Aggressive Cows Maximum Minimum Distance in DSA Java - 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

↓

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 Java

int[] stalls = {1, 2, 8, 4, 9};
int cows = 3;
Arrays.sort(stalls);
int low = 0, high = stalls[stalls.length-1] - stalls[0];
while(low <= high) {
  int mid = (low + high) / 2;
  if(canPlaceCows(stalls, cows, mid)) low = mid + 1;
  else high = mid - 1;
}
return high;

This code finds the largest minimum distance to place 3 cows in given stalls.

Execution Table

Step	Operation	low	high	mid	Can Place?	Action	Resulting Range	Max Min Distance So Far
1	Initialize	0	8	-	-	-	0 to 8	-
2	Calculate mid	0	8	4	-	-	0 to 8	-
3	Check placement with mid=4	0	8	4	Yes	low = mid + 1	5 to 8	-
4	Calculate mid	5	8	6	-	-	5 to 8	-
5	Check placement with mid=6	5	8	6	No	high = mid - 1	5 to 5	-
6	Calculate mid	5	5	5	-	-	5 to 5	-
7	Check placement with mid=5	5	5	5	Yes	low = mid + 1	6 to 5	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 7	Final
low	0	0	5	5	6	6
high	8	8	8	5	5	5
mid	-	4	6	5	5	5
Can Place?	-	-	No	-	Yes	Yes
Max Min Distance	-	-	-	-	5	5

Key Moments - 3 Insights

Why do we update low to mid + 1 when cows can be placed?

Because if cows fit with distance mid, we try to find a larger minimum distance by increasing low (see steps 3 and 7 in execution_table).

Why do we update high to mid - 1 when cows cannot be placed?

Because distance mid is too large to place all cows, so we reduce high to search smaller distances (see step 5 in execution_table).

Why is the final answer high, not low?

Loop ends when low > high; high holds the last successful distance where cows fit (step 7 in execution_table).

Visual Quiz - 3 Questions

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

A 4

B 5

C 6

D 8

💡 Hint

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

At which step does the algorithm decide to reduce the high value?

A Step 3

B Step 5

C Step 7

D Step 2

💡 Hint

Look for 'high = mid - 1' action in execution_table.

If the initial high was smaller, how would that affect the final max minimum distance?

A It would be larger

B It would not change

C It would be smaller or equal

D Algorithm would fail

💡 Hint

Refer to variable_tracker for how high limits max distance.

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, move low up to find bigger distance
- If no, move high down
- Final answer is high when low > high

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 start low at 0 and high at the maximum distance between the first and last stall. We calculate mid as the middle distance. We check if cows can be placed with at least mid distance apart. If yes, we try a bigger distance by setting low to mid plus one. If no, we try smaller distances by setting high to mid minus one. We repeat until low is greater than high. The largest minimum distance is the last successful high value. This approach efficiently finds the answer by narrowing the search space.

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 maximized

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 placed in consecutive stalls

Solution

1. Step 1: Understand the problem objective

   The problem asks to place cows in stalls such that the smallest distance between any two cows is as large as possible.

2. Step 2: Identify the key 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 maximized -> Option A

4. Quick Check:

   Maximize minimum distance = place cows far apart [OK]

Hint: Maximize the smallest gap between cows placed [OK]

Common Mistakes:

• Thinking the goal is to minimize distance
• Trying to put all cows in one stall
• Confusing maximum distance with maximum number of cows

2. Which of the following Java code snippets correctly initializes the binary search boundaries for the aggressive cows problem given an array stalls sorted in ascending order?

easy

A. int low = stalls.length; int high = stalls[0];

B. int low = 0; int high = stalls[stalls.length - 1] - stalls[0];

C. int low = 1; int high = stalls.length;

D. int low = stalls[0]; int high = stalls[stalls.length - 1];

Solution

1. Step 1: Determine the search space for distance

   The minimum distance can be 0 and maximum distance is the difference between the last and first stall positions.

