Problem Description

You are shopping in a store that has n different items, each with its own price. The store offers special bundles where you can buy multiple items together at a discounted price.

Given:

• price[i]: the individual price of the i-th item
• needs[i]: the exact quantity you need to buy of the i-th item
• special: a list of special offers, where each offer special[i] contains:
  • special[i][0] to special[i][n-1]: the quantity of each item included in this bundle
  • special[i][n]: the total price for this bundle

Your task is to find the minimum cost to buy exactly the items you need. You can:

• Buy items individually at their regular prices
• Use any special offer multiple times
• Mix and match individual purchases with special offers

However, you cannot buy more items than needed, even if it would result in a lower total cost.

For example, if you need [2, 2] of two items priced at [2, 5] each, and there's a special offer [3, 0, 5] (3 of item 0 for $5), you cannot use this offer since you only need 2 of item 0, not 3.

The goal is to determine the optimal combination of individual purchases and special offers to minimize the total cost while getting exactly the quantities specified in needs.

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. It's about finding the minimum cost to purchase items with special offers.

Need to solve for kth smallest/largest?

• No: We're not looking for the kth smallest or largest element. We need to find the minimum total cost.

Involves Linked Lists?

• No: The problem deals with arrays representing prices, needs, and special offers, not linked lists.

Does the problem have small constraints?

• Yes: The problem explicitly states that n ≤ 6 (at most 6 types of items) and each item quantity needed is at most 10. These are very small constraints that make exhaustive search feasible.

Brute force / Backtracking?

• Yes: With small constraints, we can explore all possible combinations of using special offers and buying items individually. We need to try different combinations of special offers (which can be used multiple times) and backtrack when a combination doesn't work or isn't optimal.

Conclusion: The flowchart correctly leads us to use a Backtracking approach. The small constraints (n ≤ 6 items, quantities ≤ 10) make it feasible to explore all possible ways to fulfill the shopping needs. We recursively try each special offer if applicable, backtrack, and compare with buying items individually to find the minimum cost.

Intuition

When we look at this shopping problem, we need to find the cheapest way to buy exactly what we need. At each decision point, we have multiple choices: buy items individually at regular prices, or use one of the special offers. This creates a decision tree where we need to explore different combinations.

The key insight is that this is an optimization problem with choices at each step. We can think of it as being in a certain "state" - having a specific shopping list of items we still need to buy. From this state, we can:

1. Buy all remaining items individually (base case)
2. Try each special offer that doesn't exceed our needs, creating a new state with reduced needs

Since special offers can be used multiple times, we might revisit similar states through different paths. For example, using offer A twice might lead to the same remaining needs as using offer B once. This suggests we should remember (memoize) the minimum cost for each state to avoid recalculating.

The small constraints (n ≤ 6 items, quantities ≤ 10) make another optimization possible: state compression. Instead of using a list to represent our needs, we can encode the entire shopping state into a single integer. With 4 bits per item (supporting quantities 0-15), we only need 24 bits total for 6 items. This makes memoization more efficient - we can use the compressed state as a direct key.

The backtracking nature emerges from trying each valid special offer, recursively solving the subproblem with reduced needs, then "backtracking" to try other offers. We compare all these paths plus the option of buying everything individually to find the minimum cost. The recursion naturally handles the fact that offers can be used multiple times - when we recursively call with reduced needs, that subproblem might use the same offer again if it's still valid.

Learn more about Memoization, Dynamic Programming, Backtracking and Bitmask patterns.

Solution Approach

The solution uses state compression and memoization search (DFS with caching) to efficiently explore all possible ways to fulfill the shopping needs.

State Compression: Since we have at most 6 items and each item's quantity doesn't exceed 10, we can use 4 bits to represent each item's quantity (4 bits can represent 0-15). The entire shopping list is compressed into a single integer mask where:

• Item 0's quantity is stored in bits 0-3
• Item 1's quantity is stored in bits 4-7
• Item i's quantity is stored in bits i * 4 to (i + 1) * 4 - 1

For example, if needs = [1, 2, 1], then mask = 0b0001 0010 0001 (in binary).

DFS Function: The core function dfs(cur) calculates the minimum cost to fulfill the shopping list represented by state cur. Here's how it works:

1. Base cost calculation: First, calculate the cost of buying all remaining items individually:

   ans = sum(p * (cur >> (i * bits) & 0xF) for i, p in enumerate(price))

   This extracts each item's quantity from cur using bit operations and multiplies by its price.

