Aggressive Cows Maximum Minimum Distance in DSA Go

Mental Model

We want to place cows in stalls so that the smallest distance between any two cows is as big as possible.

Analogy: Imagine placing friends in seats along a bench so that everyone has as much personal space as possible.

Stalls: [1] -> [2] -> [4] -> [8] -> [9]
Cows: C placed in some stalls
Goal: maximize minimum distance between any two C

Dry Run Walkthrough

Input: stalls: [1, 2, 4, 8, 9], cows: 3

Goal: Place 3 cows in stalls so that the smallest distance between any two cows is as large as possible

Step 1: Sort stalls and set search range low=1, high=8 (max distance between first and last stall)

Stalls: 1 -> 2 -> 4 -> 8 -> 9
Search range: low=1, high=8

Why: We need to search for the largest minimum distance between cows

Step 2: Check mid = (1+8)/2 = 4; try placing cows with at least 4 distance apart

Place cow at 1 (stall 1)
Next cow at stall ≥ 1+4=5 -> stall 8
Next cow at stall ≥ 8+4=12 -> no stall
Placed 2 cows < 3 needed

Why: Distance 4 too large, cannot place all cows

Step 3: Reduce high to mid-1 = 3; check mid = (1+3)/2 = 2; try placing cows with distance 2

Place cow at 1
Next cow at stall ≥ 3 -> stall 4
Next cow at stall ≥ 6 -> stall 8
Placed 3 cows successfully

Why: Distance 2 works, try for bigger distance

Step 4: Increase low to mid+1 = 3; check mid = (3+3)/2 = 3; try placing cows with distance 3

Place cow at 1
Next cow at stall ≥ 4 -> stall 4
Next cow at stall ≥ 7 -> stall 8
Placed 3 cows successfully

Why: Distance 3 works, try bigger distance

Step 5: Increase low to mid+1 = 4; now low=4, high=3 -> stop search

Final answer is high=3

Why: Search ends when low > high; largest minimum distance is 3

Result:

Largest minimum distance = 3

Annotated Code

DSA Go

package main

import (
	"fmt"
	"sort"
)

func canPlaceCows(stalls []int, cows int, dist int) bool {
	count := 1 // place first cow at first stall
	lastPos := stalls[0]
	for i := 1; i < len(stalls); i++ {
		if stalls[i]-lastPos >= dist {
			count++
			lastPos = stalls[i]
			if count == cows {
				return true
			}
		}
	}
	return false
}

func aggressiveCows(stalls []int, cows int) int {
	sort.Ints(stalls) // sort stalls
	low, high := 1, stalls[len(stalls)-1]-stalls[0]
	result := 0
	for low <= high {
		mid := (low + high) / 2
		if canPlaceCows(stalls, cows, mid) {
			result = mid // mid distance works
			low = mid + 1 // try bigger distance
		} else {
			high = mid - 1 // try smaller distance
		}
	}
	return result
}

func main() {
	stalls := []int{1, 2, 4, 8, 9}
	cows := 3
	maxDist := aggressiveCows(stalls, cows)
	fmt.Printf("Largest minimum distance = %d\n", maxDist)
}

sort.Ints(stalls) // sort stalls to check distances in order

sort stalls to check distances in order

low, high := 1, stalls[len(stalls)-1]-stalls[0]

set search range for minimum distance

mid := (low + high) / 2

choose middle distance to test

if canPlaceCows(stalls, cows, mid) {

check if cows can be placed with at least mid distance

result = mid // mid distance works

record current working distance

low = mid + 1 // try bigger distance

try to increase minimum distance

high = mid - 1 // try smaller distance

reduce distance if mid doesn't work

count := 1 // place first cow at first stall

start placing cows from first stall

if stalls[i]-lastPos >= dist {

place next cow only if distance condition met

if count == cows { return true }

all cows placed successfully

Output

Largest minimum distance = 3

Complexity Analysis

Time: O(n log d) because we do binary search on distance (log d) and check placement in O(n)

Space: O(1) because we use only a few extra variables, no extra data structures

vs Alternative: Naive approach tries all distances linearly O(n*d), binary search reduces it to O(n log d)

Edge Cases

Only one stall

Can place only one cow, so max distance is 0

DSA Go

count := 1 // place first cow at first stall

Number of cows equals number of stalls

Cows must occupy all stalls, so minimum distance is minimum gap between adjacent stalls

DSA Go

if stalls[i]-lastPos >= dist {

All stalls at same position

Distance between any two stalls is zero, so max minimum distance is 0

DSA Go

high := stalls[len(stalls)-1]-stalls[0]

When to Use This Pattern

When asked to maximize minimum distance between placed items in sorted positions, use binary search on distance combined with a greedy check.

Common Mistakes

Mistake: Not sorting stalls before binary search
Fix: Always sort stalls first to check distances in order

Mistake: Checking placement incorrectly by not updating last placed position
Fix: Update lastPos only when a cow is placed to maintain correct distance checks

Summary

Find the largest minimum distance to place cows in stalls so no two cows are too close.

Use when you need to space out items evenly to maximize minimum gap.

Binary search on distance combined with greedy placement check is the key insight.