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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of placing cows with minimum distance 3

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

DSA Typescript

function canPlaceCows(stalls: number[], cows: number, dist: number): boolean {
  let count = 1;
  let lastPos = stalls[0];
  for (let i = 1; i < stalls.length; i++) {
    if (stalls[i] - lastPos >= dist) {
      count++;
      lastPos = stalls[i];
      if (count === cows) return true;
    }
  }
  return false;
}

function aggressiveCows(stalls: number[], cows: number): number {
  stalls.sort((a, b) => a - b);
  let low = 1;
  let high = stalls[stalls.length - 1] - stalls[0];
  let result = 0;
  while (low <= high) {
    let mid = Math.floor((low + high) / 2);
    if (canPlaceCows(stalls, cows, mid)) {
      result = mid;
      low = mid + 1;
    } else {
      high = mid - 1;
    }
  }
  return result;
}

console.log(aggressiveCows([1, 2, 4, 8, 9], 3));

A 4

B 3

C 2

D 5

💡 Hint

Try placing cows starting with the smallest distance and increase it using binary search.

The maximum minimum distance to place 3 cows in stalls [1,2,4,8,9] is 3. This is found by binary searching the distance and checking feasibility.

🧠 Conceptual

intermediate

Why do we sort stalls before placing cows?

In the Aggressive Cows problem, why is it necessary to sort the stall positions before applying the binary search approach?

A To reduce the number of stalls to consider by removing duplicates.

B Sorting is not necessary; cows can be placed in any order.

C To ensure we can check distances in increasing order and place cows greedily.

D To make the stalls positions negative for easier calculation.

💡 Hint

Think about how distance checking works between stalls.

Sorting stalls allows us to check distances between adjacent stalls in order and place cows greedily from left to right.

🔧 Debug

advanced

Identify the error in this code snippet

What error will this TypeScript code produce when trying to find the maximum minimum distance for aggressive cows? function canPlaceCows(stalls: number[], cows: number, dist: number): boolean { let count = 1; let lastPos = stalls[0]; for (let i = 1; i < stalls.length; i++) { if (stalls[i] - lastPos > dist) { count++; lastPos = stalls[i]; if (count === cows) return true; } } return false; } // rest of code omitted for brevity

A The condition 'stalls[i] - lastPos > dist' causes incorrect placement, missing valid positions.

B SyntaxError due to missing semicolon after 'let count = 1'

C TypeError because 'stalls' is not defined

D No error, code runs correctly

💡 Hint

Check the comparison operator in the if condition.

The condition should be '>=' to allow placing cows at exactly the minimum distance, otherwise some valid placements are missed.

🚀 Application

advanced

Maximum minimum distance for large input

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 distances 10, 15, 20 and check feasibility.

The maximum minimum distance is 20 because cows can be placed at 10, 30, and 50 with minimum distance 20. It is not possible with distance 25 or more.

🧠 Conceptual

expert

Why binary search works on distance, not positions?

In the Aggressive Cows problem, why do we apply binary search on the minimum distance between cows instead of the stall positions themselves?

A Because searching positions directly would require checking all permutations of cow placements.

B Because stall positions are unsorted and cannot be searched.

C Because binary search only works on numeric arrays, not positions.

D Because the answer is a distance value, and binary searching distances narrows down the feasible minimum gap efficiently.

💡 Hint

Think about what the problem is asking to maximize and how binary search helps.

Binary search on distance helps find the largest minimum gap that can be maintained between cows, which is the problem's goal.

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

C. To minimize the total distance cows travel between stalls

D. To arrange cows so that all are in consecutive 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: Identify the correct goal

   This means maximizing the minimum distance between cows, not minimizing travel or fitting all cows in one stall.

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

Common Mistakes:

• Thinking it's about total distance traveled
• Assuming cows must be in consecutive stalls
• Confusing maximum distance with maximum number of cows

2. Which of the following TypeScript function signatures correctly represents the binary search approach used in the Aggressive Cows problem?

easy

A. function maxDistance(cows: number[], stalls: number): number

B. function placeCows(stalls: string[], cows: number): number

C. function canPlaceCows(stalls: number[], cows: number, distance: number): boolean

D. function binarySearch(distance: number[], cows: number): boolean

Solution

1. Step 1: Identify correct parameter types and purpose

   The function should check if cows can be placed in stalls with at least a given distance, so it needs stalls array, number of cows, and distance.

