(Analysis by Nick Wu)

In this problem, we have several hay bales on the number line and a few exploding cows. We can send an exploding cow to a certain location on the line and it will cause all the hay bales to explode within a given radius. We want to figure out the smallest radius that the cows can explode at so that we can force all the hay bales to explode.

Note that if there is some radius $r$ such that it is possible for all hay bales to explode, then all the hay bales will explode for any radius $R > r$. Therefore, we can binary search for the minimum radius $r$.

To check if a given radius $r$ is feasible, consider the leftmost hay bale at location $x$. We have to place a cow somewhere between $x-r$ and $x+r$ in order to make that hay bale explode. To make as many other hay bales explode, it makes sense to move the cow to the right as much as possible, because there are no hay bales to the left of $x$ so shifting the cow to the right can only include more hay bales. Therefore, we would have the hay bale explode at location $x+r$ and all hay bales in between $x$ and $x+2r$ explode. We can then find the leftmost hay bale that is at location greater than $x+2r$ and repeat this process, then count the number of cows that we placed. If the number of cows used is less than or equal to $K$, then $r$ is feasible. Otherwise, the answer must be at least $r+1$.

Here is my Java code demonstrating this.

import java.io.*;
import java.util.*;
public class angry {
public static void main(String[] args) throws IOException {
PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("angry.out")));

int n = Integer.parseInt(st.nextToken());
int k = Integer.parseInt(st.nextToken());

int[] locations = new int[n];
for(int i = 0; i < n; i++) {
}

Arrays.sort(locations);

int min = 0;
int max = 500000000;
while(min != max) {
int mid = (min+max)/2;
int used = 0;
int last = 0;
while(last < n) {
used++;
int curr = last+1;
while(curr < n && locations[curr] - locations[last] <= 2*mid) {
curr++;
}
last = curr;
}
if(used <= k) {
max = mid;
}
else {
min = mid+1;
}
}
pw.println(min);
pw.close();
}

}