Aggressive Cows Maximum Minimum Distance in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the stalls positions.

DSA Python

stalls.sort()  # Sort the stalls positions

positions = [1, 2, 8, 4, 9]
stalls = positions
stalls.[1]()

Drag options to blanks, or click blank then click option'

Apop

Breverse

Cappend

Dsort

Attempts:

3 left

💡 Hint

Common Mistakes

Using reverse() instead of sort()

Trying to use sorted() without assignment

Using append() or pop() which do not sort

We need to sort the stalls positions to apply binary search for maximum minimum distance.

2fill in blank

medium

Complete the code to check if cows can be placed with at least 'distance' apart.

DSA Python

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

Drag options to blanks, or click blank then click option'

Acount

Bcows

Cdistance

Dlast_position

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with cows count instead of distance

Using wrong variable names in comparison

We check if the gap between current stall and last placed cow is at least 'distance'.

3fill in blank

hard

Fix the error in the binary search loop condition to find maximum minimum distance.

DSA Python

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

Drag options to blanks, or click blank then click option'

A<

B<=

C>=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' which may skip checking the last possible distance

Using '>' or '>=' which breaks the loop logic

We use '<=' to include the case when low equals high to not miss the maximum distance.

4fill in blank

hard

Fill both blanks to initialize the binary search boundaries correctly.

DSA Python

def aggressive_cows(stalls, cows):
    stalls.sort()
    low = [1]
    high = [2]
    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
    return result

Drag options to blanks, or click blank then click option'

A0

Bstalls[0]

Cstalls[-1] - stalls[0]

Dlen(stalls)

Attempts:

3 left

💡 Hint

Common Mistakes

Setting low to stalls[0] which is a position, not a distance

Setting high to len(stalls) which is count, not distance

Low starts at 0 (minimum possible distance), high is max distance between first and last stall.

5fill in blank

hard

Fill all three blanks to complete the main function that reads input and prints the result.

DSA Python

def main():
    t = int(input())
    for _ in range(t):
        n, c = map(int, input().split())
        stalls = list(map(int, input().split()))
        result = aggressive_cows([1], [2])
        print([3])

if __name__ == '__main__':
    main()

Drag options to blanks, or click blank then click option'

Astalls

Bc

Cresult

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Passing n instead of c to aggressive_cows

Printing stalls or c instead of result

We pass stalls and c to aggressive_cows and print the result variable.

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