Unbounded Knapsack (Repetition of items allowed) | Set 2
Given an integer W, arrays val[] and wt[], where val[i] and wt[i] are the values and weights of the ith item, the task is to calculate the maximum value that can be obtained using weights not exceeding W.
Note: Each weight can be included multiple times.
Examples:
Input: W = 4, val[] = {6, 18}, wt[] = {2, 3}
Output: 18
Explaination: The maximum value that can be obtained is 18, by selecting the 2nd item twice.Input: W = 50, val[] = {6, 18}, wt[] = {2, 3}
Output: 294
Naive Approach: Refer to the previous post to solve the problem using traditional Unbounded Knapsack algorithm.
Time Complexity: O(N * W)
Auxiliary Space: O(W)
Efficient Approach: The above approach can be optimized based on the following observations:
- Suppose the ith index gives us the maximum value per unit weight in the given data, which can be easily found in O(n).
- For any weight X, greater than or equal to wt[i], the maximum reachable value will be dp[X – wt[i]] + val[i].
- We can calculate the values of dp[] from 0 to wt[i] using the traditional algorithm and we can also calculate the number of instances of ith item we can fit in W weight.
- So the required answer will be val[i] * (W/wt[i]) + dp[W%wt[i]].
Below is the implementation of new algorithm.
Python3
# Python Program to implement the above approach from fractions import Fraction # Function to implement optimized# Unbounded Knapsack algorithmdef unboundedKnapsackBetter(W, val, wt): # Stores most dense item maxDenseIndex = 0 # Find the item with highest unit value # (if two items have same unit value then choose the lighter item) for i in range(1, len(val)): if Fraction(val[i], wt[i]) \ > Fraction(val[maxDenseIndex], wt[maxDenseIndex]) \ or (Fraction(val[i], wt[i]) == Fraction(val[maxDenseIndex], wt[maxDenseIndex]) \ and wt[i] < wt[maxDenseIndex] ): maxDenseIndex = i dp = [0 for i in range(W + 1)] counter = 0 breaked = False for i in range(W + 1): for j in range(len(wt)): if (wt[j] <= i): dp[i] = max(dp[i], dp[i - wt[j]] + val[j]) if i - wt[maxDenseIndex] >= 0 \ and dp[i] - dp[i-wt[maxDenseIndex]] == val[maxDenseIndex]: counter += 1 if counter>= wt[maxDenseIndex]: breaked = True # print(i) break else: counter = 0 if not breaked: return dp[W] else: start = i - wt[maxDenseIndex] + 1 times = (W - start) // wt[maxDenseIndex] index = (W - start) % wt[maxDenseIndex] + start return (times * val[maxDenseIndex] + dp[index]) # Driver CodeW = 100val = [10, 30, 20]wt = [5, 10, 15] print(unboundedKnapsackBetter(W, val, wt)) |
300
Time Complexity: O( N + min(wt[i], W) * N)
Auxiliary Space: O(W)
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