Aggressive Cows Maximum Minimum Distance in DSA Javascript - 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 avoid fights between aggressive cows.

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

📋 What You'll Learn

Create an array called stalls with the exact values [1, 2, 8, 4, 9] representing stall positions.

Create a variable called cows and set it to 3 representing the number of cows to place.

Write a function called canPlaceCows that takes distance as input and returns true if cows can be placed with at least that distance apart, otherwise false.

Use binary search to find the maximum minimum distance possible between cows.

Print the maximum minimum distance found.

💡 Why This Matters

🌍 Real World

This problem models placing aggressive animals or devices in locations to maximize safety or efficiency by maintaining minimum distances.

💼 Career

Understanding binary search on answer space and greedy checking is useful in optimization problems common in software engineering and competitive programming.

1

Create stalls array

Create an array called stalls with the exact values [1, 2, 8, 4, 9] representing stall positions.

DSA Javascript

# Create the stalls array with the given positions
// Your code here

Hint

Use const stalls = [1, 2, 8, 4, 9]; to create the array.

2

Set number of cows

Create a variable called cows and set it to 3 representing the number of cows to place.

DSA Javascript

const stalls = [1, 2, 8, 4, 9];
// Create the cows variable and set it to 3
// Your code here

Hint

Use const cows = 3; to set the number of cows.

3

Write canPlaceCows function

Write a function called canPlaceCows that takes a parameter distance and returns true if it is possible to place all cows in stalls such that the minimum distance between any two cows is at least distance. Otherwise, return false. Use a greedy approach starting from the first stall.

DSA Javascript

const stalls = [1, 2, 8, 4, 9];
const cows = 3;
// Write the canPlaceCows function below
// Your code here

Hint

Sort the stalls first before using this function in the next step.

4

Find maximum minimum distance and print

Sort the stalls array. Use binary search between 1 and the maximum stall position minus minimum stall position to find the largest minimum distance where canPlaceCows(distance) returns true. Print the maximum minimum distance found using console.log.

DSA Javascript

const stalls = [1, 2, 8, 4, 9];
const cows = 3;

function canPlaceCows(distance) {
    let count = 1; // place first cow at first stall
    let lastPosition = stalls[0];
    for (let i = 1; i = distance) {
            count++;
            lastPosition = stalls[i];
            if (count === cows) {
                return true;
            }
        }
    }
    return false;
}

// Sort stalls and find max minimum distance
// Your code here

Hint

Remember to sort the stalls before binary searching. Use console.log(result); to print the answer.