Problem Description

You are making a dessert and need to calculate the cost closest to a target price. The dessert consists of:

Components:

• Ice cream base: You must choose exactly one base flavor
• Toppings: You can choose zero or more types of toppings, and for each topping type you select, you can add either 1 or 2 of that topping

Inputs:

• baseCosts: An array of length n where baseCosts[i] is the price of the i-th ice cream base
• toppingCosts: An array of length m where toppingCosts[i] is the price of one unit of the i-th topping
• target: An integer representing your desired total cost

Rules:

• Must select exactly one ice cream base
• Can select 0, 1, or 2 units of each topping type
• Total cost = cost of base + sum of selected toppings

Goal: Find the total dessert cost that is closest to the target price. If there are multiple costs with the same distance from the target (e.g., one is above and one is below by the same amount), return the lower cost.

Example: If target = 10, and you can make desserts costing 8 and 12, both are distance 2 from the target. You would return 8 (the lower cost).

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• No: The problem doesn't involve nodes and edges or graph traversal. We're dealing with selecting items (base and toppings) to reach a target cost.

Need to solve for kth smallest/largest?

• No: We're not looking for the kth smallest or largest element. We need to find the cost closest to a target value.

Involves Linked Lists?

• No: The problem uses arrays for costs, not linked list operations.

Does the problem have small constraints?

• Yes: The problem has relatively small constraints. We need to explore all possible combinations of:

  • 1 base (n choices)
  • Each topping can be taken 0, 1, or 2 times (3^m possibilities for m toppings)

  This creates a finite search space that can be explored exhaustively.

Brute force / Backtracking?

• Yes: We need to explore all possible combinations of toppings (0, 1, or 2 of each type) combined with each base. Backtracking is perfect for generating all these combinations systematically.

Conclusion: The flowchart suggests using a backtracking approach. We can use backtracking to generate all possible topping combinations (where each topping can appear 0, 1, or 2 times), then combine each of these with every base to find the total cost closest to the target.

Intuition

The key insight is that we need to explore all possible dessert combinations to find the one closest to our target price. Since each topping can be taken 0, 1, or 2 times, we have 3 choices for each topping type. This naturally leads to a backtracking approach.

Think of it like building a decision tree: for each topping, we branch into three possibilities (take 0, 1, or 2). This gives us 3^m different topping combinations. However, the provided solution takes a clever optimization approach.

Instead of generating all 3^m combinations (which could be expensive), the solution first generates all possible sums using each topping at most once (0 or 1 times). This gives us 2^m combinations stored in the arr array through the DFS function. This is more efficient because 2^m < 3^m.

But wait - we need to handle the case where toppings can be taken twice! The solution handles this elegantly: since we can take up to 2 of each topping, we can think of it as selecting two (possibly overlapping) subsets of toppings. So for any dessert, the total topping cost can be represented as the sum of two values from our arr array.

The final approach becomes:

1. Generate all possible topping subset sums (where each topping appears 0 or 1 times) - this is our arr
2. For each base cost x
3. For each possible first subset sum y from arr
4. Use binary search to find the best second subset sum from arr that makes x + y + arr[j] closest to target

The binary search optimization works because we're looking for arr[j] such that x + y + arr[j] is closest to target, which means we want arr[j] closest to target - x - y. By sorting arr first, we can use binary search to quickly find candidates around this value.

Learn more about Dynamic Programming and Backtracking patterns.

Solution Approach

The implementation uses a combination of backtracking (via DFS) and binary search to efficiently find the optimal dessert cost.

Step 1: Generate all possible topping subset sums

The dfs function generates all possible sums where each topping is taken 0 or 1 times:

def dfs(i, t):
    if i >= len(toppingCosts):
        arr.append(t)
        return
    dfs(i + 1, t)                    # Don't take topping i
    dfs(i + 1, t + toppingCosts[i])  # Take one of topping i

This creates 2^m different sums stored in arr. Starting with dfs(0, 0), we recursively decide for each topping whether to include it or not, accumulating the sum in parameter t.

Step 2: Sort the array for binary search

arr.sort()

Sorting enables us to use binary search later to quickly find values close to our target.

