Maximum Units on a Truck

▶

Steps

setup

Start with unsorted boxTypes

The algorithm begins with the original boxTypes array and the truckSize available.

💡 Seeing the initial input helps understand the starting point before sorting.

Line:

boxTypes = [[1,3],[2,2],[3,1]]
truckSize = 4

💡 Initial data setup is crucial before any processing.

setup

Sort boxTypes by units per box descending

The boxTypes array is sorted so that box types with the highest units per box come first.

💡 Sorting ensures the greedy approach picks the most valuable boxes first.

Line:

boxTypes.sort(key=lambda x: x[1], reverse=True)

💡 Sorting by units per box is the key greedy step.

setup

Initialize totalUnits to 0

Set totalUnits to zero before starting to pick boxes.

💡 This variable will accumulate the total units loaded onto the truck.

Line:

totalUnits = 0

💡 Starting with zero total units is necessary for accumulation.

traverse

Start iterating over boxTypes with i=0

Begin processing the first box type with 1 box and 3 units each.

💡 The pointer i shows which box type is currently being considered.

Line:

for boxes, units in boxTypes:
    if truckSize == 0:
        break

💡 Iteration starts from the highest units per box type.

insert

Calculate how many boxes to take from boxType 0

Calculate take = min(boxes=1, truckSize=4) = 1 to pick all boxes of this type.

💡 We pick as many boxes as possible without exceeding truck capacity.

Line:

take = min(boxes, truckSize)

💡 Taking the full amount of this box type maximizes units early.

insert

Add units from boxType 0 to totalUnits

Add take * units = 1 * 3 = 3 units to totalUnits, updating totalUnits to 3.

💡 Accumulating units as boxes are loaded is the core calculation.

Line:

totalUnits += take * units

💡 Units accumulate incrementally as boxes are chosen.

shrink

Reduce truckSize by boxes taken from boxType 0

Decrease truckSize by take = 1, updating truckSize from 4 to 3.

💡 Truck capacity decreases as boxes are loaded, limiting future picks.

Line:

truckSize -= take

💡 Tracking remaining capacity is essential for stopping early.

traverse

Move to next boxType i=1

Advance pointer i to the second box type with 2 boxes and 2 units each.

💡 Moving pointer shows progression through sorted box types.

Line:

for boxes, units in boxTypes:

💡 Each box type is processed in order of units per box.

insert

Calculate boxes to take from boxType 1

Calculate take = min(boxes=2, truckSize=3) = 2 to take all boxes of this type.

💡 We take as many boxes as possible without exceeding remaining capacity.

Line:

take = min(boxes, truckSize)

💡 Partial or full box type selection depends on remaining capacity.

insert

Add units from boxType 1 to totalUnits

Add take * units = 2 * 2 = 4 units to totalUnits, updating totalUnits from 3 to 7.

💡 Units accumulate as boxes are loaded, increasing total units.

Line:

totalUnits += take * units

💡 Incremental addition builds the final answer.

shrink

Reduce truckSize by boxes taken from boxType 1

Decrease truckSize by take = 2, updating truckSize from 3 to 1.

💡 Truck capacity decreases as boxes are loaded, limiting future picks.

Line:

truckSize -= take

💡 Remaining capacity controls how many boxes can be taken next.

traverse

Move to next boxType i=2

Advance pointer i to the third box type with 3 boxes and 1 unit each.

💡 Moving pointer shows progression through sorted box types.

Line:

for boxes, units in boxTypes:

💡 Each box type is processed in order of units per box.

insert

Calculate boxes to take from boxType 2

Calculate take = min(boxes=3, truckSize=1) = 1 to fill remaining truck capacity.

💡 We take only as many boxes as the truck can hold.

Line:

take = min(boxes, truckSize)

💡 Partial selection occurs when capacity is limited.

insert

Add units from boxType 2 to totalUnits

Add take * units = 1 * 1 = 1 unit to totalUnits, updating totalUnits from 7 to 8.

💡 Units accumulate as boxes are loaded, increasing total units.

Line:

totalUnits += take * units

💡 Incremental addition builds the final answer.

shrink

Reduce truckSize by boxes taken from boxType 2

Decrease truckSize by take = 1, updating truckSize from 1 to 0.

💡 Truck capacity decreases as boxes are loaded, limiting future picks.

Line:

truckSize -= take

💡 Truck is now full, so no more boxes can be taken.

prune

Check if truckSize is zero to break early

Since truckSize is zero, the loop breaks early without processing more box types.

