Aggressive Cows Maximum Minimum Distance in DSA C++ - 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

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Update low = mid+1

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Repeat until low > high

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Return result 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 C++

int aggressiveCows(vector<int>& stalls, int cows) {
  sort(stalls.begin(), stalls.end());
  int low = 0, high = stalls.back() - stalls.front();
  int result = 0;
  while (low <= high) {
    int mid = (low + high) / 2;

This code sorts stalls and uses binary search on distance to find max minimum distance for placing cows.

Execution Table

Step	Operation	Distance mid	Can place cows?	Update low/high	Result	Visual State (Stalls with cows)
1	Sort stalls	-	-	-	-	[1, 2, 5, 8, 9]
2	Initialize low and high	-	-	-	-	low=0, high=8
3	Calculate mid	4	-	-	-	-
4	Check placement with distance 4	4	Yes	low=5	4	Cows at positions 1, 5, 9
5	Calculate mid	6	-	-	-	-
6	Check placement with distance 6	6	No	high=5	4	Cows at positions 1, 8
7	Calculate mid	5	-	-	-	-
8	Check placement with distance 5	5	No	high=4	4	Cows at positions 1, 8
9	Loop ends	-	-	-	4	Final max minimum distance

💡 Loop ends when low (5) > high (4), returning result 4 as maximum minimum distance.

Variable Tracker

Variable	Start	After Step 2	After Step 4	After Step 6	After Step 8	Final
low	0	0	5	5	5	5
high	0	8	8	5	4	4
mid	-	-	4	6	5	-
result	0	0	4	4	4	4

Key Moments - 3 Insights

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

Because placing cows at distance mid is possible, 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?

If cows can't be placed at distance mid, we reduce the search space by setting high = mid - 1 to try smaller distances (see Steps 6 and 8).

Why do we return 'result' after the loop ends?

Result stores the largest distance where placement was possible. When low > high, no larger distance can be found, so we return result (Step 9).

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'mid' at Step 6?

A 4

B 6

C 5

D 8

💡 Hint

Check the 'Distance mid' column at Step 6 in the execution_table.

At which step does the algorithm decide that distance 5 is not possible?

A Step 8

B Step 6

C Step 4

D Step 9

💡 Hint

Look for 'Can place cows?' = No with distance mid = 5 in the execution_table.

If the stalls were [1, 2, 4, 8, 12], how would the initial 'high' value change?

A It would be 12

B It would be 8

C It would be 11

D It would be 10

💡 Hint

High is calculated as stalls.back() - stalls.front(), check the new stall positions.

Concept Snapshot

Aggressive Cows Problem:
- Sort stall positions.
- Use binary search on distance (low=0, high=max distance).
- Check if cows can be placed with mid distance.
- If yes, move low up; if no, move high down.
- Continue until low > high.
- Result is 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 try a larger distance by setting low to mid plus one. If no, we try smaller distances by setting high to mid minus one. We repeat this until low is greater than high. The result is the largest minimum distance where placement was possible.

Practice

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

easy

A. To arrange cows so that all are placed in consecutive stalls

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

C. To minimize the total distance cows travel between stalls

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

Solution

1. Step 1: Understand the problem goal

   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 objective

   This means maximizing the minimum distance between cows, not minimizing or counting cows per stall.

3. Final Answer:

   To place cows in stalls so that the minimum distance between any two cows is as large as possible -> Option B

4. Quick Check:

   Maximize minimum distance = Place cows far apart [OK]

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

Common Mistakes:

• Thinking it's about minimizing total distance
• Confusing maximum cows per stall with distance
• Assuming cows must be in consecutive stalls

2. Which of the following is the correct way to start the binary search for the minimum distance in the aggressive cows problem in C++?

easy

A. int low = min_position, high = max_position;

B. int low = 1, high = number_of_cows;

C. int low = 0, high = number_of_stalls;

D. int low = 0, high = max_position - min_position;

Solution

1. Step 1: Define search space for distance

   The minimum distance can be from 0 to the maximum possible distance between the farthest stalls, which is max_position - min_position.

2. Step 2: Set binary search boundaries

   So, low = 0 and high = max_position - min_position correctly define the search space for distance.

3. Final Answer:

   int low = 0, high = max_position - min_position; -> Option D

4. Quick Check:

   Distance range = 0 to max - min positions [OK]

Hint: Binary search distance between min and max stall positions [OK]

Common Mistakes:

• Using number of cows or stalls as high boundary
• Setting low to min_position instead of 0
• Confusing stall indices with distances

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. 3

B. 4

C. 5

D. 6

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 4: place first cow at 1, second at 8 (distance 7), third at 9 (distance 1 from 8, too close). So place third at 4 (distance 3 from 1, less than 4). Not possible. Try distance 3: place cows at 1, 4, 8 - all distances >=3 possible.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Max min distance = 3 [OK]

Hint: Sort stalls, binary search distance, check placement [OK]

Common Mistakes:

• Not sorting stalls before calculation
• Misplacing cows ignoring minimum distance
• Choosing distance too large without validation

4. Identify the error in this C++ snippet for checking if cows can be placed with given distance:

bool canPlaceCows(vector<int>& stalls, int cows, int dist) {
    int count = 1, last_pos = stalls[0];
    for (int i = 1; i < stalls.size(); i++) {
        if (stalls[i] - last_pos >= dist); {
            count++;
            last_pos = stalls[i];
        }
    }
    return count >= cows;
}

medium

A. Extra semicolon after if condition causing logic error

B. Incorrect initialization of count variable

C. Wrong comparison operator in if condition

D. Missing return statement

Solution

1. Step 1: Inspect the for loop and if statement

   There is an extra semicolon after the if condition which ends the if prematurely.

2. Step 2: Understand effect of extra semicolon

   This causes the if condition to do nothing, and the following block always executes, leading to incorrect count increment.

3. Final Answer:

   Extra semicolon after if condition causing logic error -> Option A

4. Quick Check:

   Extra semicolon breaks if logic [OK]

Hint: Watch for stray semicolons after if statements [OK]

Common Mistakes:

• Ignoring semicolon after if condition
• Assuming count starts at 0 instead of 1
• Misreading comparison operators

5. You have stalls at positions [2, 5, 11, 15, 17, 23] and 4 cows. What is the maximum minimum distance to place all cows? Choose the correct approach and answer.

hard

A. Maximum minimum distance is 5 by placing cows at 2, 5, 15, 23

B. Maximum minimum distance is 7 by placing cows at 2, 11, 18, 23

C. Maximum minimum distance is 6 by placing cows at 2, 11, 17, 23

D. Maximum minimum distance is 8 by placing cows at 2, 11, 15, 23

Solution

1. Step 1: Sort stalls and set binary search range

   Stalls sorted: [2, 5, 11, 15, 17, 23]. Distance range: low=0, high=21 (23-2).

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

   Check mid=9: place cows at 2, next at >=11 (11), next at >=20 (23), only 3 cows placed, less than 4. So reduce high to 8.

3. Step 3: Check mid=4: place cows at 2, 11, 15, 23 (4 cows placed). Try higher mid=6: place cows at 2, 11, 17, 23 (4 cows placed). Try mid=7: only 3 cows placed. So max min distance is 6.

4. Final Answer:

   Maximum minimum distance is 6 by placing cows at 2, 11, 17, 23 -> Option C

5. Quick Check:

   Binary search max distance = 6 [OK]

Hint: Binary search distances, validate placement count [OK]

Common Mistakes:

• Not validating placement for each distance
• Choosing distances too large without checking feasibility
• Ignoring sorting of stalls