2735. Collecting Chocolates - Detailed Explanation

How this optimized solution works: We maintain an array currentMin[i] which at iteration j holds the value of f[i][j] (min cost for type i with j rotations allowed). Initially (j=0) this is just nums[i]. Then for each additional rotation j, we update each currentMin[i] = min(currentMin[i], nums[(i-j+n)%n]), which implements the DP transition described. We also accumulate the sum of currentMin[i] as we update, to compute chocoCost(j) on the fly. We add the rotation cost j*x and check if this total is the best seen. By not using a 2D array, we only store the current min costs and update them in place, achieving O(n) space use.

This DP approach is effectively doing the same work as the brute force, but in a much more efficient manner by reusing previous results instead of recomputing mins from scratch for each j. It avoids the explicit inner loop over k that the brute force used to scan previous indices for each type. The result returned is the minimum total cost over all possible rotation counts.

Step-by-Step Walkthrough

Let’s walk through Example 1 (nums = [20, 1, 15], x = 5) with the DP approach to see how it finds the optimal solution:

• Initial state (0 rotations): Each type is only at its own index.

  • Type 0 cost = nums[0] = 20
  • Type 1 cost = nums[1] = 1
  • Type 2 cost = nums[2] = 15
    Sum of costs = 20 + 1 + 15 = 36. Total = 36 (since 0 rotations cost nothing).

• Allow 1 rotation: Now each type can also appear at one additional index (the previous index in circular order):

  • Type 0 appears initially at index 0 (cost 20) and after 1 rotation at index (0-1 mod 3) = 2 (cost 15). Minimum cost for type 0 with 1 rotation = min(20, 15) = 15.
  • Type 1 appears at index 1 (cost 1) and after 1 rotation at index 0 (cost 20). Min cost for type 1 = min(1, 20) = 1.
  • Type 2 appears at index 2 (cost 15) and after 1 rotation at index 1 (cost 1). Min cost for type 2 = min(15, 1) = 1.
    Now the sum of minimum costs = 15 + 1 + 1 = 17. We must add the cost of 1 rotation: 17 + 5 = 22. So with one rotation, the best total is 22.

  (To visualize the rotation: initially types were [0,1,2] at costs [20,1,15]. After one rotation, types become [1,2,0] at those indices, so type 0 moved to index 2 with cost 15, type 2 moved to index 1 with cost 1, etc. That matches our min costs above.)

• Allow 2 rotations: Now each type can appear at up to 3 indices (which in this case is actually all indices, since n=3). We extend each type’s options one more step:

  • Type 0 was at indices {0, 2} with min cost 15. After 2 rotations, it also appears at index (0-2 mod 3) = 1 (cost 1). Now type 0’s min cost = min(previous 15, new 1) = 1.
  • Type 1 was at {1, 0} with min cost 1. After another rotation, it appears at (1-2 mod 3) = 2 (cost 15). Min cost for type 1 = min(1, 15) = 1 (unchanged).
  • Type 2 was at {2, 1} with min cost 1. After another rotation, it appears at (2-2 mod 3) = 0 (cost 20). Min cost for type 2 = min(1, 20) = 1 (unchanged).
    Sum of min costs now = 1 + 1 + 1 = 3. Adding cost of 2 rotations: 3 + 2*5 = 13.

  With 2 rotations allowed, effectively each type can be taken from the cheapest of all 3 indices, which for all types happened to be cost 1 (type 0’s cheapest was at index 1, type 1’s at index 1 as well, type 2’s at index 1 or 2 – either way cost 1). The total 13 is indeed the minimum possible . The DP found that at j=2 rotations, the total cost is 13, and it will consider j=3 (which would just repeat the pattern with an extra unnecessary rotation cost) and stop, giving 13 as the answer.

This step-by-step shows how the algorithm incrementally builds the best costs. It matches the earlier explanation of Example 1: first no rotation (cost 36), then one rotation (22), then two rotations (13), which is chosen as optimal.

For a smaller example, consider Example 3 (nums = [8,2,4], x = 3):

• j=0: min costs per type = [8, 2, 4], sum = 14, total = 14.
• j=1: each type gains one more index:
  • Type0: indices {0, 2} → costs {8, 4} → min = 4
  • Type1: indices {1, 0} → costs {2, 8} → min = 2
  • Type2: indices {2, 1} → costs {4, 2} → min = 2
    Sum = 4+2+2 = 8, plus rotation cost 3 → total = 11.
• j=2: each type gains the last index:
  • Type0: {0,2,1} costs {8,4,2} → min = 2
  • Type1: {1,0,2} costs {2,8,4} → min = 2
  • Type2: {2,1,0} costs {4,2,8} → min = 2
    Sum = 2+2+2 = 6, plus rotation cost 6 → total = 12.
    The minimum total was 11 at j=1, matching our earlier reasoning. The algorithm would correctly output 11.

Common Mistakes & Edge Cases

Common Pitfalls:

• Misunderstanding the rotation: A frequent mistake is to think that rotating “moves the chocolates” or to simulate it incorrectly. Remember, the chocolates stay at the same index, but their type labels change each operation. It’s essentially a cyclic permutation of the type-to-index mapping, not a shuffle of the cost array itself. Keeping track of which index currently holds which type is crucial.

• Greedy picking in wrong order: One might greedily pick the cheapest available chocolate first, then rotate, then pick the next cheapest, and so on. This can fail because taking a slightly more expensive chocolate early might allow you to avoid an extra rotation cost. The optimal strategy may require paying the rotation cost first to make multiple chocolates cheaper, instead of grabbing one cheap chocolate immediately. That’s why a global DP approach by rotation count is safer than a greedy per-step choice.

