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

qsort(stalls, n, sizeof(int), [1]);

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Acompare

Bcompare_stalls

Ccompare_positions

Dcompare_ints

Attempts:

3 left

💡 Hint

Common Mistakes

Using a wrong function name for comparison.

Passing NULL instead of a comparison function.

The qsort function requires a comparison function pointer. 'compare_ints' is the standard name for comparing integers.

2fill in blank

medium

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

DSA C

if (count_cows >= [1]) return 1; else return 0;

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Anum_cows

Btotal_cows

Ccows

Dmax_cows

Attempts:

3 left

💡 Hint

Common Mistakes

Using a wrong variable name for total cows.

Checking count_cows > num_cows instead of >=.

The variable 'num_cows' represents the total cows to place, so we check if count_cows is at least num_cows.

3fill in blank

hard

Fix the error in the binary search loop condition to avoid infinite loop.

DSA C

while (low <= [1]) {

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Ahigh

Bmid

Cnum_cows

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid or n in the loop condition instead of high.

Using < instead of <= causing missed cases.

The binary search runs while low is less than or equal to high, which is the search space boundary.

4fill in blank

hard

Fill both blanks to update the binary search boundaries correctly.

DSA C

if (canPlaceCows(stalls, n, num_cows, mid)) low = [1]; else high = [2];

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Amid + 1

Bmid - 1

Clow + 1

Dhigh - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping low and high updates.

Using low + 1 or high - 1 incorrectly.

If cows can be placed at distance mid, we try for a bigger distance by setting low to mid + 1; else reduce high to mid - 1.

5fill in blank

hard

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

DSA C

int canPlaceCows(int stalls[], int n, int num_cows, int dist) {
    int count = 1, last_pos = stalls[0];
    for (int i = 1; i < [1]; i++) {
        if (stalls[i] - last_pos >= [2]) {
            count++;
            last_pos = stalls[i];
            if (count == [3]) return 1;
        }
    }
    return count >= [3];
}

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

An

Bdist

Cnum_cows

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong loop limit.

Comparing with wrong distance variable.

Checking count against wrong variable.

We loop through all stalls (n), check if distance between current and last placed stall is at least dist, and count cows placed until num_cows is reached.

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 given stalls such that the smallest distance between any two cows is as large as possible.

2. Step 2: Clarify what maximum minimum distance means

   This means maximizing the minimum gap between cows, not minimizing total distance or placing 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:

   Maximum minimum distance = largest smallest gap [OK]

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

Common Mistakes:

• Thinking it's about total distance traveled
• Assuming cows must be in consecutive stalls
• Confusing maximum distance with maximum minimum distance

2. Which of the following is the correct way to declare the binary search loop for finding the maximum minimum distance in C?

easy

A. do { /* binary search logic */ } while (low < high);

B. for (int i = low; i < high; i++) { /* binary search logic */ }

C. while (low <= high) { /* binary search logic */ }

D. if (low < high) { /* binary search logic */ }

Solution

1. Step 1: Identify binary search loop structure

   Binary search typically uses a while loop checking low <= high to narrow the search space.

2. Step 2: Check each option

   while (low <= high) { /* binary search logic */ } uses while(low <= high), which is correct. Options A and B use do-while and for incorrectly for binary search. if (low < high) { /* binary search logic */ } uses if, which runs only once.

3. Final Answer:

   while (low <= high) { /* binary search logic */ } -> Option C

4. Quick Check:

   Binary search loop = while(low <= high) [OK]

Hint: Binary search uses while(low <= high) loop [OK]

Common Mistakes:

• Using for loop instead of while for binary search
• Using if statement instead of loop
• Using do-while with incorrect condition

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

medium

A. 4

B. 6

C. 5

D. 3

Solution

1. Step 1: Sort stall positions

   Sorted stalls: [1, 2, 4, 8, 9]

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

   Check feasibility for mid=4: place first cow at 1, next cow at position >= 5 (8 fits), next cow at position >= 9 (9 fits). All 3 cows placed with minimum distance 4. For mid=5: place at 1, next at >=6 (8 fits), next at >=11 (no stall), only 2 cows placed. So maximum minimum distance is 4.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   Max minimum distance = 4 [OK]

Hint: Sort stalls first, then binary search distances [OK]

Common Mistakes:

• Not sorting stalls before processing
• Choosing distance too large without checking feasibility
• Miscounting number of cows placed

4. Identify the error in this C code snippet for checking if cows can be placed with given minimum distance mid:

int count = 1, last_pos = stalls[0];
for (int i = 1; i < n; i++) {
  if (stalls[i] - last_pos >= mid) {
    count++;
    last_pos = stalls[i];
  }
}
return count >= cows;

medium

A. The initial count should be 0 instead of 1

B. No error; the code correctly checks feasibility

C. The comparison should be stalls[i] - last_pos > mid instead of >= mid

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

Solution

1. Step 1: Analyze initial count and loop start

   Count starts at 1 because first cow is placed at stalls[0]. Loop starts at i=1 to check next stalls.

2. Step 2: Check distance comparison

   Using >= mid ensures cows are placed with at least mid distance apart, which is correct.

3. Final Answer:

   No error; the code correctly checks feasibility -> Option B

4. Quick Check:

   Initial count=1 and >= mid check is correct [OK]

Hint: Start count at 1 and use >= for distance check [OK]

Common Mistakes:

• Starting loop at 0 causing double count
• Using > instead of >= causing off-by-one error
• Setting count to 0 missing first cow placement

5. You have stalls at positions [2, 5, 11, 15, 17, 22] and 4 cows. What is the maximum minimum distance to place all cows? Use binary search approach.

hard

A. 5

B. 6

C. 7

D. 8

Solution

1. Step 1: Sort stalls (already sorted)

   Stalls: [2, 5, 11, 15, 17, 22]

2. Step 2: Apply binary search on distance

   Low = 1, High = 22 - 2 = 20. Feasibility check: distance 5 allows 4 cows (2, 5, 11, 17; distances 3, 6, 6). Distance 6 allows only 3 cows (2, 11, 17). Thus maximum minimum distance is 5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Max min distance = 5 [OK]

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

Common Mistakes:

• Not checking feasibility correctly for each mid
• Confusing mid calculation in binary search
• Assuming max distance is stall range without checking placement