LeetCode 2735 - Collecting Chocolates

The problem gives us an array nums, where nums[i] represents the cost of buying a chocolate currently located at index i. Initially, the chocolate at index i is also considered to be of type i. Since there are n positions, there are exactly n chocolate types.

leetcodemediumarrayenumeration

LeetCode Problem 2735

Difficulty: 🟡 Medium
Topics: Array, Enumeration

Solution

LeetCode 2735 - Collecting Chocolates

Problem Understanding

The problem gives us an array nums, where nums[i] represents the cost of buying a chocolate currently located at index i. Initially, the chocolate at index i is also considered to be of type i. Since there are n positions, there are exactly n chocolate types.

The goal is to collect one chocolate of every type while minimizing the total cost.

We are allowed to perform a special operation any number of times. Each operation costs x, and after performing it, every chocolate changes its type simultaneously:

  • The chocolate of type i becomes type (i + 1) mod n

This means the chocolate types rotate cyclically across the array.

An important observation is that the physical chocolates and their prices do not move, only their assigned types rotate. After several rotations, a type may become associated with a cheaper chocolate position than it originally had.

For every chocolate type, we want to eventually buy it at the minimum possible cost across all rotations we choose to perform. However, performing rotations is not free, because each rotation costs x.

The challenge is therefore a tradeoff:

  • More rotations may allow us to buy types at cheaper prices
  • But each rotation adds an extra fixed cost

We must compute the minimum total cost over all possible numbers of rotations.

The constraints are:

  • 1 <= n <= 1000
  • 1 <= nums[i], x <= 10^9

Since n is only 1000, an O(n^2) solution is completely feasible. However, approaches worse than quadratic may become too slow.

Several edge cases are important:

  • If rotations are extremely expensive, the optimal answer may involve zero operations.
  • If one chocolate is extremely cheap, rotating multiple times may allow every type to eventually use that cheap chocolate.
  • Arrays of size 1 should immediately return the only chocolate cost.
  • Since values can reach 10^9, integer overflow must be considered in languages like Go.

Approaches

Brute Force Approach

A naive solution would explicitly simulate every possible sequence of rotations and purchases.

For each possible number of rotations k from 0 to n - 1, we could determine the cost of purchasing every type after exactly k rotations. Then we would compute:

  • Rotation cost: k * x
  • Purchase cost for all types after those rotations

However, this interpretation is incomplete because we are allowed to buy chocolates at different moments during the sequence of rotations. A type might be bought after 1 rotation while another is bought after 3 rotations.

A true brute force approach would attempt all possible buying schedules and rotation counts. Since each type can potentially be bought after different numbers of rotations, the state space grows exponentially.

Although such an approach would eventually find the correct answer, it is far too slow.

Key Insight

The important observation is that after performing up to k rotations, each chocolate type can be purchased using the minimum cost encountered among all positions that have rotated into that type.

Suppose we focus on one target type i.

After:

  • 0 rotations, it can use nums[i]
  • 1 rotation, it can additionally use nums[(i - 1 + n) % n]
  • 2 rotations, it can additionally use nums[(i - 2 + n) % n]

and so on.

Therefore, after k rotations, the cheapest possible cost for type i is simply the minimum among those reachable positions.

This means we do not need to track purchase timing explicitly. We only need to know:

  • How many total rotations we perform
  • The minimum reachable cost for each type within those rotations

For every possible rotation count k, we compute:

$$k \cdot x + \sum_{i=0}^{n-1} \text{bestCost}[i]$$

where bestCost[i] is the minimum chocolate price that can become type i within k rotations.

Since there are only n possible rotation counts and each evaluation takes O(n), the total complexity becomes O(n^2).

Approach Comparison

Approach Time Complexity Space Complexity Notes
Brute Force Exponential Exponential Tries all purchase schedules and rotation combinations
Optimal O(n²) O(n) Tracks minimum reachable cost for each type

Algorithm Walkthrough

Step 1: Initialize the Best Costs

Create an array called minimum_costs.

Initially, without performing any rotations, the cheapest way to obtain type i is simply nums[i].

So we begin with:

minimum_costs = nums[:]

We also initialize the answer as the sum of all original costs, corresponding to zero rotations.

Step 2: Try Every Possible Rotation Count

We iterate through all possible rotation counts from 0 to n - 1.

Why only up to n - 1?

Because after n rotations, the configuration repeats cyclically, so no new possibilities appear.

Step 3: Update Reachable Minimum Costs

Suppose we are considering rotation count k.

For every type i, another chocolate becomes reachable:

nums[(i - k) % n]

We update:

minimum_costs[i] = min(minimum_costs[i], nums[(i - k) % n])

This keeps track of the cheapest chocolate that can become type i after at most k rotations.

Step 4: Compute the Total Cost

The total cost for using exactly k rotations is:

sum(minimum_costs) + k * x

The first part is the purchase cost of all types.

The second part is the cumulative rotation cost.

We compare this value against the current best answer.

