Aggressive Cows Maximum Minimum Distance in DSA Rust - Execution Trace

Concept Flow - Aggressive Cows Maximum Minimum Distance

Sort stall positions

↓

Set low = 0, high = max distance

↓

While low <= high

↓

mid = (low+high)/2

↓

Check if cows can be placed with min dist = mid

↓

If yes

→low = mid + 1

↓

If no

→high = 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 Rust

fn can_place_cows(stalls: &Vec<i32>, cows: i32, dist: i32) -> bool {
    let mut count = 1;
    let mut last_pos = stalls[0];
    for &pos in stalls.iter().skip(1) {
        if pos - last_pos >= dist {
            count += 1;
            last_pos = pos;
            if count == cows {
                return true;
            }
        }
    }
    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	Resulting Range
1	Initialize	0	9	-	-	Set low=0, high=9	[0,9]
2	Calculate mid	0	9	4	-	-	[0,9]
3	Check placement with dist=4	-	-	-	Yes	low = mid + 1 = 5	[5,9]
4	Calculate mid	5	9	7	-	-	[5,9]
5	Check placement with dist=7	-	-	-	No	high = mid - 1 = 6	[5,6]
6	Calculate mid	5	6	5	-	-	[5,6]
7	Check placement with dist=5	-	-	-	Yes	low = mid + 1 = 6	[6,6]
8	Calculate mid	6	6	6	-	-	[6,6]
9	Check placement with dist=6	-	-	-	No	high = mid - 1 = 5	[6,5]
10	Exit	6	5	-	-	-	low > high, stop

💡 Binary search ends when low (6) > high (5), max minimum distance is high = 5

Variable Tracker

Variable	Start	After Step 3	After Step 5	After Step 7	After Step 9	Final
low	0	5	5	6	6	6
high	9	9	6	6	5	5
mid	-	4	7	5	6	-
Can Place?	-	Yes	No	Yes	No	-

Key Moments - 3 Insights

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

Because if cows fit at 'mid', we try for a larger distance by setting low = mid + 1 (see step 3 and 7 in execution_table).

Why does the search stop when low > high?

Binary search ends when low surpasses high, meaning no larger minimum distance is possible (step 10). The answer is high.

How does 'can_place_cows' decide if placement is possible?

It places cows greedily in sorted stalls ensuring each is at least 'dist' apart, returning true if all cows placed (refer to execution_sample code).

Visual Quiz - 3 Questions

Look at the execution_table at step 5, what is the value of 'high' after the action?

A 9

B 6

C 5

D 7

💡 Hint

Check the 'Action' and 'Resulting Range' columns at step 5 in execution_table.

At which step does the condition 'Can Place?' become 'No' for the first time?

A Step 3

B Step 5

C Step 7

D Step 9

💡 Hint

Look at the 'Can Place?' column in execution_table to find the first 'No'.

If the initial 'high' was set to 12 instead of 9, how would the first 'mid' value change?

A It would be 6

B It would be 4

C It would be 5

D It would be 7

💡 Hint

Calculate mid as (low + high) / 2 with low=0 and high=12.

Concept Snapshot

Aggressive Cows Problem:
- Sort stall positions.
- Use binary search on distance (low=0, high=max distance).
- Check feasibility with can_place_cows function.
- If feasible, increase low to mid+1.
- Else decrease high to 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, starting with low=0 and high as the maximum distance between stalls. For each mid distance, we check if cows can be placed with at least that distance apart using a helper function. If yes, we try a larger distance by setting low to mid+1. If no, we try smaller distances by setting high to mid-1. The search ends when low exceeds high, and the answer is high. This approach efficiently finds the maximum minimum distance.

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 place cows in stalls so that the maximum distance between any two cows is as small as possible

C. To find the total number of cows that can fit in all stalls

D. To calculate the average distance between all cows placed in 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: Clarify the goal

   This means maximizing the minimum distance between cows, ensuring they are as far apart as possible.

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: Maximize the smallest gap between cows placed [OK]

Common Mistakes:

• Confusing maximum minimum distance with minimum maximum distance
• Thinking the problem is about total cows fitting
• Assuming average distance is the goal

2. Which of the following Rust code snippets correctly sorts the stall positions stored in a vector stalls for the Aggressive Cows problem?

easy

A. stalls.sort_by(|a, b| b.cmp(a));

B. stalls.sorted();

C. stalls.sort();

D. stalls.order();

Solution

1. Step 1: Identify the correct Rust method to sort a vector

   In Rust, the method sort() sorts a vector in ascending order in place.

