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.

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

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,

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

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++

// C++ program to choose item with maximum

// sum of weight under given constraint

#include

using namespace std;

// memoized recursive method to return maximum

// weight with K as ratio of weight and values

int maxWeightRec(int wt[], int val[], int K,

map& 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 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 values

int maxWeight(int wt[], int val[], int K, int N)

{

map mp;

return maxWeightRec(wt, val, K, mp, N - 1, 0);

}

// Driver code to test above methods

int main()

{

int wt[] = {4, 8, 9};

int val[] = {2, 4, 6};

int N = sizeof(wt) / sizeof(int);

int K = 2;

cout

return 0;

}

C++ 2

// 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 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)

function 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

let tmp = [last, diff]

if(mp.has(tmp))

return mp.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

mp.set(tmp, Math.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.get(tmp)

}

function 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, new Map(), N - 1, 0)

}

// Driver code

let wt = [4, 8, 9]

let val = [2, 4, 6]

let N = wt.length

let K = 2

document.write(maxWeight(wt, val, K, N),"")

// This code is contributed by shinjanpatra

Output:

12

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to [email protected] See your article appearing on the GeeksforGeeks main page and help other Geeks. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.