• Assuming all types must use the same number of rotations: It might seem like each type could be rotated a different number of times independently, but since rotations apply to all chocolates simultaneously, we use a single j for all. Some solutions go wrong by treating each type separately without accounting that rotations have a global cost and effect. Our approach balances the rotation cost across all types.

• Index arithmetic bugs: When implementing, it’s easy to get the modulo index wrong (especially dealing with negative indices). We used (i - j + n) % n to safely wrap around to a valid index for the array. Off-by-one errors in these calculations could lead to incorrect minima.

• Ignoring 0 rotations case: Always check the scenario of doing no operations – sometimes the initial configuration is already optimal (like Example 2). An algorithm that always does at least one rotation could overshoot. Our loop includes j=0 for this reason.

• Overflow issues: Summing costs of up to 1000 chocolates each up to 1e9 could reach ~1e12, which doesn’t fit in a 32-bit int. We used long (64-bit) for total cost accumulation. Not doing so could cause overflow and wrong answers for large values.

Edge Cases to consider:

• n = 1 (single chocolate): No rotations are needed (or useful) because there’s only one type. The answer is just nums[0] regardless of x. Our solution handles this: it would try j=0 (sum = nums[0], cost = nums[0]) and j doesn’t go to 1 because n-1 = 0.

• All costs equal: If nums are all the same, any rotation doesn’t change the fact you’ll pay that same cost for each type. The best answer is simply that cost * n (and never rotating, since rotations would only add cost with no benefit). The algorithm finds no improvement when checking j>0, so it sticks with j=0.

• Rotation cost extremely high: If x is very large (e.g., 1e9) compared to the variation in nums, it might never be worth rotating even once. The optimal j will be 0. Our method does evaluate j=0 and will correctly identify it as minimum.

• Rotation cost extremely low: If x is very small (e.g., 1) and there are significantly cheaper costs reachable via rotations, the optimal strategy might be to rotate as much as needed (up to n-1 times). In the extreme, if one index has a very low cost compared to others, the algorithm might rotate enough so that each type can be collected from that low-cost index at least once. (For example, if nums = [100, 1, 100] and x=1, it’s worth rotating so that the type that started at index 0 and index 2 can both move into the index1 position where cost=1, even though you pay some rotation cost.) Our solution will consider all j up to n-1 and find that optimum.

• Duplicate costs or multiple minima: The array might have multiple indices with the same minimal cost. The solution still works because each type’s minimum is computed independently. If, say, two different indices both have cost 5 which is lowest, each type will still end up with min cost 5 after enough rotations. The presence of multiple equal minima just means you might achieve the global minimum cost with fewer rotations (since not every type needs to hit the same single index; there are multiple cheap spots). The DP handles this naturally by always taking the min encountered.

Alternative Variations & Related Problems

Related LeetCode problems for practice:

• “Sum of Subarray Minimums” (LeetCode 907): Although a different scenario (finding the sum of minimums of all subarrays of an array), it also involves efficiently computing minimum values over various ranges. The technique of avoiding redundant computations via DP or a stack is similar in spirit to how we optimized scanning minima here.

• “Minimum Cost to Move Chips to The Same Position” (LeetCode 1217): A simpler problem where you have to decide whether or not to pay a cost to move items. It teaches the idea of comparing the cost of making a move versus not making it, somewhat like deciding how many rotations to perform.

• “Minimum Cost to Make Array Equal” (LeetCode 2448): A harder problem involving choosing an alignment for array values with costs. It’s not directly about rotations, but it involves minimizing total cost via a strategy (in that case, choosing a target value, often solved by a median-related approach). It’s another example of balancing individual element costs with a global decision.

• Weekly Contest 349 Problems: “Collecting Chocolates” was one of the contest problems. You might also want to try other problems from that contest (e.g., 2734 or 2736) to practice different techniques (though they cover different topics like queries and array manipulation). Solving contest problems can expose you to a variety of clever solutions and optimizations.

Variations of the problem:

• If the rotation operation could be done in the opposite direction (rotating types backward instead of forward), the problem would essentially be the same – since rotating backward by 1 is equivalent to rotating forward by n-1. The solution would not change; you’d still consider up to n-1 operations in either direction.

• If the operation cost x varied or increased with each rotation, the decision process would be more complex. For example, if the first rotation cost x, the second cost 2x, etc., then doing many rotations becomes progressively more expensive. You would need to account for a different cost for each rotation in the DP state or stop earlier if cost grows. Our current solution assumes a constant cost per rotation.

• If you could choose to rotate by more than one step in a single operation (say rotate types by 2 positions at once for a cost), you’d need to adjust the states to account for larger jumps. However, any multi-step rotation can be decomposed into single-step rotations (just maybe fewer operations), so it might reduce the number of operations needed but each operation might cost more – the strategy would then weigh doing one big rotation vs multiple small ones.

• A different twist: imagine each type had multiple chocolates (multiple indices with that type initially). Then rotating would shuffle groups of chocolates. You would then want to collect one of each type, possibly picking the cheapest among all instances of a type. That becomes a different problem – essentially you’d just pick the cheapest instance of each type initially (since rotation just permutes types, it wouldn’t help if you already have many of each type; rotation matters in our problem because there’s exactly one of each type initially).

• Another variation could limit the number of operations you can perform (e.g., at most K rotations). In that case, you would only consider j up to K (or n-1, whichever is smaller). The DP approach easily adapts to that – just stop the loop at j=K. If K < n-1, you might not be able to see all indices for each type, so the answer could be higher.

• Lastly, this problem demonstrates a technique of optimizing a decision (how many global operations to do) with a DP that precomputes consequences of that decision (min costs for each type). This pattern can appear in other problems where one has to decide how many times to apply an operation globally and measure individual effects. Always consider if the problem’s structure lets you break it down by each item (here, each type) and accumulate results, which is what made this solution manageable.