2. Step 2: Match signature to this logic

   function canPlaceCows(stalls: number[], cows: number, distance: number): boolean matches this: stalls as number array, cows count, and distance as number, returning boolean.

3. Final Answer:

   function canPlaceCows(stalls: number[], cows: number, distance: number): boolean -> Option C

4. Quick Check:

   Correct parameters and return type = function canPlaceCows(stalls: number[], cows: number, distance: number): boolean [OK]

Hint: Check parameter types match problem data (numbers, arrays) [OK]

Common Mistakes:

• Using wrong parameter types like string[] for stalls
• Returning wrong data type (e.g., number instead of boolean)
• Mixing order of parameters

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. 2

B. 4

C. 5

D. 3

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 3: place first cow at 1, second at 4 (distance 3), third at 8 (distance 4). Works.

3. Step 3: Check if distance 4 works

   Try distance 4: place first at 1, second at 8 (distance 7), third must be at least 4 away from 8, but 9 is only 1 away. Fails.

4. Final Answer:

   3 -> Option D

5. Quick Check:

   Max min distance = 3 [OK]

Hint: Sort stalls first, then binary search distances [OK]

Common Mistakes:

• Not sorting stalls before calculation
• Testing distances incorrectly
• Assuming max distance equals max stall difference

4. Consider this buggy TypeScript snippet for checking if cows can be placed:

function canPlaceCows(stalls: number[], cows: number, dist: number): boolean {
  let count = 1;
  let lastPos = stalls[0];
  for (let i = 1; i < stalls.length; i++) {
    if (stalls[i] - lastPos >= dist) {
      count++;
      lastPos = stalls[i];
    }
  }
  return count >= cows;
}

What is the main bug if stalls array is not sorted?

medium

A. The function will throw a runtime error due to index out of bounds

B. The function may incorrectly count placements because stalls are unsorted

C. The function returns a number instead of boolean

D. The function does not increment count properly

Solution

1. Step 1: Understand the function logic

   The function counts how many cows can be placed with at least 'dist' apart, assuming stalls are sorted.

2. Step 2: Identify effect of unsorted stalls

   If stalls are unsorted, the distance check stalls[i] - lastPos may be incorrect, causing wrong count of cows placed.

3. Final Answer:

   The function may incorrectly count placements because stalls are unsorted -> Option B

4. Quick Check:

   Unsorted stalls cause wrong counting = The function may incorrectly count placements because stalls are unsorted [OK]

Hint: Always sort stalls before placement checks [OK]

Common Mistakes:

• Ignoring sorting requirement
• Assuming function throws error on unsorted input
• Confusing return type or count increment logic

5. You have stalls at positions [1, 8, 15, 22, 25] and 4 cows. Using the binary search method, what is the maximum minimum distance you can place the cows apart?

hard

A. 7

B. 5

C. 8

D. 6

Solution

1. Step 1: Sort stalls (already sorted)

   Stalls: [1, 8, 15, 22, 25]

2. Step 2: Apply binary search on distance range

   Minimum distance can be 1, max distance is 25-1=24. Check mid distances:

3. Step 3: Check distance 9

   Place first cow at 1, next at 15 (distance 14), third at 25 (distance 10). Only 3 cows placed, need 4, so 9 fails.

4. Step 4: Check distance 7

   Place cows at 1, 8, 15, 22. Distances between placed cows are at least 7. 4 cows placed successfully.

5. Step 5: Check distance 8

   Place cows at 1, 15, 25. Only 3 cows placed, 4 needed, so 8 fails.

6. Final Answer:

   7 -> Option A

7. Quick Check:

   Max min distance = 7 [OK]

Hint: Test mid distances with binary search, adjust range accordingly [OK]

Common Mistakes:

• Not testing all mid values carefully
• Assuming max distance equals max stall difference
• Placing cows without checking minimum distance