💡 Early stopping improves efficiency by avoiding unnecessary work.

Line:

if truckSize == 0:
    break

💡 Early stopping is a key optimization in greedy algorithms.

reconstruct

Return totalUnits as final answer

The algorithm returns totalUnits = 8 as the maximum units that can be loaded onto the truck.

💡 Returning the accumulated total units completes the solution.

Line:

return totalUnits

💡 The final answer is the sum of units from all selected boxes.

def maximumUnits(boxTypes, truckSize):
    boxTypes.sort(key=lambda x: x[1], reverse=True)  # STEP 2
    totalUnits = 0  # STEP 3
    for boxes, units in boxTypes:  # STEP 4
        if truckSize == 0:  # STEP 16
            break
        take = min(boxes, truckSize)  # STEP 5,9,13
        totalUnits += take * units  # STEP 6,10,14
        truckSize -= take  # STEP 7,11,15
    return totalUnits  # STEP 17

if __name__ == '__main__':
    boxTypes = [[1,3],[2,2],[3,1]]
    truckSize = 4
    print(maximumUnits(boxTypes, truckSize))  # Output: 8

📊

Maximum Units on a Truck - Watch the Algorithm Execute, Step by Step

Watching the algorithm step through sorting and greedy selection reveals how early stopping and inline calculations optimize the solution efficiently.

setup

[1,3]

0

[2,2]

1

[3,1]

2

Result: 0

sort

[1,3]

0

[2,2]

1

[3,1]

2

Result: 0

initialize

[1,3]

0

[2,2]

1

[3,1]

2

Result: 0

compare

i

[1,3]

0

[2,2]

1

[3,1]

2

Result: 0

move_right

i

[1,3]

0

[2,2]

1

[3,1]

2

Result: 0

record

i

[1,3]

0

[2,2]

1

[3,1]

2

Result: 3

move_left

i

[1,3]

0

[2,2]

1

[3,1]

2

Result: 3

compare

[1,3]

0

i

[2,2]

1

[3,1]

2

Result: 3

move_right

[1,3]

0

i

[2,2]

1

[3,1]

2

Result: 3

record

[1,3]

0

i

[2,2]

1

[3,1]

2

Result: 7

move_left

[1,3]

0

i

[2,2]

1

[3,1]

2

Result: 7

compare

[1,3]

0

[2,2]

1

i

[3,1]

2

Result: 7

move_right

[1,3]

0

[2,2]

1

i

[3,1]

2

Result: 7

record

[1,3]

0

[2,2]

1

i

[3,1]

2

Result: 8

move_left

[1,3]

0

[2,2]

1

i

[3,1]

2

Result: 8

prune

[1,3]

0

[2,2]

1

i

[3,1]

2

Result: 8

return

[1,3]

0

[2,2]

1

[3,1]

2

Result: 8

Key Takeaways

✓ Sorting box types by units per box descending is the foundation of the greedy approach.

This insight is hard to see from code alone because sorting is a separate step that enables the greedy selection order.

✓ The algorithm picks as many boxes as possible from the highest unit box type before moving on.

Visualizing the pointer moving and capacity shrinking clarifies how the greedy choice is applied stepwise.

✓ Early stopping when truckSize reaches zero avoids unnecessary processing and improves efficiency.

Seeing the break condition in action helps understand the optimization beyond the basic greedy logic.

Practice

1. Consider the following code snippet implementing the peak-valley approach to maximize stock profit. What is the final returned profit when the input prices are [1, 2, 3]?

def maxProfit(prices):
    i = 0
    profit = 0
    n = len(prices)
    while i < n - 1:
        while i < n - 1 and prices[i] >= prices[i + 1]:
            i += 1
        valley = prices[i]
        while i < n - 1 and prices[i] <= prices[i + 1]:
            i += 1
        peak = prices[i]
        profit += peak - valley
    return profit

easy

A. 2

B. 0

C. 3

D. 1

Solution

1. Step 1: Trace first while loop to find valley

   i=0, prices[0]=1, prices[1]=2, 1 < 2 so inner loop skips, valley=1

2. Step 2: Trace second while loop to find peak

   i increments while prices[i] <= prices[i+1]: i=0 to 1 (2 <= 3), i=1 to 2 (3 no next), peak=3

3. Step 3: Calculate profit and return

   profit += 3 - 1 = 2, loop ends, return 2

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Profit matches sum of positive differences (2) [OK]

Hint: Sum of (3-1) = 2 profit [OK]

