Minimum Cost to Connect Sticks

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Steps

setup

Initial Check and Sort

Check if the sticks list has one or fewer elements. Since it has three, sort the sticks in ascending order.

💡 Sorting ensures the smallest sticks are at the front for the first merge.

Line:

if len(sticks) <= 1:
    return 0
sticks.sort()

💡 Sorting prepares the array for greedy merges by smallest sticks first.

delete

Remove Two Smallest Sticks

Remove the first two sticks from the sorted list: 2 and 3, the smallest sticks available.

💡 Removing the two smallest sticks is the greedy choice to minimize merge cost.

Line:

first = sticks.pop(0)
second = sticks.pop(0)

💡 Selecting the smallest two sticks ensures minimal incremental cost.

compare

Calculate Merge Cost

Calculate the cost to connect sticks 2 and 3 as 5, and add it to the total cost accumulator.

💡 Accumulating the cost after each merge tracks the total cost to connect all sticks.

Line:

cost = first + second
    total_cost += cost

💡 Each merge cost contributes to the final answer.

insert

Append Merged Stick and Sort

Append the merged stick of length 5 to the list, then sort the list to maintain ascending order.

💡 Re-sorting after insertion keeps the smallest sticks at the front for the next merge.

Line:

sticks.append(cost)
sticks.sort()

💡 Maintaining sorted order after each merge is crucial for the greedy approach.

delete

Remove Two Smallest Sticks Again

Remove the two smallest sticks from the list: 4 and 5, preparing to merge them next.

💡 Continuing to merge the smallest sticks reduces the total cost incrementally.

Line:

first = sticks.pop(0)
second = sticks.pop(0)

💡 The smallest sticks are always chosen for merging.

compare

Calculate Second Merge Cost

Calculate the cost to connect sticks 4 and 5 as 9, and add it to the total cost accumulator.

💡 Adding each merge cost accumulates the total cost to connect all sticks.

Line:

cost = first + second
    total_cost += cost

💡 The total cost accumulates all intermediate merges.

insert

Append Final Merged Stick

Append the merged stick of length 9 to the list. Since only one stick remains, no sorting is needed.

💡 Appending the merged stick prepares for the final check of the algorithm.

Line:

sticks.append(cost)
sticks.sort()

💡 Only one stick remains, so the process will end after this step.

compare

Check Loop Condition

Check if more than one stick remains. Since only one stick is left, the loop ends.

💡 Loop termination condition ensures the algorithm stops when all sticks are connected.

Line:

while len(sticks) > 1:

💡 The algorithm finishes when only one stick remains, representing the fully connected stick.

reconstruct

Prepare to Return Total Cost

Prepare to return the accumulated total cost of 14, which is the minimum cost to connect all sticks.

💡 Preparing the final answer before returning helps clarify the algorithm's completion.

Line:

return total_cost

💡 The total cost reflects the sum of all merges performed.

reconstruct

Return Total Cost Finalized

Return the accumulated total cost of 14 as the minimum cost to connect all sticks.

💡 Returning the total cost completes the algorithm and provides the final answer.

Line:

return total_cost

💡 The total cost is the final output of the algorithm.

def min_cost_sorting(sticks):
    if len(sticks) <= 1:  # STEP 1
        return 0
    sticks.sort()  # STEP 1
    total_cost = 0
    while len(sticks) > 1:  # STEP 8
        first = sticks.pop(0)  # STEP 2,5
        second = sticks.pop(0)  # STEP 2,5
        cost = first + second  # STEP 3,6
        total_cost += cost  # STEP 3,6
        sticks.append(cost)  # STEP 4,7
        sticks.sort()  # STEP 4,7
    return total_cost  # STEP 9,10

if __name__ == '__main__':
    sticks = [2, 4, 3]
    print(min_cost_sorting(sticks))  # Output: 14

📊

Minimum Cost to Connect Sticks - Watch the Algorithm Execute, Step by Step

Watching each merge and re-sort step helps you understand how the greedy approach always combines the smallest sticks first to minimize total cost.

insertmin-heap

2

0

3

1

4

2

extract_minmin-heap

4

0

peekmin-heap

4

0

Added: 5

insertmin-heap

4

0

5

1

extract_minmin-heap

empty heap

peekmin-heap

empty heap

Added: 9

insertmin-heap

9

0

peekmin-heap

9

0

peekmin-heap

9

0

peekmin-heap

9

0

Added: 14

Key Takeaways

✓ Always merging the two smallest sticks first minimizes the incremental cost and leads to the minimum total cost.

This insight is hard to see from code alone because the repeated sorting and merging steps are abstracted away.

✓ Re-sorting the array after each merge ensures the smallest sticks are always accessible for the next merge.

Visualizing the array after each sort clarifies why the greedy approach works step-by-step.

✓ The total cost accumulates the sum of all intermediate merges, not just the final stick length.

Understanding cost accumulation requires seeing each merge's contribution, which the visualization highlights.

