Aggressive Cows Maximum Minimum Distance in DSA Python - 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 stalls, then use binary search on distance to find the largest minimum distance to place all cows.

Execution Sample

DSA Python

stalls = [1, 2, 5, 4, 9]
cows = 3
# Find max min distance
stalls.sort()
low, high = 0, stalls[-1] - stalls[0]

Sort stall positions and set search range for minimum distance.

Execution Table

Step	Operation	mid (distance)	Can place cows?	low	high	Visual State (Placed cows at stalls)
1	Initial setup	-	-	0	8	Stalls sorted: [1, 2, 4, 5, 9]
2	Calculate mid	4	-	0	8	-
3	Check placement with mid=4	4	Yes	0	8	Placed cows at [1, 5, 9]
4	Update low = mid + 1	-	-	5	8	-
5	Calculate mid	6	-	5	8	-
6	Check placement with mid=6	6	No	5	8	Placed cows at [1, 9] only 2 cows
7	Update high = mid - 1	-	-	5	5	-
8	Calculate mid	5	-	5	5	-
9	Check placement with mid=5	5	No	5	5	Placed cows at [1, 9] only 2 cows
10	Update high = mid - 1	-	-	5	4	-
11	Exit loop	-	-	5	4	low > high, stop
12	Return result	-	-	-	-	Maximum minimum distance = 4

💡 Loop ends when low (5) > high (4), max minimum distance found is high = 4

Variable Tracker

Variable	Start	After Step 3	After Step 4	After Step 6	After Step 7	After Step 9	After Step 10	Final
low	0	0	5	5	5	5	5	5
high	8	8	8	8	5	5	4	4
mid	-	4	4	6	6	5	5	-
cows placed	-	3	3	2	2	2	2	-

Key Moments - 3 Insights

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

Because placing cows at distance mid is possible (Step 3), we try to find a larger minimum distance by increasing low (Step 4).

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

Because placing cows at distance mid failed (Step 6), we reduce high to mid - 1 to try smaller distances (Step 7).

Why does the loop stop when low > high?

When low exceeds high (Step 11), it means we have tested all distances; the largest minimum distance is high (Step 12).

Visual Quiz - 3 Questions

Look at the execution table, what is the value of mid at Step 6?

A 6

B 4

C 5

D 8

💡 Hint

Check the 'mid (distance)' column at Step 6 in the execution table.

At which step does the condition 'can place cows?' become No for the first time?

A Step 3

B Step 6

C Step 9

D Step 11

💡 Hint

Look at the 'Can place cows?' column in the execution table to find the first 'No'.

If the initial high was set to 10 instead of 8, how would the final maximum minimum distance change?

A It would decrease to 3

B It would increase to 5

C It would remain 4

D It would become 10

💡 Hint

Refer to the variable_tracker and execution_table showing how high is adjusted during the search.

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, increase low to mid+1
- If no, decrease high to mid-1
- Loop until low > high
- Result is high (max minimum distance)

Full Transcript

This visualization shows how to solve the Aggressive Cows problem using binary search on the minimum distance between cows. We start by sorting the stall positions. Then we set low to 0 and high to the maximum distance between the first and last stall. We repeatedly calculate mid as the average of low and high. We check if it is possible to place all cows in stalls such that the minimum distance between any two cows is at least mid. If placement is possible, we try a larger distance by setting low to mid plus one. If not, we try smaller distances by setting high to mid minus one. This continues until low exceeds high. The maximum minimum distance is then the last successful mid value stored in high. The execution table traces each step, showing the mid value, whether placement was possible, and how low and high change. The variable tracker shows how these variables evolve. Key moments clarify why we adjust low and high this way and why the loop stops. The quiz tests understanding of mid values, placement success, and the effect of initial high value. This method efficiently finds the largest minimum distance to place all cows in the stalls.

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

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

D. To minimize the total distance cows travel between 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 correct goal

   This means maximizing the minimum distance between cows, not minimizing travel or placing all cows consecutively.

3. Final Answer:

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

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
• Assuming cows must be in consecutive stalls
• Confusing maximum distance with maximum number of cows

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

easy

A. start = 0; end = max(stalls) - min(stalls)

B. start = 1; end = len(stalls)

C. start = min(stalls); end = max(stalls)

D. start = 0; end = len(stalls) - 1

Solution

1. Step 1: Determine the search space for distance

   The minimum distance can be as low as 0 and as high as the difference between the farthest stalls.

