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 min distance mid

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Update answer = mid

↓

Increase low = mid+1

↓

Repeat until low > high

↓

Return answer as max min 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 canPlaceCows(int stalls[], int n, int cows, int dist) {
    int count = 1, last_pos = stalls[0];
    for (int i = 1; i < n; i++) {
        if (stalls[i] - last_pos >= dist) {
            count++; last_pos = stalls[i];
        }
    }
    return count >= cows;
}

This function checks if cows can be placed in stalls with at least 'dist' distance apart.

Execution Table

Step	Operation	low	high	mid	Can place cows?	Answer	Visual State (Placed Cows)
1	Sort stalls	-	-	-	-	-	[1, 2, 5, 8, 9]
2	Initialize low=0, high=8 (max distance)	0	8	-	-	-	[ ]
3	Calculate mid=(0+8)/2=4	0	8	4	-	-	[ ]
4	Check if cows can be placed with dist=4	0	8	4	Yes	4	[1, 5, 9]
5	Update low=mid+1=5	5	8	-	Yes	4	[1, 5, 9]
6	Calculate mid=(5+8)/2=6	5	8	6	-	-	[ ]
7	Check if cows can be placed with dist=6	5	8	6	No	4	[1, 8]
8	Update high=mid-1=5	5	5	-	No	4	[1, 8]
9	Calculate mid=(5+5)/2=5	5	5	5	-	-	[ ]
10	Check if cows can be placed with dist=5	5	5	5	Yes	5	[1, 5, 9]
11	Update low=mid+1=6	6	5	-	Yes	5	[1, 5, 9]
12	Exit loop since low=6 > high=5	6	5	-	-	5	[1, 5, 9]

💡 Loop ends when low (6) becomes greater than high (5), final answer is 5.

Variable Tracker

Variable	Start	After Step 2	After Step 5	After Step 8	After Step 11	Final
low	-	0	5	5	6	6
high	-	8	8	5	5	5
mid	-	-	4	6	5	-
answer	-	-	4	4	5	5
placed_cows	-	-	[1,5,9]	[1,8]	[1,5,9]	[1,5,9]

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 increasing 'low' to mid+1 (see step 5 in execution_table).

Why do we update 'high' when cows cannot be placed at distance 'mid'?

If cows can't be placed at 'mid', the distance is too large, so we reduce 'high' to mid-1 to try smaller distances (see step 8 and 11).

Why do we sort the stalls first?

Sorting ensures we check distances in order and place cows greedily from left to right, which is necessary for correct placement checks (step 1).

Visual Quiz - 3 Questions

Look at the execution_table at step 4, how many cows were placed when checking distance 4?

A 3 cows

B 2 cows

C 4 cows

D 1 cow

💡 Hint

Check the 'Visual State (Placed Cows)' column at step 4.

At which step does the condition 'low > high' become true, ending the loop?

A Step 10

B Step 12

C Step 8

D Step 5

💡 Hint

Look at the 'low' and 'high' values in step 12 in the execution_table.

If the number of cows to place was increased, how would the 'answer' variable likely change?

A It would increase

B It would stay the same

C It would decrease

D It would become zero

💡 Hint

More cows require smaller minimum distances to fit, so answer decreases (see variable_tracker for answer changes).

Concept Snapshot

Aggressive Cows Problem:
- Sort stall positions.
- Use binary search on distance (low, high).
- Check feasibility with canPlaceCows function.
- Update low/high based on feasibility.
- Final answer is largest minimum distance to place all cows.

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 greedy approach. If yes, we update the answer and try a larger distance by increasing low. If no, we try smaller distances by decreasing high. We repeat until low exceeds high. The final answer is the largest minimum distance possible. This approach efficiently finds the solution by narrowing down the search space using binary search and checking placement feasibility.