Practice

1. Consider the following Python code for forming the largest number from the list [3, 30, 34, 5]. What is the final returned string?

easy

A. 534330

B. 53430

C. 534303

D. 534330

Solution

1. Step 1: Convert numbers to strings: ['3', '30', '34', '5']

   We compare pairs by concatenation: '5'+'34' vs '34'+'5' -> '534' > '345', so '5' before '34'. Similarly for others.

2. Step 2: Sort using custom comparator to get order: ['5', '34', '3', '30']

   Concatenate to get '534330'. The check for leading zero is false since first is '5'.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Concatenation order matches expected largest number [OK]

Hint: Compare concatenations 'a+b' and 'b+a' to order strings [OK]

Common Mistakes:

• Off-by-one in sorting
• Ignoring leading zero case
• Misordering '30' and '3'

2. Consider the following buggy code for candy distribution. Which line contains the subtle bug that can cause incorrect candy counts?

medium

A. Line 3: candies initialized with zeros instead of ones

B. Line 5: loop starts from 1 instead of 0

C. Line 7: candies[i] updated without checking if greater than candies[i-1]

D. Line 9: condition candies[i] <= candies[i + 1] missing

Solution

1. Step 1: Check initialization of candies array

   Line 3 initializes candies with zeros, violating the problem requirement that each child must have at least one candy.

2. Step 2: Understand impact of zero initialization

   Zero candies can cause incorrect comparisons and final sums, failing test cases where minimum is 1 per child.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Initialization must be at least 1 per child [OK]

Hint: Candies must start at 1, not 0 [OK]

Common Mistakes:

• Forgetting minimum candy per child
• Incorrect loop boundaries
• Missing update conditions

3. The following code attempts to find the largest monotone increasing digits number less than or equal to n. Identify the bug that causes incorrect results on some inputs.

def monotoneIncreasingDigits(n: int) -> int:
    digits = list(map(int, str(n)))
    marker = len(digits)
    for i in range(len(digits) - 1, 0, -1):
        if digits[i] < digits[i - 1]:
            digits[i - 1] -= 1
            marker = i
    return int(''.join(map(str, digits)))

medium

A. The code does not set digits after the marker to 9, missing the largest monotone number

B. The code decrements digits[i - 1] without checking if it causes new violations earlier

C. The code converts digits back to integer before fixing all digits, causing runtime errors

D. The code ignores the case when all digits are equal, returning incorrect output

Solution

1. Step 1: Analyze the loop effect

   The loop decrements digits[i - 1] when a violation is found and updates marker, but does not fix digits after marker.

2. Step 2: Identify missing step to set trailing digits to 9

   Without setting digits from marker to end to 9, the number may not be the largest monotone number ≤ n.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Missing trailing digit fix leads to smaller-than-necessary result [OK]

Hint: Always set trailing digits to 9 after decrement to maximize number [OK]

Common Mistakes:

• Forgetting to set trailing digits to 9
• Assuming one decrement fixes all violations
• Ignoring edge cases with equal digits

4. What is the time complexity of the optimal greedy algorithm for partitioning labels using a fixed-size array for last occurrences, given a string of length n?

medium

A. O(n) because each character is processed a constant number of times

B. O(n log n) due to sorting characters by last occurrence

C. O(n^2) due to nested scanning for last occurrences

D. O(n * 26) because of fixed alphabet size iteration

Solution

1. Step 1: Analyze last occurrence computation

   We scan the string once to record last occurrence of each character in O(n).

2. Step 2: Analyze partitioning loop

   We iterate over the string once more, updating partition end in O(1) per character.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Two linear scans over string length n -> O(n) time [OK]

Hint: Two passes over string -> O(n) time [OK]

Common Mistakes:

• Assuming nested loops cause O(n^2)
• Confusing fixed alphabet size iteration as O(n*26)
• Thinking sorting is needed

5. If tasks can be reused infinitely (i.e., unlimited supply of each task type), how should the Task Scheduler algorithm be modified to find the minimum total time with cooldown n?

hard

A. Use the same max-heap approach but reset frequencies after each full cycle

B. Calculate the minimal cycle length as (n + 1) and multiply by the number of unique tasks

C. Since tasks are infinite, schedule tasks in a fixed repeating pattern of length (n + 1) without heap

D. The problem reduces to scheduling one task repeatedly with cooldown, so total time is tasks count

Solution

1. Step 1: Understand infinite reuse implication

   With infinite supply, the scheduler can always pick a different task to fill cooldown slots.

2. Step 2: Optimal scheduling pattern

   Tasks can be scheduled in a fixed repeating pattern of length n + 1, cycling through unique tasks to avoid idle time.

3. Step 3: Algorithm modification

   No need for frequency tracking or heap; just cycle through unique tasks repeatedly.

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   Infinite tasks allow fixed pattern scheduling without heap [OK]

Hint: Infinite tasks -> fixed repeating pattern of length n+1 [OK]

Common Mistakes:

• Trying to use heap with infinite tasks
• Multiplying cycles incorrectly
• Ignoring cooldown constraints