Flipkart Coding Question – Solved

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To tackle this modified knapsack problem, it is crucial first to grasp the unique characteristics that distinguish it from the classical knapsack problem and then develop an approach that exploits these characteristics effectively.

Step 1: Understand the problem specifics and constraints
- You have n items, indexed 0 to n-1.
- Each item i has a weight of 2^i, i.e., exponential weights doubling with the index.
- Each item i has a given cost cost[i].
- You can purchase multiple units of each item.
- Your goal is to buy items (in quantities possibly greater than one) such that the total combined weight is at least minWeight.
- The objective is to minimize the total cost of purchased items.

Step 2: Reflect on classical knapsack differences
- Unlike classical knapsack problems where each item weight and cost is arbitrary, here weights are powers of two, which is a special structure.
- Also, you can buy multiple units of the same item, resembling an unbounded knapsack scenario.
- The minWeight requirement is a "greater than or equal to" constraint, meaning you can overshoot the required weight.
- Traditional dynamic programming techniques might struggle with potentially large weights, but the exponential nature offers optimization opportunities.

Step 3: Key insight - exploiting powers of two weights
- Because weights are powers of two, any total weight can be represented in binary form.
- Buying k units of item i adds k × 2^i weight.
- Multiple units of item i can sum to large weights, but you can also think in terms of combining items to cover the binary representation of the target weight.
- This suggests that the problem might be related to covering binary weight requirements efficiently.

Step 4: Consider preprocessing costs for combining items
- Since you can buy multiple units, you can think of "upgrading" the cost of buying 2 units of item i as possibly cheaper or more expensive than buying one unit of item i+1.
- For example, buying two units of item i results in weight 2 × 2^i = 2^(i+1), which is the weight of item i+1.
- However, the cost of two units of item i might be more or less than the cost of one unit of item i+1.
- So it might be beneficial to precompute a minimal cost for each weight 2^i considering possible multiple purchases of smaller items.
- This "cost optimization" step can help in choosing the best combination of items later.

Step 5: Normalize costs upwards
- Traverse from smaller weights to larger weights and update cost[i+1] = min(cost[i+1], 2 × cost[i]).
- This ensures that the cost for higher weight items is never more expensive than buying two smaller items.
- This step reduces redundant purchases and can simplify the problem.

Step 6: Construct an approach to achieve minWeight
- After cost normalization, decompose minWeight into its binary representation.
- For each bit in minWeight, if bit i is set, you must at least cover 2^i weight.
- Sum the costs of those items corresponding to the bits set in minWeight.
- However, because you can overshoot minWeight and you want to minimize cost, sometimes buying an item with higher weight but cheaper cost is better than combining multiple smaller items.

Step 7: Consider overshoot and partial purchases to minimize cost
- Starting from the highest bit position down to the lowest:
- Keep track of the accumulated cost of required items.
- Also, at each bit, consider whether buying a larger item (with weight 2^(i+1)) might reduce overall cost.
- You can maintain a running minimal cost answer while exploring these trade-offs.

Step 8: Implement a strategy to find minimum cost with potential overshoot
- Initialize a variable to store minimal cost answer as very large.
- Initialize a current sum cost as zero.
- Traverse bits from high to low (most significant to least significant).
- At each step:
- If the bit in minWeight is set, add cost of that item to current sum cost.
- Keep track of the minimal cost considering buying bigger items earlier (allowing overshoot).
- Update minimal cost answer to the minimum of itself and current sum cost plus any future costs.
- This ensures that you do not miss cheaper combinations where buying a heavier item instead of multiple lighter items reduces cost.

Step 9: Handling edge cases
- If minWeight is zero, cost is zero.
- When n is small, exhaustive checks might be easier.
- When costs vary widely, ensure cost normalization step is properly handled to avoid overpaying.

Step 10: Summary of thought process
- The problem is a twist on the unbounded knapsack with weights being powers of two.
- Preprocessing costs to ensure buying higher weight items is not more expensive than multiples of smaller ones is crucial.
- Decomposing the minimum required weight into binary form aligns well with the weights of items.
- Explore partial and overshoot strategies to minimize total cost.
- Employ a careful traversal from high to low bits, accumulating cost and checking if better options exist by overshooting.
- This approach balances the need for meeting the weight requirement and minimizing cost effectively.

In essence, the solution exploits the special property of weights being powers of two, uses cost normalization to optimize purchases, and intelligently combines items by considering binary weight decomposition with the possibility of overshooting to find a minimal cost covering the minWeight constraint.
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