Aggressive Cows Maximum Minimum Distance in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal of the Aggressive Cows problem?

To place a given number of cows in stalls such that the minimum distance between any two cows is as large as possible.

intermediate

Which algorithmic technique is commonly used to solve the Aggressive Cows problem efficiently?

Binary Search on the answer (searching the minimum distance) combined with a greedy check to place cows.

beginner

In the Aggressive Cows problem, what does the 'check' function do?

It tries to place all cows in stalls with at least the given minimum distance apart. Returns true if possible, false otherwise.

beginner

Why do we sort the stall positions before applying binary search in the Aggressive Cows problem?

Sorting allows us to efficiently check if cows can be placed with a certain minimum distance by moving linearly through stalls.

intermediate

What is the time complexity of the Aggressive Cows solution using binary search and greedy placement?

O(N log M), where N is the number of stalls and M is the range of possible distances between stalls.

What does the binary search in the Aggressive Cows problem search for?

A The total number of stalls

B The number of cows to place

C The maximum minimum distance between cows

D The minimum stall position

Binary search is used to find the largest minimum distance possible between cows.

What is the purpose of the greedy check in the Aggressive Cows problem?

A To count the total stalls

B To verify if cows can be placed with a given minimum distance

C To sort the stall positions

D To find the smallest stall position

The greedy check tries to place cows with the given minimum distance to validate feasibility.

Why must stall positions be sorted before applying the binary search approach?

A To enable linear placement checks for cows

B To calculate the average distance

C To count the number of cows

D To find the maximum stall position

Sorting allows checking placement feasibility by moving through stalls in order.

If the greedy check fails for a distance 'd', what should the binary search do next?

A Try a larger distance

B Stop searching

C Increase the number of cows

D Try a smaller distance

If placement is not possible at distance 'd', try smaller distances.

What is the initial search range for the minimum distance in the binary search?

A 0 to maximum stall position - minimum stall position

B Number of cows to number of stalls

C 1 to number of cows

D Minimum stall position to maximum stall position

The search range is from 0 to the difference between the farthest stalls.

Explain how binary search and greedy placement work together to solve the Aggressive Cows problem.

Think about searching for the answer and verifying it step-by-step.

Describe why sorting stall positions is important before solving the Aggressive Cows problem.

Consider how you would place cows one by one.

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