Aggressive Cows Maximum Minimum Distance in DSA Go - Execution Trace

Concept Flow - Aggressive Cows Maximum Minimum Distance

Sort stall positions

↓

Set low = 0, high = max distance

↓

While low <= high

↓

Calculate mid = (low+high)/2

↓

Check if cows can be placed with distance mid

↓

Update low = mid+1

↓

Repeat loop

↓

Return high as max minimum distance

We sort stall positions, then use binary search on distance to find the largest minimum distance to place all cows.

Execution Sample

DSA Go

positions := []int{1, 2, 8, 4, 9}
cows := 3
sort.Ints(positions)
maxDist := binarySearch(positions, cows)
fmt.Println(maxDist)

Sort stall positions and find max minimum distance to place 3 cows.

Execution Table

Step	Operation	low	high	mid	Can Place?	Action	Visual State (Placed Cows)
1	Sort positions	-	-	-	-	-	[1, 2, 4, 8, 9]
2	Initialize low and high	0	8	-	-	-	Positions sorted
3	Calculate mid	0	8	4	-	-	-
4	Check placement with mid=4	0	8	4	Yes	low = mid + 1 = 5	Cows at 1, 4, 8
5	Calculate mid	5	8	6	-	-	-
6	Check placement with mid=6	5	8	6	No	high = mid - 1 = 5	Cows at 1, 8
7	Calculate mid	5	5	5	-	-	-
8	Check placement with mid=5	5	5	5	Yes	low = mid + 1 = 6	Cows at 1, 4, 9
9	Calculate mid	6	5	-	-	-	-
10	Loop ends	6	5	-	-	-	-

💡 Loop ends when low (6) > high (5), max minimum distance is high = 5

Variable Tracker

Variable	Start	After Step 2	After Step 4	After Step 6	After Step 8	Final
low	undefined	0	5	5	6	6
high	undefined	8	8	5	5	5
mid	undefined	-	4	6	5	-
Can Place?	undefined	-	Yes	No	Yes	-

Key Moments - 3 Insights

Why do we update low to mid + 1 when placement is possible?

Because if cows can be placed at distance mid, we try to find a larger minimum distance by increasing low (see Step 4 and Step 8 in execution_table).

Why do we update high to mid - 1 when placement is not possible?

Because if cows cannot be placed at distance mid, we reduce the search space by decreasing high to mid - 1 (see Step 6 in execution_table).

Why does the loop stop when low > high?

Because the search space is exhausted; the maximum minimum distance is the last successful mid stored in high (see Step 10 in execution_table).

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of mid at Step 3?

A 6

B 4

C 5

D 8

💡 Hint

Check the 'mid' column at Step 3 in execution_table.

At which step does the condition 'Can Place?' become No for the first time?

A Step 6

B Step 4

C Step 8

D Step 10

💡 Hint

Look at the 'Can Place?' column in execution_table to find the first 'No'.

If the number of cows was increased to 4, how would the final max minimum distance change?

A It would increase

B It would stay the same

C It would decrease

D Cannot determine from given data

💡 Hint

More cows require placing them closer, so max minimum distance usually decreases.

Concept Snapshot

Aggressive Cows Problem:
- Sort stall positions
- Use binary search on distance (low=0, high=max distance)
- Check if cows fit with mid distance
- If yes, try larger distance (low=mid+1)
- If no, try smaller distance (high=mid-1)
- Return high as max minimum distance

Full Transcript

The Aggressive Cows problem finds the largest minimum distance to place cows in stalls. We first sort the stall positions. Then, we use binary search on the distance between cows. We set low to 0 and high to the maximum distance between the first and last stall. In each step, we calculate mid as the average of low and high. We check if cows can be placed with at least mid distance apart. If yes, we move low up to mid plus one to try a bigger distance. If no, we move high down to mid minus one to try a smaller distance. We repeat until low is greater than high. The answer is the last successful mid stored in high. This method efficiently finds the maximum minimum distance.

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