Aggressive Cows Maximum Minimum Distance in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the array of stall positions.

DSA C#

Array.[1](stalls);

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

ASort

BCopy

CClear

DReverse

Attempts:

3 left

💡 Hint

Common Mistakes

Using Array.Reverse instead of Array.Sort.

Forgetting to sort the stalls before processing.

We use Array.Sort to arrange stall positions in ascending order, which is essential for the binary search approach.

2fill in blank

medium

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

DSA C#

int count = 1;
int lastPosition = stalls[0];
for (int i = 1; i < stalls.Length; i++) {
    if (stalls[i] - lastPosition >= [1]) {
        count++;
        lastPosition = stalls[i];
    }
}
return count >= cows;

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

Astalls.Length

Bcows

Cdistance

DlastPosition

Attempts:

3 left

💡 Hint

Common Mistakes

Using cows instead of distance in the if condition.

Comparing with lastPosition instead of distance.

We check if the gap between current stall and last placed cow is at least 'distance'.

3fill in blank

hard

Fix the error in the binary search loop condition to find the maximum minimum distance.

DSA C#

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

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A<

B<=

C>=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '<=' causing missing the last check.

Using '>' or '>=' which reverses the logic.

We use 'low <= high' to ensure the binary search covers all possible distances including when low equals high.

4fill in blank

hard

Fill both blanks to initialize the binary search boundaries correctly.

DSA C#

int low = [1];
int high = [2];

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A1

B0

Cstalls[stalls.Length - 1]

Dstalls[stalls.Length - 1] - stalls[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Setting low to 0 which allows zero distance.

Setting high to the last stall position instead of the difference.

The minimum distance can start from 1, and the maximum distance is the difference between the last and first stall after sorting.

5fill in blank

hard

Fill all three blanks to complete the main function that finds the maximum minimum distance.

DSA C#

Array.Sort(stalls);
int low = [1];
int high = [2];
int result = 0;
while (low <= high) {
    int mid = (low + high) / 2;
    if ([3](stalls, cows, mid)) {
        result = mid;
        low = mid + 1;
    } else {
        high = mid - 1;
    }
}
return result;

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

A1

Bstalls[stalls.Length - 1] - stalls[0]

CCanPlaceCows

Dcows

Attempts:

3 left

💡 Hint

Common Mistakes

Using cows instead of CanPlaceCows in the if condition.

Setting low or high incorrectly.

We sort stalls, set low to 1, high to max gap, and use CanPlaceCows to check feasibility.

Practice

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

easy

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

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 place cows in stalls so that the minimum distance between any two cows is as large as possible

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 goal

   This means maximizing the minimum distance between cows, not minimizing travel or placing all cows consecutively.

3. Final Answer:

   To place cows in stalls so that the minimum distance between any two cows is as large as possible -> Option D

4. Quick Check:

   Maximize minimum distance = To place cows in stalls so that the minimum distance between any two cows is as large as possible [OK]

Hint: Focus on maximizing the smallest gap between cows [OK]

Common Mistakes:

• Thinking the goal is to minimize distance
• Confusing number of cows with distance
• Assuming cows must be in consecutive stalls

2. Which of the following C# code snippets correctly declares an array of stall positions for the Aggressive Cows problem?

easy

A. int stalls = {1, 2, 4, 8, 9};

B. int[] stalls = (1, 2, 4, 8, 9);

C. int[] stalls = new int[] {1, 2, 4, 8, 9};

D. int stalls[] = {1, 2, 4, 8, 9};

Solution

1. Step 1: Recall C# array declaration syntax

   In C#, arrays are declared with type[], then initialized with new type[] {values}.

2. Step 2: Check each option

   int[] stalls = new int[] {1, 2, 4, 8, 9}; uses correct syntax: int[] stalls = new int[] {1, 2, 4, 8, 9};. Options A, C, and D have invalid syntax.

3. Final Answer:

   int[] stalls = new int[] {1, 2, 4, 8, 9}; -> Option C

4. Quick Check:

   C# array syntax = int[] stalls = new int[] {1, 2, 4, 8, 9}; [OK]

Hint: Use 'type[] name = new type[] {values};' for arrays in C# [OK]

Common Mistakes:

• Omitting 'new' keyword
• Using parentheses instead of braces
• Incorrect array variable declaration

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

medium

A. 4

B. 5

C. 3

D. 2

Solution

1. Step 1: Sort stall positions

   The stalls are already sorted: [1, 2, 4, 8, 9].

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

   Try distance 4: place first cow at 1, second at 5 or more (4 is not a stall, next is 8), second at 8 (distance 7), third at 9 (distance 1 from 8) - third cow too close. Actually, try placing cows with distance 4: first at 1, second at 8 (distance 7), third at 9 (distance 1) - third cow too close. So distance 4 is not feasible. Try distance 3: place cows at 1, 4, 8 - all distances >=3, feasible.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   Max min distance = 4 [OK]

Hint: Try placing cows with binary search on distance [OK]

Common Mistakes:

• Not sorting stalls before placement
• Choosing distance too large without checking feasibility
• Confusing maximum and minimum distances

4. Consider this code snippet for checking if cows can be placed with a given minimum distance:

bool CanPlaceCows(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;
}

What is the error in this code?

medium

A. Extra semicolon after the if condition causing incorrect loop behavior

B. Incorrect initialization of count variable

C. Wrong return condition should be count > cows

D. Missing sorting of stalls array before placement

Solution

1. Step 1: Identify syntax issues in the loop

   There is an extra semicolon after the if condition, which ends the if statement prematurely.

2. Step 2: Understand effect of extra semicolon

   This causes the count++ and lastPos update to run every iteration, ignoring the if condition, leading to wrong placement logic.

3. Final Answer:

   Extra semicolon after the if condition causing incorrect loop behavior -> Option A

4. Quick Check:

   Extra semicolon breaks if condition [OK]

Hint: Watch for stray semicolons after if statements [OK]

Common Mistakes:

• Ignoring syntax errors like extra semicolons
• Assuming count initialization is wrong
• Forgetting to sort stalls (not shown here)

5. You have stalls at positions [2, 5, 11, 15, 17, 20] and 4 cows. Using the aggressive cows approach, what is the maximum minimum distance you can place the cows apart?

hard

A. 5

B. 4

C. 6

D. 7

Solution

1. Step 1: Sort stalls and set search range

   Stalls sorted: [2, 5, 11, 15, 17, 20]. Minimum distance can be between 1 and 18 (20-2).

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

   Try mid = 5: place first cow at 2, second at 11 (distance 9), third at 17 (distance 6), fourth at 20 (distance 3 from 17) - too close, only 3 cows placed. Try mid = 4: place cows at 2, 5, 11, 15 - distances 3, 6, 4 all >=4, but 4 cows placed. Try mid = 6: only 3 cows placed. So max distance is 4.

3. Final Answer:

   4 -> Option B

4. Quick Check:

   Max min distance = 4 [OK]

Hint: Use binary search between min and max stall positions [OK]

Common Mistakes:

• Not testing feasibility correctly for each distance
• Stopping binary search too early
• Miscounting cows placed at each step