Aggressive Cows Maximum Minimum Distance in DSA Typescript - 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

Yes No↓

low=mid+1

↓

Repeat until low > high

↓

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 Typescript

function canPlaceCows(stalls: number[], cows: number, dist: number): boolean {
  let count = 1, lastPos = stalls[0];
  for (let i = 1; i < stalls.length; i++) {
    if (stalls[i] - lastPos >= dist) {
      count++;
      lastPos = stalls[i];
    }
  }
  return count >= cows;
}

Checks if cows can be placed in stalls with at least 'dist' distance apart.

Execution Table

Step	Operation	low	high	mid	Can Place?	Action	Visual State (Placed Cows)
1	Sort stalls	-	-	-	-	-	[1, 2, 5, 8, 9]
2	Initialize low and high	0	8	-	-	-	-
3	Calculate mid	0	8	4	-	-	-
4	Check placement with dist=4	0	8	4	Yes	low = mid + 1 = 5	[1, 5, 9]
5	Calculate mid	5	8	6	-	-	-
6	Check placement with dist=6	5	8	6	No	high = mid - 1 = 5	[1, 8]
7	Calculate mid	5	5	5	-	-	-
8	Check placement with dist=5	5	5	5	Yes	low = mid + 1 = 6	[1, 5, 9]
9	Exit loop	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	-	0	5	5	6	6
high	-	8	8	5	5	5
mid	-	-	4	6	5	-
Can Place?	-	-	Yes	No	Yes	-
Placed Cows	-	-	[1,5,9]	[1,8]	[1,5,9]	-

Key Moments - 3 Insights

Why do we update 'low' when cows can be placed at distance 'mid'?

Because if cows fit at distance 'mid', we try to find a larger minimum distance by setting low = mid + 1 (see Step 4 in execution_table).

Why do we update 'high' when cows cannot be placed at distance 'mid'?

Because distance 'mid' is too large to place all cows, so we reduce the search space by setting high = mid - 1 (see Step 6).

Why do we return 'high' as the answer after the loop ends?

Because 'high' holds the largest distance where cows could still be placed, as loop ends when low > high (Step 9).

Visual Quiz - 3 Questions

Look at the execution_table at Step 4, what is the value of 'mid' and was placement possible?

A mid=4, placement not possible

B mid=4, placement possible

C mid=5, placement possible

D mid=5, placement not possible

💡 Hint

Check columns 'mid' and 'Can Place?' at Step 4 in execution_table

At which step does the condition 'canPlaceCows' return 'No' for the first time?

A Step 4

B Step 8

C Step 6

D Step 9

💡 Hint

Look at 'Can Place?' column in execution_table rows

If the stalls array was [1, 2, 4, 8, 12], how would the maximum minimum distance change compared to the current example?

A It would increase

B It would decrease

C It would stay the same

D Cannot determine without more info

💡 Hint

Consider that the max distance between stalls increases from 8 to 11, so max minimum distance can increase

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 bigger 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. 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 + 1 to try a bigger distance. If no, we move high down to mid - 1 to try a smaller distance. We repeat until low is greater than high. The answer is the last successful distance stored in high. This approach efficiently finds the maximum minimum distance for placing cows.

Practice

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

easy

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

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

C. To minimize the total distance cows travel between stalls

D. To arrange cows so that all are in consecutive stalls

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: Identify the correct goal

   This means maximizing the minimum distance between cows, not minimizing travel or fitting all 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 A

4. Quick Check:

   Maximize minimum distance = To place cows in stalls so that the minimum distance between any two cows is as large as possible [OK]

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

Common Mistakes:

• Thinking it's about total distance traveled
• Assuming cows must be in consecutive stalls
• Confusing maximum distance with maximum number of cows

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

easy

A. function maxDistance(cows: number[], stalls: number): number

B. function placeCows(stalls: string[], cows: number): number

C. function canPlaceCows(stalls: number[], cows: number, distance: number): boolean

D. function binarySearch(distance: number[], cows: number): boolean

Solution

1. Step 1: Identify correct parameter types and purpose

   The function should check if cows can be placed in stalls with at least a given distance, so it needs stalls array, number of cows, and distance.

