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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of maximum minimum distance calculation

What is the output of the Rust code that calculates the maximum minimum distance for placing 3 cows in stalls at positions [1, 2, 8, 4, 9]?

DSA Rust

fn can_place_cows(stalls: &Vec<i32>, cows: i32, dist: i32) -> bool {
    let mut count = 1;
    let mut last_position = stalls[0];
    for &stall in stalls.iter().skip(1) {
        if stall - last_position >= dist {
            count += 1;
            last_position = stall;
            if count == cows {
                return true;
            }
        }
    }
    false
}

fn aggressive_cows(stalls: &mut Vec<i32>, cows: i32) -> i32 {
    stalls.sort();
    let mut low = 0;
    let mut high = stalls[stalls.len() - 1] - stalls[0];
    let mut result = 0;
    while low <= high {
        let mid = (low + high) / 2;
        if can_place_cows(stalls, cows, mid) {
            result = mid;
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    result
}

fn main() {
    let mut stalls = vec![1, 2, 8, 4, 9];
    let cows = 3;
    let max_min_dist = aggressive_cows(&mut stalls, cows);
    println!("{}", max_min_dist);
}

A 5

B 4

C 2

D 3

💡 Hint

Think about placing cows with the largest minimum gap possible between them.

The code sorts the stalls and uses binary search to find the largest minimum distance. For stalls [1,2,4,8,9] and 3 cows, the maximum minimum distance is 3.

🧠 Conceptual

intermediate

Understanding the binary search range in Aggressive Cows problem

In the Aggressive Cows problem, what is the correct initial range for binary searching the minimum distance between cows?

A From the minimum stall position to the maximum stall position

B From 1 to the number of stalls

C From 0 to the difference between the maximum and minimum stall positions

D From 0 to the number of cows

💡 Hint

Consider the smallest and largest possible distances between cows.

The minimum distance can be as low as 0 and as high as the distance between the farthest stalls.

🔧 Debug

advanced

Identify the bug causing incorrect placement check

Given the following Rust function for checking if cows can be placed with at least 'dist' distance, which option correctly identifies the bug?

DSA Rust

fn can_place_cows(stalls: &Vec<i32>, cows: i32, dist: i32) -> bool {
    let mut count = 1;
    let mut last_position = stalls[0];
    for &stall in stalls.iter().skip(1) {
        if stall - last_position > dist {
            count += 1;
            last_position = stall;
            if count == cows {
                return true;
            }
        }
    }
    false
}

A The initial count should be 0 instead of 1

B The condition should be 'stall - last_position >= dist' instead of '>'

C The loop should start from the first stall, not skip(1)

D The function should return false when count equals cows

💡 Hint

Check the comparison operator for distance condition.

Using '>' excludes the case where distance equals 'dist', which is valid placement.

📝 Syntax

advanced

Identify the syntax error in Rust code snippet

What is the syntax error in this Rust code snippet for binary searching the maximum minimum distance?

DSA Rust

while low <= high {
    let mid = (low + high) / 2
    if can_place_cows(stalls, cows, mid) {
        result = mid;
        low = mid + 1;
    } else {
        high = mid - 1;
    }
}

A Missing semicolon after 'let mid = (low + high) / 2'

B Incorrect use of 'while' instead of 'for'

C Using 'let' inside the loop is not allowed

D Missing braces around the if-else block

💡 Hint

Check line endings in Rust.

Rust requires semicolons after statements like variable declarations.

🚀 Application

expert

Maximum minimum distance for large input stalls

Given 5 stalls at positions [10, 20, 30, 40, 50] and 3 cows, what is the maximum minimum distance to place all cows?

A 20

B 10

C 15

D 25

💡 Hint

Try placing cows with equal spacing and check feasibility.

Placing cows at 10, 30, and 50 gives minimum distance 20. The maximum minimum distance is 20, and checking feasibility confirms it works.

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 as large as possible

B. To place cows in stalls so that the maximum distance between any two cows is as small as possible

C. To find the total number of cows that can fit in all stalls

D. To calculate the average distance between all cows placed in stalls

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 the goal

   This means maximizing the minimum distance between cows, ensuring they are as far apart as possible.

3. Final Answer:

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

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: Maximize the smallest gap between cows placed [OK]

Common Mistakes:

• Confusing maximum minimum distance with minimum maximum distance
• Thinking the problem is about total cows fitting
• Assuming average distance is the goal

2. Which of the following Rust code snippets correctly sorts the stall positions stored in a vector stalls for the Aggressive Cows problem?

easy

A. stalls.sort_by(|a, b| b.cmp(a));

B. stalls.sorted();

C. stalls.sort();

D. stalls.order();

Solution

1. Step 1: Identify the correct Rust method to sort a vector

   In Rust, the method sort() sorts a vector in ascending order in place.

