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

Aggressive Cows Maximum Minimum Distance

📖 Scenario: You are helping a farmer place aggressive cows in stalls along a straight line. The farmer wants to place all cows so that the minimum distance between any two cows is as large as possible. You will write a program to find this largest minimum distance.

🎯 Goal: Build a Go program that takes stall positions and the number of cows, then calculates the maximum minimum distance possible between any two cows placed in those stalls.

📋 What You'll Learn

Create a slice called stalls with the exact values 1, 2, 4, 8, 9

Create an integer variable called cows and set it to 3

Write a function called canPlaceCows that checks if cows can be placed with at least a given distance

Use binary search to find the maximum minimum distance

Print the final maximum minimum distance

💡 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 and greedy algorithms is essential for software engineering roles involving optimization and algorithmic problem solving.

1

Create the stalls slice

Create a slice called stalls with the exact values 1, 2, 4, 8, 9.

DSA Go

# Create the stalls slice with values 1, 2, 4, 8, 9
# Your code here

Hint

Use := to create and assign the slice stalls with the given numbers.

2

Add the cows variable

Create an integer variable called cows and set it to 3.

DSA Go

package main

func main() {
    stalls := []int{1, 2, 4, 8, 9}
    // Create cows variable and set to 3
    // Your code here
}

Hint

Use cows := 3 inside main to create the variable.

3

Write canPlaceCows function

Write a function called canPlaceCows that takes stalls []int, cows int, and distance int. It returns bool. The function should check if it is possible to place all cows in the stalls so that the minimum distance between any two cows is at least distance. Use a greedy approach starting from the first stall.

DSA Go

package main

func main() {
    stalls := []int{1, 2, 4, 8, 9}
    cows := 3
}

// Write canPlaceCows function below
// Your code here

Hint

Start placing the first cow at the first stall. Then place next cows only if the distance condition is met. Return true if all cows placed, else false.

4

Find and print maximum minimum distance

Use binary search between low = 1 and high = stalls[len(stalls)-1] - stalls[0] to find the maximum minimum distance. Use canPlaceCows to check feasibility. Print the final maximum minimum distance using fmt.Println.

DSA Go

package main

import "fmt"

func canPlaceCows(stalls []int, cows int, distance int) bool {
    count := 1
    lastPosition := stalls[0]
    for i := 1; i = distance {
            count++
            lastPosition = stalls[i]
            if count == cows {
                return true
            }
        }
    }
    return false
}

func main() {
    stalls := []int{1, 2, 4, 8, 9}
    cows := 3

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

Hint

Use binary search on distance. If canPlaceCows returns true, try bigger distance. Else try smaller.

Practice

1. What is the main goal of the Aggressive Cows Maximum Minimum Distance problem?

easy

A. To arrange cows in increasing order of their stall numbers

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

C. To minimize the distance between all cows placed in the stalls

D. To place cows in stalls so that the minimum distance between any two cows is as large as possible

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, not minimizing or counting 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 D

4. Quick Check:

   Maximize minimum distance = Place cows far apart [OK]

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

Common Mistakes:

• Thinking the goal is to minimize distance
• Confusing number of cows with distance
• Assuming cows can share stalls

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

easy

A. func canPlaceCows(stalls []int, dist int) bool

B. func canPlaceCows(cows int, dist int, stalls []int) int

C. func canPlaceCows(stalls []int, cows int, dist int) bool

D. func canPlaceCows(stalls []int, cows int) int

Solution

1. Step 1: Identify required parameters

   The helper function needs stalls array, number of cows, and distance to check feasibility.

2. Step 2: Match correct signature

   func canPlaceCows(stalls []int, cows int, dist int) bool has all parameters in correct order and returns bool indicating if placement is possible.

