Aggressive Cows Maximum Minimum Distance in DSA Rust - Build from Scratch

Aggressive Cows Maximum Minimum Distance

📖 Scenario: You are managing a farm with several stalls placed along a straight line. You want to place cows in these stalls such that the minimum distance between any two cows is as large as possible. This helps prevent fights between cows.

🎯 Goal: Build a Rust program that finds the largest minimum distance possible between cows placed in given stalls.

📋 What You'll Learn

Create a vector called stalls with exact values [1, 2, 8, 4, 9]

Create an integer variable called cows and set it to 3

Write a function can_place_cows that checks if cows can be placed with at least a given minimum distance

Use binary search to find the maximum minimum distance

Print the maximum minimum distance found

💡 Why This Matters

🌍 Real World

This problem models placing resources or devices in locations to maximize minimum spacing, useful in network design, agriculture, and facility planning.

💼 Career

Understanding binary search on answer space and greedy feasibility checks is important for software engineering interviews and optimization problems.

1

Create the stalls vector

Create a vector called stalls with the exact values [1, 2, 8, 4, 9].

DSA Rust

# Create the stalls vector with values [1, 2, 8, 4, 9]
// Your code here

Hint

Use let mut stalls = vec![1, 2, 8, 4, 9]; to create the vector.

2

Add the cows variable

Add an integer variable called cows and set it to 3 inside the main function after the stalls vector.

DSA Rust

fn main() {
    let mut stalls = vec![1, 2, 8, 4, 9];
    // Add cows variable set to 3
    // Your code here
}

Hint

Use let cows = 3; to create the variable.

3

Write the can_place_cows function and sort stalls

Write a function called can_place_cows that takes stalls: &Vec<i32>, cows: i32, and distance: i32 and returns bool. It should check if it is possible to place all cows in stalls such that the minimum distance between any two cows is at least distance. Also, sort the stalls vector inside main before calling this function.

DSA Rust

fn main() {
    let mut stalls = vec![1, 2, 8, 4, 9];
    let cows = 3;

    // Sort stalls vector
    // Your code here
}

// Define can_place_cows function here
// Your code here

Hint

Sort the stalls using stalls.sort();. The function should count how many cows can be placed with at least the given distance.

4

Find and print the maximum minimum distance

Use binary search in main to find the maximum minimum distance to place all cows in stalls. Use the can_place_cows function to check feasibility. Print the maximum minimum distance found using println!("{}", max_distance);.

DSA Rust

fn main() {
    let mut stalls = vec![1, 2, 8, 4, 9];
    let cows = 3;

    stalls.sort();

    // Use binary search to find max minimum distance
    // Your code here
}

fn can_place_cows(stalls: &Vec, cows: i32, distance: i32) -> bool {
    let mut count = 1;
    let mut last_position = stalls[0];

    for &stall in stalls.iter().skip(1) {
        if stall - last_position >= distance {
            count += 1;
            last_position = stall;
            if count == cows {
                return true;
            }
        }
    }
    false
}

Hint

Use binary search between 1 and the max distance between first and last stall. Update max_distance when 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