Aggressive Cows Maximum Minimum Distance in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of placing cows with minimum distance 3

Given the stalls at positions [1, 2, 4, 8, 9] and 3 cows, what is the maximum minimum distance between any two cows?

DSA Java

int[] stalls = {1, 2, 4, 8, 9};
int cows = 3;
// Function call
int result = maxMinDistance(stalls, cows);
System.out.println(result);

A 5

B 4

C 3

D 2

💡 Hint

Try placing cows starting from the first stall and check if minimum distance 3 is possible.

The maximum minimum distance is 3 because cows can be placed at positions 1, 4, and 8.

🧠 Conceptual

intermediate

Why sort the stalls before placing cows?

In the Aggressive Cows problem, why is it necessary to sort the stall positions before trying to place cows?

A To reduce the number of cows needed

B To increase the maximum distance arbitrarily

C To randomly place cows in stalls

D To ensure we check stall positions in increasing order for correct distance calculation

💡 Hint

Think about how distance between cows is calculated.

Sorting helps to check stalls in order and calculate distances correctly for binary search.

🔧 Debug

advanced

Identify the error in this distance check function

What error does this code produce when checking if cows can be placed with given minimum distance? boolean canPlace(int[] stalls, int cows, int dist) { int count = 1; int lastPos = stalls[0]; for (int i = 1; i < stalls.length; i++) { if (stalls[i] - lastPos >= dist) { count++; lastPos = stalls[i]; } } return count >= cows; }

A No error, returns correct boolean

B ArrayIndexOutOfBoundsException

C Always returns false

D Infinite loop

💡 Hint

Check loop and conditions carefully.

The function correctly counts cows placed with minimum distance and returns true if count meets cows.

❓ Predict Output

advanced

Output of maximum minimum distance for large input

Given stalls at positions [1, 2, 4, 8, 9, 12, 16, 20] and 4 cows, what is the maximum minimum distance?

DSA Java

int[] stalls = {1, 2, 4, 8, 9, 12, 16, 20};
int cows = 4;
int result = maxMinDistance(stalls, cows);
System.out.println(result);

A 6

B 7

C 5

D 4

💡 Hint

Try placing cows with distance 4 and check feasibility.

Cows can be placed at 1, 8, 12, and 20 with minimum distance 6.

🚀 Application

expert

Maximum minimum distance to place 5 cows in given stalls

You have stalls at positions [2, 5, 11, 15, 17, 20, 25] and need to place 5 cows. What is the maximum minimum distance between any two cows?

A 4

B 5

C 7

D 6

💡 Hint

Use binary search between 1 and max distance (23) to find feasible minimum distance.

The maximum minimum distance is 5 by placing cows at 2, 5, 11, 17, 25.

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