a weight limit w.
    2. Initialization: Set dp[0][w] to 0 for all w, since no items means no value.
    3. Fill the Table: For each item i and weight w, decide whether to include the item. If included, add its value to the best solution for the remaining capacity (w minus the item’s weight). Update dp[i][w] with the maximum value between including and not including the item.

    The final solution, dp[n][W], gives the maximum value for n items and knapsack capacity W. This method ensures all combinations are considered, providing the optimal solution efficiently.

  2. Pravallika
    Begginer
    This answer was edited.

    Defination: Maximize total value of items without exceeding weight limit using given weights and values.

    def knapsack(values,weights,capacity):

   n=len(values)
   dp=[[0 for _ in range(capacity+1)] for _ in range(n+1)]
   #Fill the dp array
   for i in range(1,n+1):
      for w in range(1,capacity+1):
          if weights[i-1]<=w:
             dp[i][w]=max(dp[i-1][w],dp[i-1][w-weights[i-1]]+values[i-1])
          else:
             dp[i][w]=dp[i-1][w]
   #The maximum value is in the bottom-right corner of the dp array
   return dp[n][capacity]
values=[60,100,120]
weights=[10,20,30]
capacity=50
max_value=knapsack(values,weights,capacity)
print(f"The maximum value that can be obtained is: {max_value}")

    Explanation:

    1. Initialization:
      • We initialize a 2D list dp where dp[i][w] represents the maximum value that can be obtained using the first i items with a total weight not exceeding w.
    2. Filling the DP Table:
      • For each item i (from 1 to n), and for each weight w (from 1 to capacity):
        • If the weight of the item i (weights[i-1]) is less than or equal to w, we have two choices:
          • Exclude the item i: The value is dp[i-1][w].
          • Include the item i: The value is dp[i-1][w-weights[i-1]] + values[i-1].
        • We take the maximum of these two choices.
        • If the weight of the item i is greater than w, we exclude the item i.
    3. Result:
      • The maximum value that can be obtained is found in dp[n][capacity].

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