2. Try each special offer: For each special offer, check if it's valid (doesn't exceed current needs):

   for offer in special:
       nxt = cur
       for j in range(len(needs)):
           if (cur >> (j * bits) & 0xF) < offer[j]:
               break  # Offer exceeds needs, invalid
           nxt -= offer[j] << (j * bits)  # Subtract offer quantities

3. Recursive calculation: If the offer is valid (the loop completes without breaking), recursively calculate the cost with the reduced needs:

   else:  # Python's for-else: executes if loop didn't break
       ans = min(ans, offer[-1] + dfs(nxt))

   The total cost is the offer's price plus the minimum cost for the remaining needs.

4. Memoization: The @cache decorator automatically memoizes results, preventing redundant calculations for states we've already solved.

Initialization: The algorithm starts by converting the initial needs list into the compressed state mask:

bits, mask = 4, 0
for i, need in enumerate(needs):
    mask |= need << (i * bits)

Then calls dfs(mask) to find the minimum cost for the initial shopping list.

The beauty of this approach is that it naturally handles:

• Using offers multiple times (through recursion)
• Finding the optimal combination of offers and individual purchases
• Avoiding redundant calculations through memoization
• Efficient state representation through bit manipulation

Example Walkthrough

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

Example:

• price = [2, 5] (Item 0 costs $2, Item 1 costs$5)
• needs = [3, 2] (Need 3 of Item 0, 2 of Item 1)
• special = [[1, 1, 3], [2, 0, 4]]
  • Offer 1: Buy 1 of Item 0 + 1 of Item 1 for $3
  • Offer 2: Buy 2 of Item 0 for $4

Step 1: State Compression Convert needs = [3, 2] into a bitmask:

• Item 0 (quantity 3): stored in bits 0-3 as 0011
• Item 1 (quantity 2): stored in bits 4-7 as 0010
• Initial mask = 0010 0011 (binary) = 35 (decimal)

Step 2: DFS Exploration from dfs(35)

Starting with state 35 (needs [3, 2]):

1. Calculate base cost (buying individually):

   • Extract quantities: Item 0 = 3, Item 1 = 2
   • Cost = 3 × $2 + 2 ×$5 = $16

2. Try Offer 1 [1, 1, 3]:

   • Check validity: Need 3 ≥ 1 (Item 0) ✓, Need 2 ≥ 1 (Item 1) ✓
   • New state after using offer:
     • Item 0: 3 - 1 = 2 (bits 0-3: 0010)
     • Item 1: 2 - 1 = 1 (bits 4-7: 0001)
     • New mask = 0001 0010 = 18
   • Cost = $3 + dfs(18)

3. Try Offer 2 [2, 0, 4]:

   • Check validity: Need 3 ≥ 2 (Item 0) ✓, Need 2 ≥ 0 (Item 1) ✓
   • New state after using offer:
     • Item 0: 3 - 2 = 1 (bits 0-3: 0001)
     • Item 1: 2 - 0 = 2 (bits 4-7: 0010)
     • New mask = 0010 0001 = 33
   • Cost = $4 + dfs(33)

Step 3: Recursive Calls

For dfs(18) (needs [2, 1]):

1. Base cost: 2 × $2 + 1 ×$5 = $9
2. Try Offer 1: Valid → New state = 1 (needs [1, 0]) → Cost = $3 + dfs(1)
3. Try Offer 2: Valid → New state = 16 (needs [0, 1]) → Cost = $4 + dfs(16)
4. Return minimum

For dfs(33) (needs [1, 2]):

1. Base cost: 1 × $2 + 2 ×$5 = $12
2. Try Offer 1: Valid → New state = 16 (needs [0, 1]) → Cost = $3 + dfs(16)
3. Try Offer 2: Invalid (need 1 < 2 for Item 0)
4. Return minimum

Step 4: Base Cases

Eventually reaching simpler states:

• dfs(1) (needs [1, 0]): Base cost = $2, no valid offers
• dfs(16) (needs [0, 1]): Base cost = $5, no valid offers

Step 5: Backtracking and Result

Working backward with memoized results:

• dfs(1) = $2
• dfs(16) = $5
• dfs(18) = min($9,$3 + $2,$4 + $5) = min($9, $5,$9) = $5
• dfs(33) = min($12,$3 + $5) = min($12, $8) =$8
• dfs(35) = min($16,$3 + $5,$4 + $8) = min($16, $8,$12) = $8