2. Step 2: Match signature to this logic

   function canPlaceCows(stalls: number[], cows: number, distance: number): boolean matches this: stalls as number array, cows count, and distance as number, returning boolean.

3. Final Answer:

   function canPlaceCows(stalls: number[], cows: number, distance: number): boolean -> Option C

4. Quick Check:

   Correct parameters and return type = function canPlaceCows(stalls: number[], cows: number, distance: number): boolean [OK]

Hint: Check parameter types match problem data (numbers, arrays) [OK]

Common Mistakes:

• Using wrong parameter types like string[] for stalls
• Returning wrong data type (e.g., number instead of boolean)
• Mixing order of parameters

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

medium

A. 2

B. 4

C. 5

D. 3

Solution

1. Step 1: Sort stall positions

   Sorted stalls: [1, 2, 4, 8, 9]

2. Step 2: Use binary search to find max minimum distance

   Try distance 3: place first cow at 1, second at 4 (distance 3), third at 8 (distance 4). Works.

3. Step 3: Check if distance 4 works

   Try distance 4: place first at 1, second at 8 (distance 7), third must be at least 4 away from 8, but 9 is only 1 away. Fails.

4. Final Answer:

   3 -> Option D

5. Quick Check:

   Max min distance = 3 [OK]

Hint: Sort stalls first, then binary search distances [OK]

Common Mistakes:

• Not sorting stalls before calculation
• Testing distances incorrectly
• Assuming max distance equals max stall difference

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

function canPlaceCows(stalls: number[], cows: number, dist: number): boolean {
  let count = 1;
  let lastPos = stalls[0];
  for (let i = 1; i < stalls.length; i++) {
    if (stalls[i] - lastPos >= dist) {
      count++;
      lastPos = stalls[i];
    }
  }
  return count >= cows;
}

What is the main bug if stalls array is not sorted?

medium

A. The function will throw a runtime error due to index out of bounds

B. The function may incorrectly count placements because stalls are unsorted

C. The function returns a number instead of boolean

D. The function does not increment count properly

Solution

1. Step 1: Understand the function logic

   The function counts how many cows can be placed with at least 'dist' apart, assuming stalls are sorted.

2. Step 2: Identify effect of unsorted stalls

   If stalls are unsorted, the distance check stalls[i] - lastPos may be incorrect, causing wrong count of cows placed.

3. Final Answer:

   The function may incorrectly count placements because stalls are unsorted -> Option B

4. Quick Check:

   Unsorted stalls cause wrong counting = The function may incorrectly count placements because stalls are unsorted [OK]

Hint: Always sort stalls before placement checks [OK]

Common Mistakes:

• Ignoring sorting requirement
• Assuming function throws error on unsorted input
• Confusing return type or count increment logic

5. You have stalls at positions [1, 8, 15, 22, 25] and 4 cows. Using the binary search method, what is the maximum minimum distance you can place the cows apart?

hard

A. 7

B. 5

C. 8

D. 6

Solution

1. Step 1: Sort stalls (already sorted)

   Stalls: [1, 8, 15, 22, 25]

2. Step 2: Apply binary search on distance range

   Minimum distance can be 1, max distance is 25-1=24. Check mid distances:

3. Step 3: Check distance 9

   Place first cow at 1, next at 15 (distance 14), third at 25 (distance 10). Only 3 cows placed, need 4, so 9 fails.

4. Step 4: Check distance 7

   Place cows at 1, 8, 15, 22. Distances between placed cows are at least 7. 4 cows placed successfully.

5. Step 5: Check distance 8

   Place cows at 1, 15, 25. Only 3 cows placed, 4 needed, so 8 fails.

6. Final Answer:

   7 -> Option A

7. Quick Check:

   Max min distance = 7 [OK]

Hint: Test mid distances with binary search, adjust range accordingly [OK]

Common Mistakes:

• Not testing all mid values carefully
• Assuming max distance equals max stall difference
• Placing cows without checking minimum distance