Fruit Basket Packing - Google Top Interview Questions


Problem Statement :


You are given a two-dimensional list of integers fruits. 
Each fruits[i] contains [cost, size, total], meaning that fruit i costs cost each, each one has size of size, and there are total of total of them. 
You're also given k number of fruit baskets of capacity capacity.

You want to fill the fruit baskets with the following constraints in this order:

Each basket can only contain fruits of the same kind

Each basket should be as full as possible

Each basket should be as cheap as possible

Return the minimum cost required to fill as many baskets as possible.



Constraints



n ≤ 100,000 where n is the length of fruits

0 ≤ k, capacity < 2 ** 31

Example 1

Input

fruits = [
    [4, 2, 3],
    [5, 3, 2],
    [1, 3, 2]
]

k = 2

capacity = 4

Output

9


Explanation

We use two fruit 0s since it makes the first basket as full as possible for total size of 4, which costs 8. 
Then, we use one of fruit 2 even though packing fruit 1 would make it just as full because it's also 
cheaper. This costs 1 unit.


Solution :



title-img



                        Solution in C++ :

int solve(vector<vector<int>>& fruits, int k, int cap) {
    // Points to consider:
    // x -> remaining capacity of bucket after filling the i'th fruit
    // 1. x = (cap % size) if(size * total < cap) x = cap - (total*size)
    // value of x should be minimum -> as we want to fill the basket as full as possible
    // 2. cost per unit size should be minimum
    multiset<tuple<int, double, vector<int>>> mul;
    for (auto v : fruits) {
        double cost = v[0];
        double size = v[1];
        double costratio = (cost / size);
        int rem = cap % v[1];
        if (v[1] * v[2] < cap) rem = cap - (v[1] * v[2]);
        if (v[1] <= cap) mul.insert({rem, costratio, v});
    }
    int ans = 0;
    while (!mul.empty() && k > 0) {
        auto [rem, costratio, v] = *mul.begin();
        mul.erase(mul.begin());
        int fieb = min(v[2], (int)(cap / v[1]));  // fruit in each basket
        int bf = min(k, (int)(v[2] / fieb));      // baskets filled
        ans += fieb * bf * v[0];  // cost of adding the 'fieb' fruits to 'bf' baskets
        v[2] -= fieb * bf;        // remaining number of current fruits
        k -= bf;                  // remaining baskets
        int r = cap % v[1];
        if (v[1] * v[2] < cap) r = cap - (v[1] * v[2]);
        if (v[2] > 0)
            mul.insert({r, costratio,
                        v});  // Inserting the remaining fruits of this kind into the multiset
    }
    return ans;
}
                    

                        Solution in Java :

import java.util.*;

class Solution {
    class BasketItem {
        public int cost, cap, idx, canTake;
        BasketItem(int cost, int cap, int idx, int canTake) {
            this.cost = cost;
            this.cap = cap;
            this.idx = idx;
            this.canTake = canTake;
        }
    }

    public int solve(int[][] fruits, int k, int capacity) {
        PriorityQueue<BasketItem> pq =
            new PriorityQueue<BasketItem>((BasketItem a, BasketItem b) -> {
                if (a.cap != b.cap)
                    return b.cap - a.cap; // big cap first
                return a.cost - b.cost; // small cost second
            });

        for (int i = 0; i < fruits.length; i++) {
            int take = Math.min((int) Math.floor(capacity / fruits[i][1]), fruits[i][2]);
            if (take > 0) {
                int canTake = (int) Math.floor(fruits[i][2] / take);
                BasketItem item =
                    new BasketItem(take * fruits[i][0], take * fruits[i][1], i, canTake);
                pq.add(item);
                fruits[i][2] -= canTake * take;
            }
        }
        int ans = 0;
        while (k > 0 && !pq.isEmpty()) {
            BasketItem topItem = pq.remove();
            if (topItem.canTake >= k) {
                ans += k * topItem.cost;
                k = 0;
                break;
            } else {
                ans += topItem.canTake * topItem.cost;
                k -= topItem.canTake;
            }
            int i = topItem.idx;
            int take = Math.min((int) Math.floor(capacity / fruits[i][1]), fruits[i][2]);
            if (take > 0) {
                int canTake = (int) Math.floor(fruits[i][2] / take);
                BasketItem item =
                    new BasketItem(take * fruits[i][0], take * fruits[i][1], i, canTake);
                pq.add(item);
                fruits[i][2] -= canTake * take;
            }
        }
        return ans;
    }
}
                    

                        Solution in Python : 
                            
class Solution:
    def solve(self, fruits, k, size):
        baskets = []

        for fruit_cost, fruit_size, fruit_count in fruits:
            if fruit_size > size:
                continue

            full_basket_count = size // fruit_size
            full_baskets = 0 if not full_basket_count else fruit_count // full_basket_count
            full_basket_size = full_basket_count * fruit_size
            if full_basket_count:
                baskets.append((full_basket_size, full_basket_count * fruit_cost, full_baskets))

            rest_basket_count = fruit_count - full_baskets * full_basket_count
            rest_basket_size = rest_basket_count * fruit_size
            if rest_basket_count:
                baskets.append((rest_basket_size, rest_basket_count * fruit_cost, 1))

        total_cost = 0

        for _basket_size, basket_cost, baskets_available in sorted(
            baskets, key=lambda b: (-b[0], b[1])
        ):
            if k == 0:
                break
            baskets_used = min(k, baskets_available)
            total_cost += baskets_used * basket_cost
            k -= baskets_used

        return total_cost
                    

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