Given weights and values of n items and a value k. We need to choose a subset of these items in such a way that ratio of the sum of weight and sum of values of chosen items is K and sum of weight is maximum among all possible subset choices.
Input : weight = [4, 8, 9] values = [2, 4, 6] K = 2 Output : 12 We can choose only first and second item only, because (4 + 8) / (2 + 4) = 2 which is equal to K we can't include third item with weight 9 because then ratio condition won't be satisfied so result will be (4 + 8) = 12
We can solve this problem using dynamic programming. We can make a 2 state dp where dp(i, j) will store maximum possible sum of weights under given conditions when total items are N and required ratio is K.
Now in two states of dp, we will store the last item chosen and the difference between sum of weight and sum of values. We will multiply item values by K so that second state of dp will actually store (sum of weight – K*(sum of values)) for chosen items. Now we can see that our answer will be stored in dp(N-1, 0) because as last item is (N-1)th so all items are being considered and difference between sum of weight and K*(sum of values) is 0 that means sum of weight and sum of values has a ratio K.
After defining above dp state we can write transition among states simply as shown below,
dp(last, diff) = max (dp(last - 1, diff), dp(last-1, diff + wt[last] - val[last]*K)) dp(last – 1, diff) represents the condition when current item is not chosen and dp(last – 1, diff + wt[last] – val[last] * K)) represents the condition when current item is chosen so difference is updated with weight and value of current item.
In below code a top-down approach is used for solving this dynamic programming and for storing dp states a map is used because the difference can be negative also and the 2D array can create problem in that case and special care need to be taken.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
This article is attributed to GeeksforGeeks.org