**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 :

` ````
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
```

## View More Similar Problems

## Palindromic Subsets

Consider a lowercase English alphabetic letter character denoted by c. A shift operation on some c turns it into the next letter in the alphabet. For example, and ,shift(a) = b , shift(e) = f, shift(z) = a . Given a zero-indexed string, s, of n lowercase letters, perform q queries on s where each query takes one of the following two forms: 1 i j t: All letters in the inclusive range from i t

View Solution →## Counting On a Tree

Taylor loves trees, and this new challenge has him stumped! Consider a tree, t, consisting of n nodes. Each node is numbered from 1 to n, and each node i has an integer, ci, attached to it. A query on tree t takes the form w x y z. To process a query, you must print the count of ordered pairs of integers ( i , j ) such that the following four conditions are all satisfied: the path from n

View Solution →## Polynomial Division

Consider a sequence, c0, c1, . . . , cn-1 , and a polynomial of degree 1 defined as Q(x ) = a * x + b. You must perform q queries on the sequence, where each query is one of the following two types: 1 i x: Replace ci with x. 2 l r: Consider the polynomial and determine whether is divisible by over the field , where . In other words, check if there exists a polynomial with integer coefficie

View Solution →## Costly Intervals

Given an array, your goal is to find, for each element, the largest subarray containing it whose cost is at least k. Specifically, let A = [A1, A2, . . . , An ] be an array of length n, and let be the subarray from index l to index r. Also, Let MAX( l, r ) be the largest number in Al. . . r. Let MIN( l, r ) be the smallest number in Al . . .r . Let OR( l , r ) be the bitwise OR of the

View Solution →## The Strange Function

One of the most important skills a programmer needs to learn early on is the ability to pose a problem in an abstract way. This skill is important not just for researchers but also in applied fields like software engineering and web development. You are able to solve most of a problem, except for one last subproblem, which you have posed in an abstract way as follows: Given an array consisting

View Solution →## Self-Driving Bus

Treeland is a country with n cities and n - 1 roads. There is exactly one path between any two cities. The ruler of Treeland wants to implement a self-driving bus system and asks tree-loving Alex to plan the bus routes. Alex decides that each route must contain a subset of connected cities; a subset of cities is connected if the following two conditions are true: There is a path between ever

View Solution →