Minimum Cost to Connect Sticks

▶

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. Identify the bug in the following code snippet for the Minimum Domino Rotations problem:

def minDominoRotations(A, B):
    def check(x):
        rotations_a = rotations_b = 0
        for i in range(len(A)):
            if A[i] != x and B[i] != x:
                return 0  # Bug here
            elif A[i] != x:
                rotations_a += 1
            elif B[i] != x:
                rotations_b += 1
        return min(rotations_a, rotations_b)

    rotations = check(A[0])
    if rotations != -1:
        return rotations
    else:
        return check(B[0])

medium

A. Incorrect initialization of rotations_a and rotations_b

B. Not incrementing rotations when both sides equal x

C. Returning rotations without checking both candidates

D. The return value 0 instead of -1 when no domino can be rotated to x

Solution

1. Step 1: Analyze early return condition

   If neither side matches candidate x, the function should return -1 to indicate failure, not 0.

2. Step 2: Impact of returning 0

   Returning 0 falsely indicates zero rotations needed, causing incorrect positive results.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Returning -1 signals no solution; 0 misleads caller [OK]

Hint: Return -1 on failure, not 0 [OK]

Common Mistakes:

• Returning 0 instead of -1 on failure
• Overcounting rotations when both sides equal candidate

2. What is the time complexity of the optimal max heap approach to reorganize a string of length n with k unique characters?

medium

A. O(n²) because each character insertion may require scanning the entire string

B. O(n log k) because each of the n characters is pushed and popped from a heap of size k

C. O(k log n) because the heap operations depend on the string length

D. O(n) because each character is processed once without extra overhead

Solution

1. Step 1: Identify heap operations per character

   Each character is pushed and popped at most once per occurrence, total n operations.

2. Step 2: Analyze heap size and operation cost

   Heap size is at most k (unique chars), each push/pop is O(log k), so total O(n log k).

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Heap operations dominate, not scanning entire string [OK]

Hint: Heap ops cost O(log k) per character [OK]

Common Mistakes:

• Confusing n and k in complexity
• Assuming linear time without heap cost
• Mistaking quadratic due to nested loops

3. What is the time complexity of the optimal Task Scheduler algorithm using a max-heap for t total tasks and m unique tasks?

medium

A. O(t log m) because each task is pushed and popped from a heap of size up to m

B. O(t + m) because counting frequencies and scheduling are linear

C. O(m log t) because heap operations depend on total tasks

D. O(t * m) because each task may be compared with all unique tasks

Solution

1. Step 1: Analyze heap operations

   Heap size is at most m (unique tasks). Each task is pushed and popped at most once per execution.

2. Step 2: Calculate total operations

   For t tasks, each heap operation costs O(log m), so total is O(t log m).

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Heap size depends on unique tasks, not total tasks [OK]

Hint: Heap operations scale with unique tasks, not total tasks [OK]

Common Mistakes:

• Confusing total tasks and unique tasks
• Assuming linear heap operations
• Ignoring log factor in heap push/pop

4. What is the time complexity of the optimal greedy algorithm for the wiggle subsequence problem, and why might some candidates mistakenly think it is higher?

medium

A. O(n) because it scans the list once, updating counters based on difference signs

B. O(2^n) because it explores all subsequences recursively

C. O(n log n) due to sorting or binary search steps involved

D. O(n^2) because it compares each element with all previous elements

Solution

1. Step 1: Analyze algorithm operations

   The greedy algorithm iterates through the list once, computing differences and updating counters in O(1) time per element.

2. Step 2: Address common misconceptions

   Some candidates confuse it with brute force or DP approaches, thinking it compares pairs or explores subsequences exponentially, leading to O(n^2) or O(2^n) assumptions.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Single pass with constant work per element -> O(n) [OK]

Hint: Single pass with constant updates -> O(n)

Common Mistakes:

• Confusing with brute force exponential time
• Assuming nested loops for comparisons
• Thinking sorting is involved

5. Suppose the Jump Game problem is modified so that you can jump backward as well as forward (i.e., jumps can be negative or positive). Which of the following approaches correctly determines if you can reach the last index from the first index under this new constraint?

hard

A. Use the original greedy approach tracking max reachable index, ignoring backward jumps

B. Use a breadth-first search (BFS) or graph traversal to explore all reachable indices including backward jumps

C. Use dynamic programming with memoization to recursively check reachability from each index

D. Sort the array and apply binary search to find reachable indices efficiently

Solution

1. Step 1: Understand the impact of backward jumps

   Backward jumps mean the problem is no longer monotonic; maxReach tracking fails as reachable indices can decrease.

2. Step 2: Identify suitable approach

   BFS or graph traversal explores all reachable indices in any direction, correctly handling negative jumps.

3. Step 3: Explain why other options fail

   Greedy fails due to backward jumps; DP recursion is possible but less efficient; sorting is irrelevant.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   BFS explores all reachable nodes regardless of jump direction [OK]

Hint: Backward jumps break greedy; BFS needed to explore all reachable indices [OK]

Common Mistakes:

• Trying to apply greedy despite backward jumps
• Assuming sorting helps reachability