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.

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