Minimum cost to fill given weight in a bag

You are given a bag of size w kg and you are provided costs of packets different weights of oranges in array cost[], find the minimum total cost to buy exactly w kg oranges.

• The cost of 1 kg orange is present at index 0 and in general arr[i] has cost of (i+1) kg orange..
• cost[i] = -1 means that 'i+1' kg packet of orange is unavailable
• If it is not possible to buy exactly w kg oranges then print -1. It may be assumed that there is an infinite supply of all available packet types.

Examples:

Input : w = 5, cost[] = [20, 10, 4, 50, 100]
Output : 14
Explanation: We can choose two oranges to minimize cost. First orange of 2 Kg and cost 10. Second orange of 3 Kg and cost 4.

Input : w = 5, cost[] = [1, 10, 4, 50, 100]
Output : 5
Explanation : We can choose five oranges of weight 1 kg.

Input : w = 5, cost[] = [1, 2, 3, 4, 5]
Output : 5
Explanation : Costs of 1, 2, 3, 4 and 5 kg packets are 1, 2, 3, 4 and 5 Rs respectively. We choose packet of 5 kg having cost 5 for minimum cost to get 5 Kg oranges.

Input : w = 5, cost[] = [-1, -1, 4, 5, -1 ]
Output : -1
Explanation: Packets of size 1, 2 and 5 kg are unavailable because they have cost -1. Cost of 3 kg packet is 4 Rs and of 4 kg is 5 Rs. Here we have only weights 3 and 4 so by using these two we can not make exactly W kg weight, therefore answer is -1.

Using Unbounded Knapsack - O(n * w) Time and O(n * w) Space

This problem is can be reduced to Unbounded Knapsack. So in the cost array, we first ignore those packets which are not available i.e; cost is -1 and then traverse the cost array and create two array val[] for storing the cost of 'i' kg packet of orange and wt[] for storing weight of the corresponding packet. Suppose cost[i] = 50 so the weight of the packet will be i and the cost will be 50. 

• Create matrix min_cost[n+1][W+1], where n is number of distinct weighted packets of orange and W is the maximum capacity of the bag.
• Initialize the 0th row with INF (infinity) and 0th Column with 0.
• Now fill the matrix: if wt[i-1] > j then min_cost[i][j] = min_cost[i-1][j]. Else min_cost[i][j] = min(min_cost[i-1][j], val[i-1] + min_cost[i][j-wt[i-1]]);
• If min_cost[n][W] == INF then output will be -1 because this means that we can not make weight W by using these weights else output will be min_cost[n][W].

Below is the implementation of the above algorithm:

5

Space Optimization (1D DP - Unbounded Knapsack) - O(n * w) Time and O(w) Space

If we take a closer look at this problem, we may notice that this is a variation of Rod Cutting Problem. Instead of doing maximization, here we need to do minimization.

• dp[i] : minimum cost to get exactly weight i
• For every weight i, try all packets with weights less than it.
• If packet is available and usable, update minimum cost
• This is Unbounded Knapsack because we can reuse packets

14