Aggressive Cows Maximum Minimum Distance in DSA Java - Build from Scratch

Aggressive Cows Maximum Minimum Distance

📖 Scenario: You are managing a farm with several stalls placed along a straight line. You want to place cows in these stalls such that the minimum distance between any two cows is as large as possible. This helps reduce fights between cows.

🎯 Goal: Build a program that finds the largest minimum distance possible between cows placed in the stalls.

📋 What You'll Learn

Create an array called stalls with given stall positions

Create an integer variable cows for the number of cows to place

Write a function canPlaceCows to check if cows can be placed with a given minimum distance

Use binary search to find the maximum minimum distance

Print the maximum minimum distance found

💡 Why This Matters

🌍 Real World

This problem models placing resources or devices in locations to maximize minimum spacing, useful in network design, agriculture, and facility planning.

💼 Career

Understanding binary search on answer space and greedy algorithms is important for software engineering interviews and optimization problems.

1

Create the stalls array and cows variable

Create an integer array called stalls with these exact values: 1, 2, 8, 4, 9. Create an integer variable called cows and set it to 3.

DSA Java

// Create the stalls array and cows variable
// Your code here

Hint

Use int[] stalls = {1, 2, 8, 4, 9}; to create the array and int cows = 3; for the number of cows.

2

Sort the stalls array

Sort the stalls array in ascending order using Arrays.sort(stalls);.

DSA Java

int[] stalls = {1, 2, 8, 4, 9};
int cows = 3;
// Sort the stalls array
// Your code here

Hint

Use Arrays.sort(stalls); to sort the array.

3

Write the canPlaceCows function

Write a function called canPlaceCows that takes int[] stalls, int cows, and int distance as parameters. It returns true if it is possible to place all cows in the stalls such that the minimum distance between any two cows is at least distance. Otherwise, it returns false. Use a greedy approach to place cows starting from the first stall.

DSA Java

import java.util.Arrays;

int[] stalls = {1, 2, 8, 4, 9};
int cows = 3;
Arrays.sort(stalls);

// Write the canPlaceCows function below
// Your code here

Hint

Start placing the first cow at stalls[0]. Then place next cows only if the distance from last placed cow is at least distance.

4

Find and print the maximum minimum distance

Use binary search to find the largest minimum distance between cows. Set low to 0 and high to the difference between the last and first stall positions. While low is less than or equal to high, calculate mid as the middle distance. Use canPlaceCows to check if cows can be placed with mid. If yes, update result to mid and set low to mid + 1. Otherwise, set high to mid - 1. Finally, print the result.

DSA Java

import java.util.Arrays;

int[] stalls = {1, 2, 8, 4, 9};
int cows = 3;
Arrays.sort(stalls);

public static boolean canPlaceCows(int[] stalls, int cows, int distance) {
    int count = 1; // place first cow at first stall
    int lastPosition = stalls[0];
    for (int i = 1; i = distance) {
            count++;
            lastPosition = stalls[i];
            if (count == cows) {
                return true;
            }
        }
    }
    return false;
}

// Use binary search to find maximum minimum distance and print it
// Your code here

Hint

Use binary search between 0 and max stall distance. Use canPlaceCows to check feasibility. Print the final result.

Practice

1. What is the main goal of the Aggressive Cows Maximum Minimum Distance problem?

easy

A. To place cows in stalls so that the minimum distance between any two cows is maximized

B. To find the maximum number of cows that can fit in a single stall

C. To minimize the total distance cows travel between stalls

D. To arrange cows so that all are placed in consecutive stalls

Solution

1. Step 1: Understand the problem objective

   The problem asks to place cows in stalls such that the smallest distance between any two cows is as large as possible.

2. Step 2: Identify the key goal

   This means maximizing the minimum distance between cows, not minimizing or counting cows in one stall.

3. Final Answer:

   To place cows in stalls so that the minimum distance between any two cows is maximized -> Option A

4. Quick Check:

   Maximize minimum distance = place cows far apart [OK]

Hint: Maximize the smallest gap between cows placed [OK]

Common Mistakes:

• Thinking the goal is to minimize distance
• Trying to put all cows in one stall
• Confusing maximum distance with maximum number of cows

2. Which of the following Java code snippets correctly initializes the binary search boundaries for the aggressive cows problem given an array stalls sorted in ascending order?

easy

A. int low = stalls.length; int high = stalls[0];

B. int low = 0; int high = stalls[stalls.length - 1] - stalls[0];

C. int low = 1; int high = stalls.length;

D. int low = stalls[0]; int high = stalls[stalls.length - 1];

Solution

1. Step 1: Determine the search space for distance

   The minimum distance can be 0 and maximum distance is the difference between the last and first stall positions.