2. Step 2: Check other options

   sorted() does not exist on vectors, sort_by with reversed order sorts descending, and order() is invalid.

3. Final Answer:

   stalls.sort(); -> Option C

4. Quick Check:

   Use sort() to sort vector in Rust [OK]

Hint: Use stalls.sort() to sort vector ascending in Rust [OK]

Common Mistakes:

• Using non-existent methods like sorted()
• Sorting in descending order by mistake
• Trying to assign result of sort() which returns ()

3. Given the Rust code below, what is the output printed?

let stalls = vec![1, 2, 8, 4, 9];
let cows = 3;

let mut sorted_stalls = stalls.clone();
sorted_stalls.sort();

fn can_place_cows(stalls: &Vec<i32>, cows: i32, dist: i32) -> bool {
    let mut count = 1;
    let mut last_pos = stalls[0];
    for &pos in stalls.iter().skip(1) {
        if pos - last_pos >= dist {
            count += 1;
            last_pos = pos;
            if count == cows {
                return true;
            }
        }
    }
    false
}

let mut low = 1;
let mut high = sorted_stalls[sorted_stalls.len() - 1] - sorted_stalls[0];
let mut result = 0;

while low <= high {
    let mid = (low + high) / 2;
    if can_place_cows(&sorted_stalls, cows, mid) {
        result = mid;
        low = mid + 1;
    } else {
        high = mid - 1;
    }
}

println!("{}", result);

medium

A. 4

B. 6

C. 5

D. 3

Solution

1. Step 1: Sort stalls and set search range

   Sorted stalls: [1, 2, 4, 8, 9]. Search distance range low=1, high=8 (9-1).

2. Step 2: Binary search for max minimum distance

   mid=4: places 1,8 (count=2<3) fails, high=3.
   mid=2: places 1,4,8 succeeds, result=2, low=3.
   mid=3: places 1,4,8 (dists 3,4 >=3) succeeds, result=3, low=4>3 end. Prints 3.

3. Final Answer:

   3 -> Option D

4. Quick Check:

   Max min distance = 3 [OK]

Hint: Binary search distances, check feasibility with can_place_cows [OK]

Common Mistakes:

• Not sorting stalls before binary search
• Incorrect binary search boundaries
• Miscounting cows placed in can_place_cows

4. Identify the bug in the following Rust function used in the Aggressive Cows problem and select the correct fix:

fn can_place_cows(stalls: &Vec<i32>, cows: i32, dist: i32) -> bool {
    let mut count = 0;
    let mut last_pos = stalls[0];
    for &pos in stalls.iter() {
        if pos - last_pos >= dist {
            count += 1;
            last_pos = pos;
            if count == cows {
                return true;
            }
        }
    }
    false
}

medium

A. Initialize count to cows instead of 0

B. Both A and B

C. Start loop from stalls.iter().skip(1) instead of stalls.iter()

D. No bug, function is correct

Solution

1. Step 1: Analyze initial count and loop start

   Count should start at 1 because first cow is placed at stalls[0]. Loop should skip first stall to avoid counting it twice.

2. Step 2: Fix both initialization and loop

   Set count = 1 and loop with stalls.iter().skip(1) to correctly count cows placed.

3. Final Answer:

   Both A and B -> Option B

4. Quick Check:

   Count=1 and skip first stall in loop [OK]

Hint: Start count at 1 and skip first stall in loop [OK]

Common Mistakes:

• Starting count at 0 causes off-by-one error
• Looping from first stall counts first cow twice
• Returning false incorrectly due to wrong count

5. You have stalls at positions [2, 5, 11, 15, 17, 20] and want to place 4 cows. Using the Aggressive Cows approach, what is the maximum minimum distance you can achieve between any two cows?

hard

A. 4

B. 5

C. 7

D. 8

Solution

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

   Stalls sorted: [2, 5, 11, 15, 17, 20]. Low=1, High=18 (20-2).

2. Step 2: Binary search for max minimum distance

   mid=9: place cows at 2,11,20 (3 cows <4) fails, high=8.
   mid=4: place at 2,11,15,20 (dists 9,4,5 >=4, 4 cows) succeeds, result=4, low=5.
   mid=6: 2,11,17 (9,6 >=6, 20-17=3<6, 3 cows) fails, high=5.
   mid=5: 2,11,17 (9,6>=5, 20-17=3<5, 3 cows) fails, high=4.
   Max distance is 4.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   Max min distance = 4 [OK]

Hint: Use binary search and feasibility check to find max distance [OK]

Common Mistakes:

• Not updating binary search bounds correctly
• Misplacing cows during feasibility check
• Assuming max distance equals max stall gap