## Hackerrank - Bonetrousle Solution

Here's a *humerus* joke:

Why did Papyrus the skeleton go to the store by himself? Because he hadno bodyto go with him!

Did you like it? Don't worry, I've got a ton more. A skele-*ton*.

Once upon a time, Papyrus the skeleton went to buy some pasta from the store. The store's inventory is *bare-bones* and they only sell one thing — boxes of uncooked spaghetti! The store always stocks exactly boxes of pasta, and each box is numbered sequentially from to . This box number also corresponds to the number of sticks of spaghetti in the box, meaning the first box contains stick, the second box contains sticks, the third box contains sticks, …, and the box contains sticks. Because they only stock one box of each kind, the store has a *tendon*-cy to sell out of spaghetti.

During each trip to the store, Papyrus likes to buy exactly sticks of spaghetti by purchasing exactly boxes (no more, no less). Not sure *which* boxes to purchase, Papyrus calls *Sherlock Bones* for help but he's also stumped! Do you have the *guts* to solve this puzzle?

Given the values of , , and for trips to the store, determine which boxes Papyrus must purchase during each trip. For each trip, print a single line of distinct space-separated integers denoting the box number for each box of spaghetti Papyrus purchases (recall that the store only has *one* box of each kind). If it's not possible to buy sticks of spaghetti by purchasing boxes, print `-1`

instead.

For example, Papyrus wants to purchase sticks of spaghetti in boxes and the store has different box sizes. He can buy boxes of sizes , , and other combinations. Any of the combinations will work.

**Function Description**

Complete the *bonetrousle* function in the editor below. It should return an array of integers.

bonetrousle has the following parameter(s):

*n*: the integer number of sticks to buy*k*: the integer number of box sizes the store carries*b*: the integer number of boxes to buy

**Input Format**

The first line contains a single integer , the number of trips to the store.

Each of the next lines contains three space-separated integers , and , the number of sticks to buy, the number of boxes for sale and the number of boxes to buy on this trip to the store.

**Constraints**

**Output Format**

For each trip to the store:

- If there is no solution, print
`-1`

on a new line. - If there is a solution, print a single line of distinct space-separated integers where each integer denotes the numbers of noodles in each box that Papyrus must purchase.

If there are multiple possible solutions, *you can print any one of them.* Do not print any leading or trailing spaces or extra newlines.

**Sample Input**

```
4
12 8 3
10 3 3
9 10 2
9 10 2
```

**Sample Output**

```
2 3 7
-1
5 4
1 8
```

**Explanation**

Papyrus makes the following trips to the store:

He wants to buy exactly boxes of spaghetti and have a total number of sticks. During this trip, the store has boxes of spaghetti sticks where the first box has stick, the second box has sticks, the third box has sticks, and so on. One possible solution would be the following:

Papyrus can buy the -stick, -stick, and -stick boxes for the total of sticks. *Note that this is not the only valid solution; other valid solutions are acceptable.*

- He wants to buy exactly boxes of spaghetti and have a total number of sticks. Because the store only has three boxes in stock containing , , and sticks of spaghetti, it's not possible for Papyrus to buy sticks of spaghetti as buying all three boxes would only yield sticks (which is less than the that he wanted to purchase). Thus, we print
`-1`

on a new line. - The third and fourth trips to the store both contain the same values (, , ). This illustrates that there may be multiple solutions for any given trip to the store and any valid solution is acceptable.

### Solution in Python

```
def Sn(n,a,d=1):
return (n*(2*a+(n-1)*d))//2
def bonetrousle(n, k, b):
lower = Sn(b,1)
upper = Sn(b,k-b+1)
if n<lower or n>upper:
return [-1]
total = Sn(b,1)
arr = list(range(1,b+1))
while total<n:
add = min(k-arr[b-1],n-total)
arr[b-1]+=add
total+=add
b-=1
k-=1
return (i for i in arr)
for x in range(int(input())):
n,k,b = map(int,input().split())
print((*(bonetrousle(n,k,b))))
```

*The explanation given in editorial is already good. So I won't add my explanation for this question. Just submit answer once and refresh the page to unlock editorial*

Editorial's Answer