Final Answer: $8

The optimal strategy is to use Offer 1 twice:

• First use: [1, 1, 3] → reduces needs from [3, 2] to [2, 1]
• Second use: [1, 1, 3] → reduces needs from [2, 1] to [1, 0]
• Buy remaining 1 of Item 0 individually for $2
• Total: $3 +$3 + $2 =$8

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × k × m^n)

The algorithm uses dynamic programming with memoization to explore different purchasing options. The state is encoded as an integer where each item's quantity occupies 4 bits. With n types of items and maximum demand m for each item, there are at most m^n possible states (combinations of quantities). For each state, the algorithm:

• Calculates the direct purchase cost: O(n) operations
• Iterates through k special offers, and for each offer, checks if it's applicable by examining all n items: O(k × n) operations

Therefore, the overall time complexity is O(m^n × (n + k × n)) = O(n × k × m^n).

Space Complexity: O(n × m^n)

The space is primarily consumed by:

• The memoization cache which stores up to m^n states (all possible combinations of item quantities)
• Each recursive call uses O(n) space for local variables and iteration
• The recursion depth can be at most O(m^n) in the worst case

Since the cache dominates the space usage and each cached entry requires O(n) bits to represent the state, the total space complexity is O(n × m^n).

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

Common Pitfalls

1. Incorrect Bit Extraction and Manipulation

One of the most common mistakes is incorrectly extracting or updating item quantities in the bit-packed state. Developers often make errors with bit shift operations or masks.

Pitfall Example:

# Wrong: Using wrong bit mask or shift amount
quantity = current_state >> (item_idx * 4) & 0xFF  # Using 0xFF instead of 0xF
# or
quantity = current_state & (0xF << (item_idx * 4))  # Incorrect order of operations

Solution: Always use parentheses to ensure correct order of operations and use the right mask (0xF for 4 bits):

quantity = (current_state >> (item_idx * BITS_PER_ITEM)) & 0xF

2. Integer Overflow with Large State Values

When dealing with 6 items, the state requires 24 bits. If not careful with bit operations, especially subtraction, you might encounter unexpected behavior.

Pitfall Example:

# Potential issue if subtraction makes state negative
next_state = current_state
for j in range(len(needs)):
    next_state -= offer[j] << (j * BITS_PER_ITEM)  # Could go negative if not checked

Solution: Always validate that the offer can be applied before performing subtraction:

# Check feasibility first
for item_idx in range(len(needs)):
    if (current_state >> (item_idx * BITS_PER_ITEM)) & 0xF < offer[item_idx]:
        can_apply_offer = False
        break
# Only subtract if offer is valid
if can_apply_offer:
    for item_idx in range(len(needs)):
        next_state -= offer[item_idx] << (item_idx * BITS_PER_ITEM)

3. Forgetting to Filter Invalid Special Offers

Some special offers might be inherently invalid (e.g., offering more items than maximum needs). Not filtering these can lead to unnecessary computation.

Pitfall Example:

# Processing all offers without validation
for offer in special:
    # Directly trying to apply without checking if it's ever usable

Solution: Pre-filter offers that exceed any item's needs:

# Filter out offers that are never applicable
valid_specials = []
for offer in special:
    is_valid = all(offer[i] <= needs[i] for i in range(len(needs)))
    if is_valid:
        valid_specials.append(offer)

4. Not Handling Edge Cases

Missing edge cases like empty needs, no special offers, or zero-priced items can cause issues.

Pitfall Example:

# Not handling empty needs or special offers
if not needs:  # This check is missing
    return 0

Solution: Add proper edge case handling:

# Handle edge cases
if not needs or all(n == 0 for n in needs):
    return 0
if not special:
    return sum(p * n for p, n in zip(price, needs))

5. Inefficient State Representation

Using more bits than necessary or not considering the constraints can lead to memory issues.

Pitfall Example:

# Using 8 bits per item when 4 would suffice
BITS_PER_ITEM = 8  # Wasteful if max quantity is 10

Solution: Calculate the minimum bits needed based on constraints:

# Calculate bits needed based on max quantity
max_quantity = max(needs) if needs else 0
BITS_PER_ITEM = max_quantity.bit_length() if max_quantity > 0 else 1
# But for this problem, 4 bits is sufficient since max is 10
BITS_PER_ITEM = 4