AGGRCOW - Aggressive cows


Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,...,xN (0 <= xi <= 1,000,000,000).

His C (2 <= C <= N) cows don't like this barn layout and become aggressive towards each other once put into a stall. To prevent the cows from hurting each other, FJ wants to assign the cows to the stalls, such that the minimum distance between any two of them is as large as possible. What is the largest minimum distance?

Input

t – the number of test cases, then t test cases follows.
* Line 1: Two space-separated integers: N and C
* Lines 2..N+1: Line i+1 contains an integer stall location, xi

Output

For each test case output one integer: the largest minimum distance.

Example

Input:

1
5 3
1
2
8
4
9

Output:

3

Output details:

FJ can put his 3 cows in the stalls at positions 1, 4 and 8,
resulting in a minimum distance of 3.


hide comments
mohamedhamada: 2018-09-10 15:00:51

Got AC :D

timmymbek: 2018-08-12 04:41:31

hi

Last edit: 2018-08-12 06:09:57
king_tsar: 2018-07-27 10:35:56

Last edit: 2018-07-27 10:41:23
nitishyadav169: 2018-07-20 05:53:27

Last edit: 2018-08-17 22:06:54
ironhulk: 2018-07-12 08:23:56

why 1 4 9 is not the answer of given example??

alucard_01: 2018-07-03 07:18:36

Don't see any kind of solution.

foool: 2018-06-27 09:34:57

Good problem ..Must for beginners...

nesdi: 2018-06-26 13:17:16

Great one, If you know what is binary search but you aren't able to crack the trick.
Check Geeksforgeeks tutorial.

Last edit: 2018-06-26 13:17:54
sharansh12: 2018-06-17 06:44:11

AC in one go : )
P.S. i already knew method

karan_yadav: 2018-06-16 20:22:48

0.01s
Initially, I was getting 0.07s but after following optimizations reached 0.01s
1. std::ios::sync_with_stdio(false)
2. Traversed the array once and found the least difference between two consecutive array elements. Initialised the lo of binary search with this difference.
3. The max possible solution to this problem for any given case can be (n - c)/(c - 1). This formula can be easily formulated using pigeonhole principle. Initialise the hi of binary search with this value + 1 (just to be safe, integer division can cause truncation).

Also, I was using vector instead of an array. Switched to array and shaved off 0.02s.

Last edit: 2018-06-16 20:24:06

Added by:Roman Sol
Date:2005-02-16
Time limit:2s
Source limit:10000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All
Resource:USACO February 2005 Gold Division