A11594. Points on the line
编程题
普及/提高-
知识点
题目描述
We've got no test cases. A big olympiad is coming up. But the problemsetters' number one priority should be adding another problem to the round.
The diameter of a multiset of points on the line is the largest distance between two points from this set. For example, the diameter of the multiset ${1,3,2,1}$ is 2.
Diameter of multiset consisting of one point is 0.
You are given $n$ points on the line. What is the minimum number of points you have to remove, so that the diameter of the multiset of the remaining points will not exceed $d$ ?
The diameter of a multiset of points on the line is the largest distance between two points from this set. For example, the diameter of the multiset ${1,3,2,1}$ is 2.
Diameter of multiset consisting of one point is 0.
You are given $n$ points on the line. What is the minimum number of points you have to remove, so that the diameter of the multiset of the remaining points will not exceed $d$ ?
输入格式
The first line contains two integers $n$ and $d$ ( $1<=n<=100,0<=d<=100$ ) — the amount of points and the maximum allowed diameter respectively.
The second line contains $n$ space separated integers ( $1<=x_{i}<=100$ ) — the coordinates of the points.
The second line contains $n$ space separated integers ( $1<=x_{i}<=100$ ) — the coordinates of the points.
输出格式
Output a single integer — the minimum number of points you have to remove.
输入输出样例
输入 #1
3 1 2 1 4
输出 #1
1
输入 #2
3 0 7 7 7
输出 #2
0
输入 #3
6 3 1 3 4 6 9 10
输出 #3
3
说明/提示
In the first test case the optimal strategy is to remove the point with coordinate $4$ . The remaining points will have coordinates $1$ and $2$ , so the diameter will be equal to $2-1=1$ .
In the second test case the diameter is equal to $0$ , so its is unnecessary to remove any points.
In the third test case the optimal strategy is to remove points with coordinates $1$ , $9$ and $10$ . The remaining points will have coordinates $3$ , $4$ and $6$ , so the diameter will be equal to $6-3=3$ .
In the second test case the diameter is equal to $0$ , so its is unnecessary to remove any points.
In the third test case the optimal strategy is to remove points with coordinates $1$ , $9$ and $10$ . The remaining points will have coordinates $3$ , $4$ and $6$ , so the diameter will be equal to $6-3=3$ .