2. Step 2: Set binary search boundaries

   So, start = 0 and end = max(stalls) - min(stalls) correctly sets the search range for distances.

3. Final Answer:

   start = 0; end = max(stalls) - min(stalls) -> Option A

4. Quick Check:

   Distance range = 0 to max-min stall position [OK]

Hint: Distance search range is from 0 to max-min stall position [OK]

Common Mistakes:

• Using stall count as distance range
• Setting start or end to stall positions directly
• Confusing stall indices with distances

3. What will be the output of the following code snippet for stalls = [1, 2, 8, 4, 9] and cows = 3?

def can_place_cows(stalls, cows, dist):
    count = 1
    last_position = stalls[0]
    for i in range(1, len(stalls)):
        if stalls[i] - last_position >= dist:
            count += 1
            last_position = stalls[i]
        if count == cows:
            return True
    return False

stalls.sort()
low, high = 0, stalls[-1] - stalls[0]
result = 0
while low <= high:
    mid = (low + high) // 2
    if can_place_cows(stalls, cows, mid):
        result = mid
        low = mid + 1
    else:
        high = mid - 1
print(result)

medium

A. 4

B. 3

C. 5

D. 6

Solution

1. Step 1: Sort stalls and set search range

   Sorted stalls = [1, 2, 4, 8, 9], low=0, high=8 (9-1)

2. Step 2: Binary search for max minimum distance

   mid=4: can_place False (1 then 8 only, count=2), high=3.
   mid=1: True (1,2,4), result=1, low=2.
   mid=2: True (1,4,8), result=2, low=3.
   mid=3: True (1,4,8), result=3, low=4 > high.
   Loop ends, result=3.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Max min distance found = 3 [OK]

Hint: Use binary search on distance, check feasibility with can_place_cows [OK]

Common Mistakes:

• Not sorting stalls before processing
• Incorrectly updating low and high in binary search
• Miscounting cows placed in can_place_cows function

4. Identify the bug in the following code snippet for the aggressive cows problem:

def can_place_cows(stalls, cows, dist):
    count = 0
    last_position = stalls[0]
    for i in range(1, len(stalls)):
        if stalls[i] - last_position >= dist:
            count += 1
            last_position = stalls[i]
        if count == cows:
            return True
    return False

stalls.sort()
low, high = 0, stalls[-1] - stalls[0]
result = 0
while low <= high:
    mid = (low + high) // 2
    if can_place_cows(stalls, cows, mid):
        result = mid
        low = mid + 1
    else:
        high = mid - 1
print(result)

medium

A. The binary search boundaries are incorrect

B. last_position should be initialized to 0

C. The stalls list should not be sorted

D. count should start at 1, not 0

Solution

1. Step 1: Check initial count of placed cows

   Count starts at 0, but first cow is placed at stalls[0], so count should start at 1.

2. Step 2: Understand impact of count initialization

   Starting count at 0 causes undercounting cows placed, leading to incorrect feasibility checks.

3. Final Answer:

   count should start at 1, not 0 -> Option D

4. Quick Check:

   First cow placed at stalls[0], so count=1 initially [OK]

Hint: Initialize count with first cow placed at first stall [OK]

Common Mistakes:

• Starting count at 0 instead of 1
• Not sorting stalls before binary search
• Incorrect binary search range setup

5. Given stalls at positions [2, 4, 7, 11, 14, 17, 23] and 4 cows, what is the maximum minimum distance to place all cows? Use the aggressive cows approach.

hard

A. 6

B. 7

C. 8

D. 9

Solution

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

   Stalls are already sorted: [2, 4, 7, 11, 14, 17, 23].
   Minimum distance search range: low=0, high=23-2=21.

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

   Check mid=9: place cows at 2, 11, 23 (only 3 cows), not enough.
   High=8.
   Check mid=4: place cows at 2, 7, 14, 23 (4 cows), possible.
   Result=4, low=5.
   Check mid=6: place cows at 2, 11, 17, 23 (4 cows), possible.
   Result=6, low=7.
   Check mid=7: place cows at 2, 11, 23 (3 cows), not enough.
   High=6.
   Loop ends with result=6.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   Max min distance found = 6 [OK]

Hint: Use binary search and feasibility check to find max min distance [OK]

Common Mistakes:

• Not checking feasibility correctly for mid distances
• Misplacing cows at non-stall positions
• Incorrectly updating binary search bounds