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++
// 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
// Java program to choose item with maximum// sum of weight under given constraintimport 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
# Python3 program to choose item with maximum# sum of weight under given constraintINT_MIN = -9999999999def 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 |
C#
// C# code for the above approachusing System;using System.Collections.Generic;class GFG{ // memoized recursive method to return maximum // weight with K as ratio of weight and values static int maxWeightRec(int[] wt, int[] val, int K, Dictionary<Tuple<int, int>, int> hm, int last, int diff) { // base cases : if no item is remaining if (last == -1) { if (diff == 0) return 0; else return int.MinValue; } // first make pair with last chosen item and // difference between weight and values Tuple<int, int> tmp = new Tuple<int, int>(last, diff); if (hm.ContainsKey(tmp)) return hm[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[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[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) { Dictionary<Tuple<int, int>, int> hm = new Dictionary<Tuple<int, int>, int>(); 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; Console.WriteLine(maxWeight(wt, val, K, wt.Length)); }}// This code is contributed by lokeshpotta20. |
Javascript
<script>// JavaScript program to choose item with maximum// sum of weight under given constraintconst INT_MIN = -9999999999function 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 codelet wt = [4, 8, 9]let val = [2, 4, 6]let N = wt.lengthlet K = 2document.write(maxWeight(wt, val, K, N),"</br>")// This code is contributed by shinjanpatra</script> |
Output:
12
The time complexity of the above code is O(N2), where N is the size of the array. This is because the map stores the results of the subproblems, which is done using a recursive approach.
The space complexity is also O(N2) as the map is used to store the intermediate results.
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