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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the stall positions before placing cows.

DSA C++

#include <algorithm>
#include <vector>

void placeCows(std::vector<int>& stalls) {
    std::sort(stalls.begin(), stalls.[1]);
}

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

Arend

Brbegin

Cbegin

Dend

Attempts:

3 left

💡 Hint

Common Mistakes

Using reverse iterators like rbegin or rend which are incorrect for std::sort.

Using begin twice which does not specify the full range.

The std::sort function requires the range to be sorted from begin() to end(). Using stalls.end() correctly specifies the end of the range.

2fill in blank

medium

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

DSA C++

bool canPlaceCows(const std::vector<int>& stalls, int cows, int dist) {
    int count = 1;
    int last_position = stalls[0];
    for (int i = 1; i < stalls.size(); i++) {
        if (stalls[i] - last_position >= [1]) {
            count++;
            last_position = stalls[i];
        }
    }
    return count >= cows;
}

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Acows

Bdist

Clast_position

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using the number of cows instead of the distance.

Using the last position variable incorrectly.

The condition checks if the distance between the current stall and the last placed cow is at least dist. This ensures the minimum distance requirement.

3fill in blank

hard

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

DSA C++

int aggressiveCows(std::vector<int>& stalls, int cows) {
    std::sort(stalls.begin(), stalls.end());
    int low = 1;
    int high = stalls.back() - stalls.front();
    int result = 0;
    while (low [1] high) {
        int mid = low + (high - low) / 2;
        if (canPlaceCows(stalls, cows, mid)) {
            result = mid;
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    return result;
}

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A<=

B<

C>=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' which may skip checking the last possible distance.

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

The binary search loop should continue while low is less than or equal to high to ensure all possible distances are checked.

4fill in blank

hard

Fill both blanks to correctly update the binary search bounds based on placement feasibility.

DSA C++

if (canPlaceCows(stalls, cows, mid)) {
    result = 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 the updates of low and high.

Using incorrect arithmetic on mid.

If cows can be placed with distance mid, try for a bigger distance by moving low up. Otherwise, reduce high to try smaller distances.

5fill in blank

hard

Fill all three blanks to complete the main function that reads input and outputs the result.

DSA C++

#include <iostream>
#include <vector>

int main() {
    int n, c;
    std::cin >> [1] >> [2];
    std::vector<int> stalls([3]);
    for (int i = 0; i < n; i++) {
        std::cin >> stalls[i];
    }
    std::cout << aggressiveCows(stalls, c) << std::endl;
    return 0;
}

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

An

Bc

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up the order of inputs.

Using wrong variable for vector size.

The program reads the number of stalls n and number of cows c. Then it creates a vector of size n to store stall positions.

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

C. To minimize the total distance cows travel between stalls

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

Solution

1. Step 1: Understand the problem goal

   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: Identify the objective

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

3. Final Answer:

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

4. Quick Check:

   Maximize minimum distance = Place cows far apart [OK]

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

Common Mistakes:

• Thinking it's about minimizing total distance
• Confusing maximum cows per stall with distance
• Assuming cows must be in consecutive stalls

2. Which of the following is the correct way to start the binary search for the minimum distance in the aggressive cows problem in C++?

easy

A. int low = min_position, high = max_position;

B. int low = 1, high = number_of_cows;

C. int low = 0, high = number_of_stalls;

D. int low = 0, high = max_position - min_position;

Solution

1. Step 1: Define search space for distance

   The minimum distance can be from 0 to the maximum possible distance between the farthest stalls, which is max_position - min_position.

2. Step 2: Set binary search boundaries

   So, low = 0 and high = max_position - min_position correctly define the search space for distance.

3. Final Answer:

   int low = 0, high = max_position - min_position; -> Option D

4. Quick Check:

   Distance range = 0 to max - min positions [OK]

Hint: Binary search distance between min and max stall positions [OK]

Common Mistakes:

• Using number of cows or stalls as high boundary
• Setting low to min_position instead of 0
• Confusing stall indices with distances

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

medium

A. 3

B. 4

C. 5

D. 6

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

   Try distance 4: place first cow at 1, second at 8 (distance 7), third at 9 (distance 1 from 8, too close). So place third at 4 (distance 3 from 1, less than 4). Not possible. Try distance 3: place cows at 1, 4, 8 - all distances >=3 possible.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Max min distance = 3 [OK]

Hint: Sort stalls, binary search distance, check placement [OK]

Common Mistakes:

• Not sorting stalls before calculation
• Misplacing cows ignoring minimum distance
• Choosing distance too large without validation

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

bool canPlaceCows(vector<int>& stalls, int cows, int dist) {
    int count = 1, last_pos = stalls[0];
    for (int i = 1; i < stalls.size(); i++) {
        if (stalls[i] - last_pos >= dist); {
            count++;
            last_pos = stalls[i];
        }
    }
    return count >= cows;
}

medium

A. Extra semicolon after if condition causing logic error

B. Incorrect initialization of count variable

C. Wrong comparison operator in if condition

D. Missing return statement

Solution

1. Step 1: Inspect the for loop and if statement

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

2. Step 2: Understand effect of extra semicolon

   This causes the if condition to do nothing, and the following block always executes, leading to incorrect count increment.

3. Final Answer:

   Extra semicolon after if condition causing logic error -> Option A

4. Quick Check:

   Extra semicolon breaks if logic [OK]

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

Common Mistakes:

• Ignoring semicolon after if condition
• Assuming count starts at 0 instead of 1
• Misreading comparison operators

5. You have stalls at positions [2, 5, 11, 15, 17, 23] and 4 cows. What is the maximum minimum distance to place all cows? Choose the correct approach and answer.

hard

A. Maximum minimum distance is 5 by placing cows at 2, 5, 15, 23

B. Maximum minimum distance is 7 by placing cows at 2, 11, 18, 23

C. Maximum minimum distance is 6 by placing cows at 2, 11, 17, 23

D. Maximum minimum distance is 8 by placing cows at 2, 11, 15, 23

Solution

1. Step 1: Sort stalls and set binary search range

   Stalls sorted: [2, 5, 11, 15, 17, 23]. Distance range: low=0, high=21 (23-2).

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

   Check mid=9: place cows at 2, next at >=11 (11), next at >=20 (23), only 3 cows placed, less than 4. So reduce high to 8.

3. Step 3: Check mid=4: place cows at 2, 11, 15, 23 (4 cows placed). Try higher mid=6: place cows at 2, 11, 17, 23 (4 cows placed). Try mid=7: only 3 cows placed. So max min distance is 6.

4. Final Answer:

   Maximum minimum distance is 6 by placing cows at 2, 11, 17, 23 -> Option C

5. Quick Check:

   Binary search max distance = 6 [OK]

Hint: Binary search distances, validate placement count [OK]

Common Mistakes:

• Not validating placement for each distance
• Choosing distances too large without checking feasibility
• Ignoring sorting of stalls