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 minimize the total distance cows travel between stalls

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

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 what maximum minimum distance means

   This means maximizing the minimum gap between cows, not minimizing total distance or placing 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 D

4. Quick Check:

   Maximum minimum distance = largest smallest gap [OK]

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

Common Mistakes:

• Thinking it's about total distance traveled
• Assuming cows must be in consecutive stalls
• Confusing maximum distance with maximum minimum distance

2. Which of the following is the correct way to declare the binary search loop for finding the maximum minimum distance in C?

easy

A. do { /* binary search logic */ } while (low < high);

B. for (int i = low; i < high; i++) { /* binary search logic */ }

C. while (low <= high) { /* binary search logic */ }

D. if (low < high) { /* binary search logic */ }

Solution

1. Step 1: Identify binary search loop structure

   Binary search typically uses a while loop checking low <= high to narrow the search space.

2. Step 2: Check each option

   while (low <= high) { /* binary search logic */ } uses while(low <= high), which is correct. Options A and B use do-while and for incorrectly for binary search. if (low < high) { /* binary search logic */ } uses if, which runs only once.

3. Final Answer:

   while (low <= high) { /* binary search logic */ } -> Option C

4. Quick Check:

   Binary search loop = while(low <= high) [OK]

Hint: Binary search uses while(low <= high) loop [OK]

Common Mistakes:

• Using for loop instead of while for binary search
• Using if statement instead of loop
• Using do-while with incorrect condition

3. Given stalls at positions [1, 2, 8, 4, 9] and 3 cows, what is the maximum minimum distance between any two cows after running the algorithm?

medium

A. 4

B. 6

C. 5

D. 3

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

   Check feasibility for mid=4: place first cow at 1, next cow at position >= 5 (8 fits), next cow at position >= 9 (9 fits). All 3 cows placed with minimum distance 4. For mid=5: place at 1, next at >=6 (8 fits), next at >=11 (no stall), only 2 cows placed. So maximum minimum distance is 4.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   Max minimum distance = 4 [OK]

Hint: Sort stalls first, then binary search distances [OK]

Common Mistakes:

• Not sorting stalls before processing
• Choosing distance too large without checking feasibility
• Miscounting number of cows placed

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

int count = 1, last_pos = stalls[0];
for (int i = 1; i < n; i++) {
  if (stalls[i] - last_pos >= mid) {
    count++;
    last_pos = stalls[i];
  }
}
return count >= cows;

medium

A. The initial count should be 0 instead of 1

B. No error; the code correctly checks feasibility

C. The comparison should be stalls[i] - last_pos > mid instead of >= mid

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

Solution

1. Step 1: Analyze initial count and loop start

   Count starts at 1 because first cow is placed at stalls[0]. Loop starts at i=1 to check next stalls.

2. Step 2: Check distance comparison

   Using >= mid ensures cows are placed with at least mid distance apart, which is correct.

3. Final Answer:

   No error; the code correctly checks feasibility -> Option B

4. Quick Check:

   Initial count=1 and >= mid check is correct [OK]

Hint: Start count at 1 and use >= for distance check [OK]

Common Mistakes:

• Starting loop at 0 causing double count
• Using > instead of >= causing off-by-one error
• Setting count to 0 missing first cow placement

5. You have stalls at positions [2, 5, 11, 15, 17, 22] and 4 cows. What is the maximum minimum distance to place all cows? Use binary search approach.

hard

A. 5

B. 6

C. 7

D. 8

Solution

1. Step 1: Sort stalls (already sorted)

   Stalls: [2, 5, 11, 15, 17, 22]

2. Step 2: Apply binary search on distance

   Low = 1, High = 22 - 2 = 20. Feasibility check: distance 5 allows 4 cows (2, 5, 11, 17; distances 3, 6, 6). Distance 6 allows only 3 cows (2, 11, 17). Thus maximum minimum distance is 5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Max min distance = 5 [OK]

Hint: Use binary search between min and max stall positions [OK]

Common Mistakes:

• Not checking feasibility correctly for each mid
• Confusing mid calculation in binary search
• Assuming max distance is stall range without checking placement