Aggressive Cows Problem GFG

 You are given an array with unique elements of stalls[], which denote the position of a stall. You are also given an integer k which denotes the number of aggressive cows. Your task is to assign stalls to k cows such that the minimum distance between any two of them is the maximum possible.


Examples :


Input: stalls[] = [1, 2, 4, 8, 9], k = 3

Output: 3

Explanation: The first cow can be placed at stalls[0], 

the second cow can be placed at stalls[2] and 

the third cow can be placed at stalls[3]. 

The minimum distance between cows, in this case, is 3, which also is the largest among all possible ways.

Input: stalls[] = [10, 1, 2, 7, 5], k = 3

Output: 4

Explanation: The first cow can be placed at stalls[0],

the second cow can be placed at stalls[1] and

the third cow can be placed at stalls[4].

The minimum distance between cows, in this case, is 4, which also is the largest among all possible ways.

Input: stalls[] = [2, 12, 11, 3, 26, 7], k = 5

Output: 1

Explanation: Each cow can be placed in any of the stalls, as the no. of stalls are exactly equal to the number of cows.

The minimum distance between cows, in this case, is 1, which also is the largest among all possible ways.



// { Driver Code Starts

// Initial Template for Java


import java.io.*;

import java.util.*;


class GFG {

    public static void main(String[] args) throws IOException {

        // Input reader setup

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));


        // Read number of test cases

        int tc = Integer.parseInt(br.readLine().trim());


        // Run each test case

        while (tc-- > 0) {


            // Read the stall positions as a space-separated string and convert to int array

            String[] str = br.readLine().trim().split(" ");

            int[] a = new int[str.length];

            for (int i = 0; i < str.length; i++) {

                a[i] = Integer.parseInt(str[i]);

            }


            // Read number of cows to place

            String[] nk = br.readLine().trim().split(" ");

            int k = Integer.parseInt(nk[0]);


            // Call the solution method

            Solution sln = new Solution();

            int ans = sln.aggressiveCows(a, k);


            // Print result for each test case

            System.out.println(ans);

            System.out.println("~");

        }

    }

}

// } Driver Code Ends


// User function Template for Java

class Solution {

    public static int aggressiveCows(int[] stalls, int k) {

        // Step 1: Sort the stall positions

        Arrays.sort(stalls);

        int n = stalls.length;


        // Step 2: Binary search for the largest minimum distance

        int start = 1; // Smallest possible distance between cows

        int end = stalls[n - 1] - stalls[0]; // Maximum distance between farthest stalls

        int ans = 0;


        // Step 3: Binary search loop

        while (start <= end) {

            int mid = start + (end - start) / 2; // Try this distance as the minimum


            int cowcount = 1; // Place the first cow in the first stall

            int cowposition = stalls[0]; // Last placed cow's position


            // Step 4: Try placing the rest of the cows

            for (int i = 1; i < n; i++) {

                // If the current stall is at least `mid` distance away from last cow

                if (stalls[i] >= cowposition + mid) {

                    cowcount++; // Place a new cow

                    cowposition = stalls[i]; // Update last cow's position

                }

            }


            // Step 5: Check if all cows can be placed with at least `mid` distance

            if (cowcount < k) {

                // Couldn't place all cows → try smaller distance

                end = mid - 1;

            } else {

                // Cows placed successfully → try for a bigger distance

                ans = mid;

                start = mid + 1;

            }

        }


        // Step 6: Return the best found distance

        return ans;

    }

}



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