3. Final Answer:

   func canPlaceCows(stalls []int, cows int, dist int) bool -> Option C

4. Quick Check:

   Parameters: stalls, cows, dist; returns bool [OK]

Hint: Helper checks feasibility, needs stalls, cows, distance [OK]

Common Mistakes:

• Wrong parameter order
• Missing cows or distance parameter
• Returning int instead of bool

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

medium

A. 3

B. 4

C. 5

D. 2

Solution

1. Step 1: Sort stalls and set search range

   Sorted stalls: [1, 2, 4, 8, 9]. Minimum distance = 1, maximum = 8 (9-1).

2. Step 2: Binary search for max minimum distance

   Check mid=4: place first at 1 (last=1), skip 2 (1<4), skip 4 (3<4), place 8 (7>=4, count=2, last=8), skip 9 (1<4). Only 2 cows, fail.

3. Step 3: Check mid=3

   Place first at 1 (last=1), skip 2 (1<3), place 4 (3>=3, count=2, last=4), place 8 (4>=3, count=3, last=8), skip 9 (1<3). 3 cows placed, success.

4. Step 4: Final answer

   Maximum minimum distance is 3 (e.g., 1,4,8).

5. Final Answer:

   3 -> Option A

6. Quick Check:

   Max min distance = 3 [OK]

Hint: Sort stalls, binary search distance, check feasibility [OK]

Common Mistakes:

• Not sorting stalls before processing
• Incorrect binary search mid calculation
• Miscounting cows placed at given distance

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

func canPlaceCows(stalls []int, cows int, dist int) bool {
    count := 1
    lastPos := stalls[0]
    for i := 1; i < len(stalls); i++ {
        if stalls[i]-lastPos >= dist {
            count++
            lastPos = stalls[i]
        }
    }
    return count >= cows
}

What is the bug if stalls array is not sorted before calling this function?

medium

A. The function may incorrectly count placements because stalls are unordered

B. The function will cause a runtime panic due to index out of range

C. The function will always return false

D. No bug; sorting is not required

Solution

1. Step 1: Understand function logic

   The function assumes stalls are sorted to place cows greedily at increasing positions.

2. Step 2: Effect of unsorted stalls

   If stalls are unordered, distance checks become invalid, causing wrong count of cows placed.

3. Final Answer:

   The function may incorrectly count placements because stalls are unordered -> Option A

4. Quick Check:

   Unsorted stalls break greedy placement logic [OK]

Hint: Always sort stalls before placement check [OK]

Common Mistakes:

• Ignoring sorting requirement
• Assuming function panics without sorting
• Thinking function always fails without sorting

5. You have stalls at positions [5, 10, 15, 20, 25, 30] and need to place 4 cows. If the minimum distance between cows must be at least 7, which is the correct maximum minimum distance achievable?

hard

A. 7

B. 5

C. 15

D. 10

Solution

1. Step 1: Analyze stall positions

   Stalls: [5,10,15,20,25,30] (spaced by 5), 4 cows.

2. Step 2: Check dist=10

   Greedy: place at 5 (last=5), skip 10 (5<10), place 15 (10>=10), skip 20 (5<10), place 25 (10>=10), skip 30 (5<10). 3 cows placed, fail.

3. Step 3: Check dist=7

   Greedy: place at 5 (last=5), skip 10 (5<7), place 15 (10>=7), skip 20 (5<7), place 25 (10>=7), place 30 (5>=7? No, 5<7), so only 3 cows placed, fail.

4. Step 4: Check dist=6

   Similar to above, 3 cows placed, fail.

5. Step 5: Check dist=5

   Greedy: place at 5, 10, 15, 20 (all distances >=5), 4 cows placed, success.

6. Step 6: Final answer

   Maximum minimum distance is 5, since 7 or more fails to place 4 cows.

7. Final Answer:

   5 -> Option B

8. Quick Check:

   Max min distance = 5 [OK]

Hint: Test distances with binary search and feasibility check [OK]

Common Mistakes:

• Assuming max distance is stall range without checks
• Ignoring minimum distance constraint
• Misplacing cows ignoring stall positions