2. Step 2: Set binary search boundaries accordingly

   So, low = 0 and high = stalls[stalls.length - 1] - stalls[0] correctly sets the search range.

3. Final Answer:

   int low = 0; int high = stalls[stalls.length - 1] - stalls[0]; -> Option B

4. Quick Check:

   Distance range = 0 to max stall gap [OK]

Hint: Low = 0, High = max stall position difference [OK]

Common Mistakes:

• Setting low or high to stall positions instead of distances
• Using stall count as high boundary
• Reversing low and high values

3. Given the sorted stalls array [1, 2, 8, 12, 17] and 3 cows, what is the maximum minimum distance between any two cows?

medium

A. 5

B. 6

C. 8

D. 7

Solution

1. Step 1: Check feasibility for distances

   Try placing cows with minimum distance 7: Place first cow at 1, second at 8 (distance 7), third at 17 (distance 9). All distances >= 7, so 7 is possible.

2. Step 2: Check if larger distance is possible

   Try 8: Place first at 1, second at 8 (distance 7) fails, so 8 not possible. Therefore, 7 is possible but we need to check if 6 is also possible.

3. Step 3: Check 6

   Try 6: Place first at 1, second at 8 (distance 7), third at 17 (distance 9). All distances >= 6, so 6 is possible.

4. Step 4: Maximum minimum distance is 7, but since 7 is possible, the maximum minimum distance is 7.

5. Final Answer:

   7 -> Option D

6. Quick Check:

   Max min distance = 7 [OK]

Hint: Test distances by placing cows greedily [OK]

Common Mistakes:

• Choosing the largest stall difference directly
• Not verifying placement feasibility
• Confusing maximum distance with stall count

4. Identify the error in this Java method snippet for checking if cows can be placed with at least dist distance:

boolean canPlace(int[] stalls, int cows, int dist) {
  int count = 1, lastPos = stalls[0];
  for (int i = 1; i < stalls.length; i++) {
    if (stalls[i] - lastPos >= dist) {
      count++;
      lastPos = stalls[i];
    }
  }
  return count >= cows;
}

medium

A. No error; the method correctly checks placement feasibility

B. The initial count should be 0 instead of 1

C. The loop should start from i = 0 instead of i = 1

D. The return condition should be count > cows instead of count >= cows

Solution

1. Step 1: Analyze initial count and last position

   Count starts at 1 because first cow is placed at stalls[0]. lastPos is set to stalls[0] correctly.

2. Step 2: Check loop and return condition

   Loop starts at 1 to check next stalls. If distance condition meets, count increments. Return true if count >= cows means enough cows placed.

3. Final Answer:

   No error; the method correctly checks placement feasibility -> Option A

4. Quick Check:

   Initial count 1 and correct loop start [OK]

Hint: First cow placed before loop; count starts at 1 [OK]

Common Mistakes:

• Starting count at 0 causing off-by-one error
• Loop starting at 0 causing redundant check
• Using > instead of >= in return condition

5. You have stalls at positions [0, 7, 14, 21, 28] and 4 cows. Using binary search, what is the maximum minimum distance you can place the cows apart?

hard

A. 5

B. 6

C. 7

D. 8

Solution

1. Step 1: Set binary search boundaries

   Minimum distance low = 0, high = 28 - 0 = 28.

2. Step 2: Use binary search to find max min distance

   Check mid = 7: Place cows at 0, 7 (distance 7), 14 (7), 21 (7). Cows placed: 4 successfully.

3. Step 3: Try higher distance 8

   Place cows at 0, 14 (14), 28 (14) only 3 cows placed, fails.

4. Final Answer:

   7 -> Option C

5. Quick Check:

   Max min distance = 7 by binary search [OK]

Hint: Use binary search on distance, check feasibility each step [OK]

Common Mistakes:

• Not updating binary search boundaries correctly
• Miscounting cows placed at each distance
• Choosing distance without feasibility check