Optimal way of filling 2 knapsacks?

Question

The dynamic programming algorithm to optimally fill a knapsack works well in the case of one knapsack. But is there an efficient known algorithm that will optimally fill 2 k

Answer 1

I will assume each of the n items can only be used once, and you must maximize your profit.

Original knapsack is dp[i] = best profit you can obtain for weight i

for i = 1 to n do
  for w = maxW down to a[i].weight do
    if dp[w] < dp[w - a[i].weight] + a[i].gain
      dp[w] = dp[w - a[i].weight] + a[i].gain

Now, since we have two knapsacks, we can use dp[i, j] = best profit you can obtain for weight i in knapsack 1 and j in knapsack 2

for i = 1 to n do
  for w1 = maxW1 down to a[i].weight do
    for w2 = maxW2 down to a[i].weight do
      dp[w1, w2] = max
                   {
                       dp[w1, w2], <- we already have the best choice for this pair
                       dp[w1 - a[i].weight, w2] + a[i].gain <- put in knapsack 1
                       dp[w1, w2 - a[i].weight] + a[i].gain <- put in knapsack 2
                   }

Time complexity is O(n * maxW1 * maxW2), where maxW is the maximum weight the knapsack can carry. Note that this isn't very efficient if the capacities are large.