### 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;
double size = v;
double costratio = (cost / size);
int rem = cap % v;
if (v * v < cap) rem = cap - (v * v);
if (v <= 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, (int)(cap / v));  // fruit in each basket
int bf = min(k, (int)(v / fieb));      // baskets filled
ans += fieb * bf * v;  // cost of adding the 'fieb' fruits to 'bf' baskets
v -= fieb * bf;        // remaining number of current fruits
k -= bf;                  // remaining baskets
int r = cap % v;
if (v * v < cap) r = cap - (v * v);
if (v > 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 {
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) {
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]), fruits[i]);
if (take > 0) {
int canTake = (int) Math.floor(fruits[i] / take);
new BasketItem(take * fruits[i], take * fruits[i], i, canTake);
fruits[i] -= canTake * take;
}
}
int ans = 0;
while (k > 0 && !pq.isEmpty()) {
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]), fruits[i]);
if (take > 0) {
int canTake = (int) Math.floor(fruits[i] / take);
new BasketItem(take * fruits[i], take * fruits[i], i, canTake);
fruits[i] -= canTake * take;
}
}
return ans;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, fruits, k, size):

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

total_cost = 0

):
if k == 0:
break

```

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