Choose maximum weight with given weight and value ratio – Code Tip

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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.

1. Input : weight[] = [4, 8, 9]
2. values[] = [2, 4, 6]
3. K = 2
4. Output : 12
5. We can choose only first and second item only,
6. because (4 + 8) / (2 + 4) = 2 which is equal to K
7. we can't include third item with weight 9 because
8. then ratio condition won't be satisfied so result
9. will be (4 + 8) = 12

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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,

1. dp(last, diff) = max (dp(last - 1, diff),
2. dp(last-1, diff + wt[last] - val[last]*K))
3. dp(last – 1, diff) represents the condition when current
4. item is not chosen and
5. dp(last – 1, diff + wt[last] – val[last] * K)) represents
6. the condition when current item is chosen so difference
7. 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.

## C++

1. // C++ program to choose item with maximum
2. // sum of weight under given constraint
3. #include
4. using namespace std;
5. // memoized recursive method to return maximum
6. // weight with K as ratio of weight and values
7. int maxWeightRec(int wt[], int val[], int K,
8. map& mp,
9. int last, int diff)
10. {
11. // base cases : if no item is remaining
12. if (last == -1)
13. {
14. if (diff == 0)
15. return 0;
16. else
17. return INT_MIN;
18. }
19. // first make pair with last chosen item and
20. // difference between weight and values
21. pair tmp = make_pair(last, diff);
22. if (mp.find(tmp) != mp.end())
23. return mp[tmp];
24. /* choose maximum value from following two
25. 1) not selecting the current item and calling
26. recursively
27. 2) selection current item, including the weight
28. and updating the difference before calling
29. recursively */
30. mp[tmp] = max(maxWeightRec(wt, val, K, mp, last - 1, diff),
31. wt[last] + maxWeightRec(wt, val, K, mp,
32. last - 1, diff + wt[last] - val[last] * K));
33. return mp[tmp];
34. }
35. // method returns maximum sum of weight with K
36. // as ration of sum of weight and their values
37. int maxWeight(int wt[], int val[], int K, int N)
38. {
39. map mp;
40. return maxWeightRec(wt, val, K, mp, N - 1, 0);
41. }
42. // Driver code to test above methods
43. int main()
44. {
45. int wt[] = {4, 8, 9};
46. int val[] = {2, 4, 6};
47. int N = sizeof(wt) / sizeof(int);
48. int K = 2;
49. cout
50. return 0;
51. }

## C++ 2

1. // Java program to choose item with maximum
2. // sum of weight under given constraint
3. import java.awt.Point;
4. import java.util.HashMap;
5. class Test
6. {
7. // memoized recursive method to return maximum
8. // weight with K as ratio of weight and values
9. static int maxWeightRec(int wt[], int val[], int K,
10. HashMap hm,
11. int last, int diff)
12. {
13. // base cases : if no item is remaining
14. if (last == -1)
15. {
16. if (diff == 0)
17. return 0;
18. else
19. return Integer.MIN_VALUE;
20. }
21. // first make pair with last chosen item and
22. // difference between weight and values
23. Point tmp = new Point(last, diff);
24. if (hm.containsKey(tmp))
25. return hm.get(tmp);
26. /* choose maximum value from following two
27. 1) not selecting the current item and calling
28. recursively
29. 2) selection current item, including the weight
30. and updating the difference before calling
31. recursively */
32. hm.put(tmp,Math.max(maxWeightRec(wt, val, K, hm, last - 1, diff),
33. wt[last] + maxWeightRec(wt, val, K, hm,
34. last - 1, diff + wt[last] - val[last] * K)));
35. return hm.get(tmp);
36. }
37. // method returns maximum sum of weight with K
38. // as ration of sum of weight and their values
39. static int maxWeight(int wt[], int val[], int K, int N)
40. {
41. HashMap hm = new HashMap();
42. return maxWeightRec(wt, val, K, hm, N - 1, 0);
43. }
44. // Driver method
45. public static void main(String args[])
46. {
47. int wt[] = {4, 8, 9};
48. int val[] = {2, 4, 6};
49. int K = 2;
50. System.out.println(maxWeight(wt, val, K, wt.length));
51. }
52. }
53. // This code is contributed by Gaurav Miglani

## Python3

1. # Python3 program to choose item with maximum
2. # sum of weight under given constraint
3. INT_MIN = -9999999999
4. def maxWeightRec(wt, val, K, mp, last, diff):
5. # memoized recursive method to return maximum
6. # weight with K as ratio of weight and values
7. # base cases : if no item is remaining
8. if last == -1:
9. if diff == 0:
10. return 0
11. else:
12. return INT_MIN
13. # first make pair with last chosen item and
14. # difference between weight and values
15. tmp = (last, diff)
16. if tmp in mp:
17. return mp[tmp]
18. # choose maximum value from following two
19. # 1) not selecting the current item and
20. # calling recursively
21. # 2) selection current item, including
22. # the weight and updating the difference
23. # before calling recursively
24. mp[tmp] = max(maxWeightRec(wt, val, K, mp,
25. last - 1, diff), wt[last] +
26. maxWeightRec(wt, val, K, mp,
27. last - 1, diff +
28. wt[last] - val[last] * K))
29. return mp[tmp]
30. def maxWeight(wt, val, K, N):
31. # method returns maximum sum of weight with K
32. # as ration of sum of weight and their values
33. return maxWeightRec(wt, val, K, {}, N - 1, 0)
34. # Driver code
35. if __name__ == "__main__":
36. wt = [4, 8, 9]
37. val = [2, 4, 6]
38. N = len(wt)
39. K = 2
40. print(maxWeight(wt, val, K, N))
41. # This code is contributed
42. # by vibhu4agarwal

## Javascript

1. // JavaScript program to choose item with maximum
2. // sum of weight under given constraint
3. const INT_MIN = -9999999999
4. function maxWeightRec(wt, val, K, mp, last, diff){
5. // memoized recursive method to return maximum
6. // weight with K as ratio of weight and values
7. // base cases : if no item is remaining
8. if(last == -1){
9. if(diff == 0)
10. return 0
11. else
12. return INT_MIN
13. }
14. // first make pair with last chosen item and
15. // difference between weight and values
16. let tmp = [last, diff]
17. if(mp.has(tmp))
18. return mp.get(tmp)
19. // choose maximum value from following two
20. // 1) not selecting the current item and
21. // calling recursively
22. // 2) selection current item, including
23. // the weight and updating the difference
24. // before calling recursively
25. mp.set(tmp, Math.max(maxWeightRec(wt, val, K, mp,
26. last - 1, diff), wt[last] +
27. maxWeightRec(wt, val, K, mp,
28. last - 1, diff +
29. wt[last] - val[last] * K)))
30. return mp.get(tmp)
31. }
32. function maxWeight(wt, val, K, N){
33. // method returns maximum sum of weight with K
34. // as ration of sum of weight and their values
35. return maxWeightRec(wt, val, K, new Map(), N - 1, 0)
36. }
37. // Driver code
38. let wt = [4, 8, 9]
39. let val = [2, 4, 6]
40. let N = wt.length
41. let K = 2
42. document.write(maxWeight(wt, val, K, N),"")
43. // This code is contributed by shinjanpatra

Output:

`12`

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