Step 3: Try all combinations

Initialize tracking variables:

• d = inf: tracks the minimum distance from target
• ans = inf: stores the best cost found

For each base x and each first topping subset sum y:

for x in baseCosts:
    for y in arr:

Step 4: Binary search for the optimal third component

We need to find arr[j] such that x + y + arr[j] is closest to target. This means finding arr[j] closest to target - x - y:

i = bisect_left(arr, target - x - y)

bisect_left returns the insertion point where target - x - y would go in the sorted array. We check both positions i and i-1 because:

• arr[i] is the smallest value ≥ target - x - y
• arr[i-1] is the largest value < target - x - y

One of these will give us the closest sum.

Step 5: Update the answer

for j in (i, i - 1):
    if 0 <= j < len(arr):
        t = abs(x + y + arr[j] - target)
        if d > t or (d == t and ans > x + y + arr[j]):
            d = t
            ans = x + y + arr[j]

We update our answer if:

• We found a closer cost to target (smaller distance t)
• We found an equal distance but lower cost (tiebreaker rule)

The time complexity is O(n * 4^m + m * 2^m * log(2^m)) where:

• O(2^m) to generate all topping subsets
• O(2^m * log(2^m)) to sort the array
• O(n * 2^m * 2^m * log(2^m)) = O(n * 4^m * m) for the triple nested loop with binary search

Since we're using each value in arr twice (as y and as arr[j]), this effectively covers all cases where each topping can be taken 0, 1, or 2 times.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Given:

• baseCosts = [1, 7]
• toppingCosts = [3, 4]
• target = 10

Step 1: Generate all topping subset sums using DFS

We'll trace through the DFS to generate all possible sums where each topping is taken 0 or 1 times:

dfs(0, 0):
  ├─ dfs(1, 0):        # Don't take topping 0 (cost=3)
  │   ├─ dfs(2, 0):    # Don't take topping 1 (cost=4)
  │   │   └─ arr.append(0)
  │   └─ dfs(2, 4):    # Take topping 1 (cost=4)
  │       └─ arr.append(4)
  └─ dfs(1, 3):        # Take topping 0 (cost=3)
      ├─ dfs(2, 3):    # Don't take topping 1 (cost=4)
      │   └─ arr.append(3)
      └─ dfs(2, 7):    # Take topping 1 (cost=4)
          └─ arr.append(7)

Result: arr = [0, 4, 3, 7]

Step 2: Sort the array

arr = [0, 3, 4, 7] (sorted)

Step 3: Try all combinations

Initialize: d = inf, ans = inf

For each base and each first topping subset:

Base = 1:

• When y = 0: Need to find arr[j] closest to 10 - 1 - 0 = 9

  • Binary search finds insertion point at index 4 (after 7)
  • Check j = 3: cost = 1 + 0 + 7 = 8, distance = |8 - 10| = 2
  • Update: d = 2, ans = 8

• When y = 3: Need to find arr[j] closest to 10 - 1 - 3 = 6

  • Binary search finds insertion point at index 3 (between 4 and 7)
  • Check j = 3: cost = 1 + 3 + 7 = 11, distance = |11 - 10| = 1
  • Check j = 2: cost = 1 + 3 + 4 = 8, distance = |8 - 10| = 2
  • Update: d = 1, ans = 11

• When y = 4: Need to find arr[j] closest to 10 - 1 - 4 = 5

  • Binary search finds insertion point at index 3
  • Check j = 3: cost = 1 + 4 + 7 = 12, distance = |12 - 10| = 2
  • Check j = 2: cost = 1 + 4 + 4 = 9, distance = |9 - 10| = 1
  • No update (same distance, but 11 > 9, so we keep ans = 11)

• When y = 7: Need to find arr[j] closest to 10 - 1 - 7 = 2

  • Binary search finds insertion point at index 1
  • Check j = 1: cost = 1 + 7 + 3 = 11, distance = |11 - 10| = 1
  • Check j = 0: cost = 1 + 7 + 0 = 8, distance = |8 - 10| = 2
  • No update (already have distance 1)

Base = 7:

• When y = 0: Need to find arr[j] closest to 10 - 7 - 0 = 3
  • Binary search finds insertion point at index 1
  • Check j = 1: cost = 7 + 0 + 3 = 10, distance = |10 - 10| = 0
  • Update: d = 0, ans = 10 ✓

Since we found an exact match (distance = 0), we have our answer: 10

This corresponds to choosing:

• Base with cost 7
• One unit of the first topping (cost 3)
• Zero units of the second topping
• Total: 7 + 3 = 10

Solution Implementation

Time and Space Complexity

Time Complexity: O(3^m + n * 4^m * log(3^m)) where n is the length of baseCosts and m is the length of toppingCosts.

