Aggressive Cows Maximum Minimum Distance in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the array of stall positions.

DSA Java

Arrays.[1](stalls);

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Asort

Breverse

CcopyOf

Dfill

Attempts:

3 left

💡 Hint

Common Mistakes

Using reverse instead of sort causes wrong order.

Using fill or copyOf does not sort the array.

The Arrays.sort() method sorts the array in ascending order, which is necessary before placing cows to maximize minimum distance.

2fill in blank

medium

Complete the code to check if cows can be placed with at least given distance.

DSA Java

if (stalls[i] - lastPos >= [1]) {

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Amid

Bdistance

Ccount

DmaxDist

Attempts:

3 left

💡 Hint

Common Mistakes

Using count or maxDist instead of mid leads to incorrect checks.

Using distance without defining it causes errors.

The variable mid represents the current minimum distance we are testing between cows.

3fill in blank

hard

Fix the error in the binary search loop condition.

DSA Java

while (low [1] high) {

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A>

B>=

C<

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using > or >= causes the loop to never run or run incorrectly.

Using < might miss the optimal value when low == high.

The standard binary search loop condition is while (low <= high) to ensure the case when low equals high is handled.

4fill in blank

hard

Fill both blanks to update the search range after checking feasibility.

DSA Java

if (canPlaceCows(stalls, cows, [1])) {
    result = [2];
    low = mid + 1;
} else {
    high = mid - 1;
}

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Amid

Blow

Dhigh

Attempts:

3 left

💡 Hint

Common Mistakes

Using low or high instead of mid causes wrong updates.

Mixing variables leads to incorrect binary search.

If cows can be placed with distance mid, we update result to mid and try for a bigger distance by setting low = mid + 1.

5fill in blank

hard

Fill all three blanks to complete the method that checks if cows can be placed with given distance.

DSA Java

int count = 1;
int lastPos = stalls[0];
for (int i = 1; i < stalls.length; i++) {
    if (stalls[i] - lastPos >= [1]) {
        count++;
        lastPos = stalls[i];
        if (count == [2]) {
            return [3];
        }
    }
}
return false;

Drag options to blanks, or click blank then click option'

Amid

Bcows

Ctrue

Dfalse

Attempts:

3 left

💡 Hint

Common Mistakes

Returning false too early or using wrong variables.

Using count instead of cows for comparison.

The method checks if cows can be placed at least mid distance apart. When count reaches cows, it returns true.

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