Aggressive Cows Maximum Minimum Distance in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the stall positions.

DSA Rust

stalls.sort_by(|a, b| a.[1](b));

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Acmp

B-

C+

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using arithmetic operators instead of a comparator function.

Forgetting to sort the stalls before placing cows.

The sort_by method requires a comparator function like cmp to sort the vector.

2fill in blank

medium

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

DSA Rust

if stalls[i] - last_position >= [1] {

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Amax_dist

Bcount

Cdistance

Dmin_dist

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong variable for distance comparison.

Mixing up count and distance variables.

The variable distance represents the minimum required distance between cows.

3fill in blank

hard

Fix the error in the binary search loop condition.

DSA Rust

while low [1] high {

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A>

B<

C>=

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using < instead of <= causing infinite loops or missed checks.

Using > or >= which reverses the logic.

The binary search loop should continue while low is less than or equal to high.

4fill in blank

hard

Fill both blanks to update the binary search bounds correctly.

DSA Rust

if can_place_cows(&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 mid instead of mid ± 1 causing infinite loops.

If cows can be placed at distance mid, increase low to mid + 1 to try for a bigger distance. Otherwise, decrease high to mid - 1.

5fill in blank

hard

Fill all three blanks to complete the function signature and return statement.

DSA Rust

fn can_place_cows(stalls: &Vec<i32>, cows: [1], distance: [2]) -> [3] {
    let mut count = 1;
    let mut last_position = stalls[0];
    for i in 1..stalls.len() {
        if stalls[i] - last_position >= distance {
            count += 1;
            last_position = stalls[i];
            if count == cows {
                return true;
            }
        }
    }
    false
}

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

Ai32

Busize

Cbool

Du32

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong types for parameters causing type errors.

Returning integer instead of boolean.

The number of cows is a usize (index type), distance is i32, and the function returns a bool indicating if placement is possible.

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