Common Mistakes:

• Off-by-one error missing last peak
• Confusing valley and peak assignments
• Returning zero if no decreasing sequence found

2. You are given a list of non-negative integers and need to arrange them to form the largest possible number when concatenated. Which algorithmic approach guarantees an optimal solution for this problem?

easy

A. Dynamic Programming to find the maximum concatenation by exploring all subsequences

B. Sorting the numbers as strings using a custom comparator that compares concatenations

C. Greedy approach by always picking the largest integer first

D. Brute force generating all permutations and selecting the maximum concatenation

Solution

1. Step 1: Understand the problem requires ordering numbers to maximize concatenation

   The key is to compare pairs by concatenating in both possible orders and deciding which order yields a larger combined string.

2. Step 2: Recognize that sorting with a custom comparator based on concatenation comparisons guarantees optimal order

   This approach ensures the final concatenation is lexicographically largest, unlike greedy or DP which do not handle pairwise ordering correctly.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Custom comparator sorting is the standard solution for this problem [OK]

Hint: Compare concatenations as strings to decide order [OK]

Common Mistakes:

• Assuming greedy pick of largest integer works
• Using DP which is unnecessary
• Brute force is correct but inefficient

3. Consider the following buggy code for assigning cookies to children. Which line contains the subtle bug that can cause incorrect results?

medium

A. Line 5: Missing increment of j when cookie is assigned

B. Line 3: Incorrect loop condition including count < len(g)

C. Line 1: Missing sorting of g and s arrays

D. Line 7: Incrementing j only in else block

Solution

1. Step 1: Identify missing pointer increment

   When a cookie satisfies a child (s[j] ≥ g[i]), j should increment to avoid reusing the same cookie.

2. Step 2: Check code lines

   Line 5 increments count and i but does not increment j, causing the same cookie to be assigned multiple times.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Without j increment on assignment, cookie reuse occurs [OK]

Hint: Check pointer increments on both branches [OK]

Common Mistakes:

• Forgetting to sort arrays
• Incorrect loop conditions
• Not incrementing both pointers on assignment

4. The following code attempts to implement the optimal wiggle subsequence algorithm. Identify the line containing the subtle bug that causes incorrect results on inputs with consecutive equal elements.

def wiggleMaxLength(nums):
    if not nums:
        return 0
    count = 1
    last_diff = 0
    for i in range(1, len(nums)):
        diff = nums[i] - nums[i - 1]
        if (diff > 0 and last_diff < 0) or (diff < 0 and last_diff > 0):
            count += 1
            last_diff = diff
    return count

medium

A. Line 6: Using strict inequalities (last_diff < 0 and last_diff > 0) instead of inclusive (<= 0 and >= 0)

B. Line 3: Initializing count to 1 instead of 0

C. Line 7: Updating last_diff inside the if condition

D. Line 2: Returning 0 if nums is empty

Solution

1. Step 1: Understand the condition for counting wiggles

   The condition must allow last_diff to be zero or equal to zero to handle initial or equal consecutive elements correctly.

2. Step 2: Identify the bug

   Using strict inequalities excludes cases where last_diff is zero, causing the algorithm to skip valid wiggles after equal elements, leading to incorrect counts.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Inclusive inequalities fix counting on equal consecutive elements [OK]

Hint: Use <= 0 and >= 0 to handle zero last_diff correctly

Common Mistakes:

• Using strict inequalities causing missed wiggles
• Incorrect initialization of counters
• Updating last_diff outside condition

5. Suppose the problem is modified so that characters can be reused unlimited times (infinite supply), and you want to generate a string of length n with no two adjacent characters the same. Which modification to the max heap approach is necessary to handle this variant correctly?

hard

A. No change needed; the original heap approach works as is for infinite reuse

B. Track only the last used character and pick any other character from the set for the next position

C. Use a queue instead of a heap to cycle through characters ensuring no adjacency

D. Remove frequency counts and always push characters back immediately after use to allow reuse

Solution

1. Step 1: Understand infinite reuse implication

   Since characters can be reused infinitely, frequency counts are irrelevant; characters must be available again immediately after use.

2. Step 2: Modify heap usage

   Remove frequency decrement logic and always push characters back immediately after use to allow reuse while avoiding adjacency.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Immediate pushback enables infinite reuse without adjacency violation [OK]

Hint: Infinite reuse means no frequency decrement [OK]

Common Mistakes:

• Assuming original approach works unchanged
• Using queue without frequency logic
• Ignoring adjacency constraints