2. Step 2: Check other options

   sorted() does not exist on vectors, sort_by with reversed order sorts descending, and order() is invalid.

3. Final Answer:

   stalls.sort(); -> Option C

4. Quick Check:

   Use sort() to sort vector in Rust [OK]

Hint: Use stalls.sort() to sort vector ascending in Rust [OK]

Common Mistakes:

• Using non-existent methods like sorted()
• Sorting in descending order by mistake
• Trying to assign result of sort() which returns ()

3. Given the Rust code below, what is the output printed?

let stalls = vec![1, 2, 8, 4, 9];
let cows = 3;

let mut sorted_stalls = stalls.clone();
sorted_stalls.sort();

fn can_place_cows(stalls: &Vec<i32>, cows: i32, dist: i32) -> bool {
    let mut count = 1;
    let mut last_pos = stalls[0];
    for &pos in stalls.iter().skip(1) {
        if pos - last_pos >= dist {
            count += 1;
            last_pos = pos;
            if count == cows {
                return true;
            }
        }
    }
    false
}

let mut low = 1;
let mut high = sorted_stalls[sorted_stalls.len() - 1] - sorted_stalls[0];
let mut result = 0;

while low <= high {
    let mid = (low + high) / 2;
    if can_place_cows(&sorted_stalls, cows, mid) {
        result = mid;
        low = mid + 1;
    } else {
        high = mid - 1;
    }
}

println!("{}", result);

medium

A. 4

B. 6

C. 5

D. 3

Solution

1. Step 1: Sort stalls and set search range

   Sorted stalls: [1, 2, 4, 8, 9]. Search distance range low=1, high=8 (9-1).

2. Step 2: Binary search for max minimum distance

   mid=4: places 1,8 (count=2<3) fails, high=3.
   mid=2: places 1,4,8 succeeds, result=2, low=3.
   mid=3: places 1,4,8 (dists 3,4 >=3) succeeds, result=3, low=4>3 end. Prints 3.

3. Final Answer:

   3 -> Option D

4. Quick Check:

   Max min distance = 3 [OK]

Hint: Binary search distances, check feasibility with can_place_cows [OK]

Common Mistakes:

• Not sorting stalls before binary search
• Incorrect binary search boundaries
• Miscounting cows placed in can_place_cows

4. Identify the bug in the following Rust function used in the Aggressive Cows problem and select the correct fix:

fn can_place_cows(stalls: &Vec<i32>, cows: i32, dist: i32) -> bool {
    let mut count = 0;
    let mut last_pos = stalls[0];
    for &pos in stalls.iter() {
        if pos - last_pos >= dist {
            count += 1;
            last_pos = pos;
            if count == cows {
                return true;
            }
        }
    }
    false
}

medium

A. Initialize count to cows instead of 0

B. Both A and B

C. Start loop from stalls.iter().skip(1) instead of stalls.iter()

D. No bug, function is correct

Solution

1. Step 1: Analyze initial count and loop start

   Count should start at 1 because first cow is placed at stalls[0]. Loop should skip first stall to avoid counting it twice.

2. Step 2: Fix both initialization and loop

   Set count = 1 and loop with stalls.iter().skip(1) to correctly count cows placed.

3. Final Answer:

   Both A and B -> Option B

4. Quick Check:

   Count=1 and skip first stall in loop [OK]

Hint: Start count at 1 and skip first stall in loop [OK]

Common Mistakes:

• Starting count at 0 causes off-by-one error
• Looping from first stall counts first cow twice
• Returning false incorrectly due to wrong count

5. You have stalls at positions [2, 5, 11, 15, 17, 20] and want to place 4 cows. Using the Aggressive Cows approach, what is the maximum minimum distance you can achieve between any two cows?

hard

A. 4

B. 5

C. 7

D. 8

Solution

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

   Stalls sorted: [2, 5, 11, 15, 17, 20]. Low=1, High=18 (20-2).

2. Step 2: Binary search for max minimum distance

   mid=9: place cows at 2,11,20 (3 cows <4) fails, high=8.
   mid=4: place at 2,11,15,20 (dists 9,4,5 >=4, 4 cows) succeeds, result=4, low=5.
   mid=6: 2,11,17 (9,6 >=6, 20-17=3<6, 3 cows) fails, high=5.
   mid=5: 2,11,17 (9,6>=5, 20-17=3<5, 3 cows) fails, high=4.
   Max distance is 4.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   Max min distance = 4 [OK]

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

Common Mistakes:

• Not updating binary search bounds correctly
• Misplacing cows during feasibility check
• Assuming max distance equals max stall gap