Choose maximum weight with given weight and value ratio - GeeksforGeeks

17 Jun, 2021
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. 
C++
Java
Python3
// C++ program to choose item with maximum// sum of weight under given constraint#include <bits/stdc++.h>using namespace std; // memoized recursive method to return maximum// weight with K as ratio of weight and valuesint maxWeightRec(int wt[], int val[], int K,                  map<pair<int, int>, int>& mp,                            int last, int diff){    //  base cases : if no item is remaining    if (last == -1)    {        if (diff == 0)            return 0;        else            return INT_MIN;    }     // first make pair with last chosen item and    // difference between weight and values    pair<int, int> tmp = make_pair(last, diff);    if (mp.find(tmp) != mp.end())        return mp[tmp];     /*  choose maximum value from following two        1) not selecting the current item and calling           recursively        2) selection current item, including the weight           and updating the difference before calling           recursively */    mp[tmp] = max(maxWeightRec(wt, val, K, mp, last - 1, diff),                   wt[last] + maxWeightRec(wt, val, K, mp,                   last - 1, diff + wt[last] - val[last] * K));     return mp[tmp];} // method returns maximum sum of weight with K// as ration of sum of weight and their valuesint maxWeight(int wt[], int val[], int K, int N){    map<pair<int, int>, int> mp;    return maxWeightRec(wt, val, K, mp, N - 1, 0);} //  Driver code to test above methodsint main(){    int wt[] = {4, 8, 9};    int val[] = {2, 4, 6};    int N = sizeof(wt) / sizeof(int);    int K = 2;     cout << maxWeight(wt, val, K, N);    return 0;}
// Java program to choose item with maximum// sum of weight under given constraint import java.awt.Point;import java.util.HashMap; class Test{    // memoized recursive method to return maximum    // weight with K as ratio of weight and values    static int maxWeightRec(int wt[], int val[], int K,                      HashMap<Point, Integer> hm,                                int last, int diff)    {        //  base cases : if no item is remaining        if (last == -1)        {            if (diff == 0)                return 0;            else                return Integer.MIN_VALUE;        }              // first make pair with last chosen item and        // difference between weight and values        Point tmp = new Point(last, diff);        if (hm.containsKey(tmp))            return hm.get(tmp);              /*  choose maximum value from following two            1) not selecting the current item and calling               recursively            2) selection current item, including the weight               and updating the difference before calling               recursively */       hm.put(tmp,Math.max(maxWeightRec(wt, val, K, hm, last - 1, diff),                       wt[last] + maxWeightRec(wt, val, K, hm,                       last - 1, diff + wt[last] - val[last] * K)));              return hm.get(tmp);    }          // method returns maximum sum of weight with K    // as ration of sum of weight and their values    static int maxWeight(int wt[], int val[], int K, int N)    {        HashMap<Point, Integer> hm = new HashMap<>();        return maxWeightRec(wt, val, K, hm, N - 1, 0);    }         // Driver method    public static void main(String args[])    {        int wt[] = {4, 8, 9};        int val[] = {2, 4, 6};                 int K = 2;              System.out.println(maxWeight(wt, val, K, wt.length));    }}// This code is contributed by Gaurav Miglani
# Python3 program to choose item with maximum# sum of weight under given constraintINT_MIN = -9999999999 def maxWeightRec(wt, val, K, mp, last, diff):         # memoized recursive method to return maximum    # weight with K as ratio of weight and values     # base cases : if no item is remaining    if last == -1:        if diff == 0:            return 0        else:            return INT_MIN     # first make pair with last chosen item and    # difference between weight and values    tmp = (last, diff)    if tmp in mp:        return mp[tmp]     # choose maximum value from following two    # 1) not selecting the current item and    #    calling recursively    # 2) selection current item, including    #    the weight and updating the difference    #    before calling recursively     mp[tmp] = max(maxWeightRec(wt, val, K, mp,                               last - 1, diff), wt[last] +                  maxWeightRec(wt, val, K, mp,                               last - 1, diff +                               wt[last] - val[last] * K))    return mp[tmp] def maxWeight(wt, val, K, N):         # method returns maximum sum of weight with K    # as ration of sum of weight and their values    return maxWeightRec(wt, val, K, {}, N - 1, 0) # Driver codeif __name__ == "__main__":    wt = [4, 8, 9]    val = [2, 4, 6]    N = len(wt)    K = 2    print(maxWeight(wt, val, K, N)) # This code is contributed# by vibhu4agarwal
Output:  
12
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vibhu4agarwal
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Dynamic Programming
Dynamic Programming
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