The analysis breaks down as follows:

• The dfs function generates all possible topping combinations. Since each topping can be chosen 0, 1, or 2 times (the code shows 0 or 1 time per recursive call, but the outer loop with y effectively allows selecting the same combination twice), there are 3^m possible combinations.
• Generating all combinations takes O(3^m) time.
• Sorting the array takes O(3^m * log(3^m)) time.
• The main nested loops iterate through:
  • n base costs
  • 3^m values in arr for variable y
  • For each combination, bisect_left performs binary search in O(log(3^m)) time
  • Checking at most 2 positions (indices i and i-1)
• Total for the nested loops: O(n * 3^m * log(3^m))

Since the code iterates through arr twice (once for y and once for binary search target), the effective complexity becomes O(n * 9^m * log(3^m)) which simplifies to O(n * 4^m * log(3^m)) considering the dominant terms.

Space Complexity: O(3^m) where m is the length of toppingCosts.

The space usage comes from:

• The arr list storing all possible topping combinations: O(3^m)
• The recursion stack depth for DFS: O(m)
• Since 3^m >> m, the overall space complexity is O(3^m)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Topping Combination Generation

A common mistake is generating topping combinations where each topping can only be used 0 or 1 times, forgetting that the problem allows using up to 2 units of each topping.

Incorrect approach:

def generate_combinations(index, current_cost):
    if index >= len(toppingCosts):
        combinations.append(current_cost)
        return
    # Only considering 0 or 1 of each topping
    generate_combinations(index + 1, current_cost)
    generate_combinations(index + 1, current_cost + toppingCosts[index])

Correct approach:

def generate_combinations(index, current_cost):
    if index >= len(toppingCosts):
        combinations.append(current_cost)
        return
    # Consider 0, 1, or 2 of each topping
    generate_combinations(index + 1, current_cost)
    generate_combinations(index + 1, current_cost + toppingCosts[index])
    generate_combinations(index + 1, current_cost + 2 * toppingCosts[index])

2. Exponential Memory Usage Without Optimization

The current solution generates 3^m combinations (where m is the number of toppings), which can cause memory issues for large m. The provided solution actually uses a clever optimization by generating 2^m combinations and using them twice, but developers might miss this optimization.

Memory-intensive approach:

# Generates 3^m combinations directly
topping_combinations = []
generate_all_3_way_combinations(0, 0)  # This creates 3^m entries

Optimized approach (as in the solution):

# Generate 2^m combinations and reuse them
# The double loop (for y in arr, for arr[j]) effectively covers all 3^m cases

3. Incorrect Tie-Breaking Logic

When two costs have the same distance from the target, the problem requires returning the lower cost. A common mistake is to return the higher cost or not handle ties at all.

Incorrect:

if current_difference < min_difference:
    min_difference = current_difference
    closest_cost = total_cost
# Missing the tie-breaker condition

Correct:

if (current_difference < min_difference or 
    (current_difference == min_difference and total_cost < closest_cost)):
    min_difference = current_difference
    closest_cost = total_cost

4. Off-by-One Errors in Binary Search

When using bisect_left, developers often forget to check both the insertion point and the previous element, missing potentially optimal solutions.

Incorrect:

insertion_point = bisect_left(topping_combinations, remaining)
if insertion_point < len(topping_combinations):
    # Only checking one position
    total_cost = base_cost + topping_sum + topping_combinations[insertion_point]

Correct:

insertion_point = bisect_left(topping_combinations, remaining)
# Check both positions: the insertion point and the element before it
for candidate_index in (insertion_point, insertion_point - 1):
    if 0 <= candidate_index < len(topping_combinations):
        total_cost = base_cost + topping_sum + topping_combinations[candidate_index]

5. Not Handling Edge Cases

Forgetting to handle cases where the topping array is empty or when all combinations exceed the target.

Solution: Always initialize tracking variables properly and ensure the algorithm works even when toppingCosts is empty (the combination array will just contain [0]).