2. Step 2: Set binary search boundaries accordingly

   So, low = 0 and high = stalls[stalls.length - 1] - stalls[0] correctly sets the search range.

3. Final Answer:

   int low = 0; int high = stalls[stalls.length - 1] - stalls[0]; -> Option B

4. Quick Check:

   Distance range = 0 to max stall gap [OK]

Hint: Low = 0, High = max stall position difference [OK]

Common Mistakes:

• Setting low or high to stall positions instead of distances
• Using stall count as high boundary
• Reversing low and high values

3. Given the sorted stalls array [1, 2, 8, 12, 17] and 3 cows, what is the maximum minimum distance between any two cows?

medium

A. 5

B. 6

C. 8

D. 7

Solution

1. Step 1: Check feasibility for distances

   Try placing cows with minimum distance 7: Place first cow at 1, second at 8 (distance 7), third at 17 (distance 9). All distances >= 7, so 7 is possible.

2. Step 2: Check if larger distance is possible

   Try 8: Place first at 1, second at 8 (distance 7) fails, so 8 not possible. Therefore, 7 is possible but we need to check if 6 is also possible.

3. Step 3: Check 6

   Try 6: Place first at 1, second at 8 (distance 7), third at 17 (distance 9). All distances >= 6, so 6 is possible.

4. Step 4: Maximum minimum distance is 7, but since 7 is possible, the maximum minimum distance is 7.

5. Final Answer:

   7 -> Option D

6. Quick Check:

   Max min distance = 7 [OK]

Hint: Test distances by placing cows greedily [OK]

Common Mistakes:

• Choosing the largest stall difference directly
• Not verifying placement feasibility
• Confusing maximum distance with stall count

4. Identify the error in this Java method snippet for checking if cows can be placed with at least dist distance:

boolean canPlace(int[] stalls, int cows, int dist) {
  int count = 1, lastPos = stalls[0];
  for (int i = 1; i < stalls.length; i++) {
    if (stalls[i] - lastPos >= dist) {
      count++;
      lastPos = stalls[i];
    }
  }
  return count >= cows;
}

medium

A. No error; the method correctly checks placement feasibility

B. The initial count should be 0 instead of 1

C. The loop should start from i = 0 instead of i = 1

D. The return condition should be count > cows instead of count >= cows

Solution

1. Step 1: Analyze initial count and last position

   Count starts at 1 because first cow is placed at stalls[0]. lastPos is set to stalls[0] correctly.

2. Step 2: Check loop and return condition

   Loop starts at 1 to check next stalls. If distance condition meets, count increments. Return true if count >= cows means enough cows placed.

3. Final Answer:

   No error; the method correctly checks placement feasibility -> Option A

4. Quick Check:

   Initial count 1 and correct loop start [OK]

Hint: First cow placed before loop; count starts at 1 [OK]

Common Mistakes:

• Starting count at 0 causing off-by-one error
• Loop starting at 0 causing redundant check
• Using > instead of >= in return condition

5. You have stalls at positions [0, 7, 14, 21, 28] and 4 cows. Using binary search, what is the maximum minimum distance you can place the cows apart?

hard

A. 5

B. 6

C. 7

D. 8

Solution

1. Step 1: Set binary search boundaries

   Minimum distance low = 0, high = 28 - 0 = 28.

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

   Check mid = 7: Place cows at 0, 7 (distance 7), 14 (7), 21 (7). Cows placed: 4 successfully.

3. Step 3: Try higher distance 8

   Place cows at 0, 14 (14), 28 (14) only 3 cows placed, fails.

4. Final Answer:

   7 -> Option C

5. Quick Check:

   Max min distance = 7 by binary search [OK]

Hint: Use binary search on distance, check feasibility each step [OK]

Common Mistakes:

• Not updating binary search boundaries correctly
• Miscounting cows placed at each distance
• Choosing distance without feasibility check