Step 5: Return the Minimum Answer

After checking all possible rotation counts, the smallest computed total is the optimal answer.

Why It Works

For a fixed number of rotations k, every type i can only be matched with chocolates that rotate into it within those k operations. The algorithm maintains exactly the minimum price among all such reachable chocolates.

Since the algorithm evaluates every possible number of rotations and computes the optimal purchasing cost for that scenario, the global minimum over all rotation counts must be the optimal answer.

Python Solution

from typing import List

class Solution:
    def minCost(self, nums: List[int], x: int) -> int:
        n = len(nums)

        minimum_costs = nums[:]
        answer = sum(nums)

        for rotations in range(1, n):
            for i in range(n):
                minimum_costs[i] = min(
                    minimum_costs[i],
                    nums[(i - rotations) % n]
                )

            total_cost = sum(minimum_costs) + rotations * x
            answer = min(answer, total_cost)

        return answer

The implementation begins by copying nums into minimum_costs. This array continuously stores the cheapest reachable chocolate cost for each type.

The outer loop iterates over all possible rotation counts. For each rotation count, we update every type's minimum reachable cost using the newly available chocolate position.

The modulo operation handles the circular rotation correctly.

After updating all minimum costs, we compute the total cost by combining:

  • The current sum of minimum purchase prices
  • The accumulated rotation expense

The answer is updated whenever a smaller total is found.

Finally, the minimum possible total cost is returned.

Go Solution

package main

func minCost(nums []int, x int) int64 {
	n := len(nums)

	minimumCosts := make([]int, n)
	copy(minimumCosts, nums)

	var answer int64 = 0
	for _, value := range nums {
		answer += int64(value)
	}

	for rotations := 1; rotations < n; rotations++ {
		for i := 0; i < n; i++ {
			candidate := nums[(i-rotations+n)%n]

			if candidate < minimumCosts[i] {
				minimumCosts[i] = candidate
			}
		}

		var currentSum int64 = 0
		for _, value := range minimumCosts {
			currentSum += int64(value)
		}

		totalCost := currentSum + int64(rotations*x)

		if totalCost < answer {
			answer = totalCost
		}
	}

	return answer
}

The Go implementation follows the same logic as the Python version.

The primary difference is integer handling. Since values may grow beyond 32 bit integer range, the final answer and accumulated sums use int64.

Slices are used instead of Python lists, and the built in copy function duplicates the initial array.

The modulo expression includes + n to avoid negative indices:

(i - rotations + n) % n

Worked Examples

Example 1

Input:

nums = [20, 1, 15]
x = 5

Initial state:

Type Minimum Cost
0 20
1 1
2 15

Initial total:

20 + 1 + 15 = 36

Current answer:

36

Rotation Count = 1

For each type:

Type Current Min New Candidate Updated Min
0 20 nums[2] = 15 15
1 1 nums[0] = 20 1
2 15 nums[1] = 1 1

Updated minimum costs:

[15, 1, 1]

Purchase cost:

15 + 1 + 1 = 17

Rotation cost:

1 * 5 = 5

Total:

17 + 5 = 22

Updated answer:

22

Rotation Count = 2

For each type:

Type Current Min New Candidate Updated Min
0 15 nums[1] = 1 1
1 1 nums[2] = 15 1
2 1 nums[0] = 20 1

Updated minimum costs:

[1, 1, 1]

Purchase cost:

1 + 1 + 1 = 3

Rotation cost:

2 * 5 = 10

Total:

13

Updated answer:

13

Final answer:

13

Example 2

Input:

nums = [1, 2, 3]
x = 4

Initial total:

1 + 2 + 3 = 6

Rotation Count = 1

Updated minimums:

[1, 1, 2]

Purchase cost:

4

Rotation cost:

4

Total:

8

This is worse than 6.

Rotation Count = 2

Updated minimums:

[1, 1, 1]

Purchase cost:

3

Rotation cost:

8

Total:

11

Still worse than 6.

Final answer:

6

Complexity Analysis

Measure Complexity Explanation
Time O(n²) We try all n rotation counts and update all n types each time
Space O(n) The minimum_costs array stores one value per type

The algorithm uses two nested loops, each running at most n times. Since n <= 1000, the quadratic complexity is efficient enough.

The additional memory usage is linear because we only maintain a single auxiliary array.

Test Cases

from typing import List

class Solution:
    def minCost(self, nums: List[int], x: int) -> int:
        n = len(nums)

        minimum_costs = nums[:]
        answer = sum(nums)

        for rotations in range(1, n):
            for i in range(n):
                minimum_costs[i] = min(
                    minimum_costs[i],
                    nums[(i - rotations) % n]
                )

            answer = min(
                answer,
                sum(minimum_costs) + rotations * x
            )

        return answer

solution = Solution()

assert solution.minCost([20, 1, 15], 5) == 13  # provided example 1
assert solution.minCost([1, 2, 3], 4) == 6  # provided example 2

assert solution.minCost([5], 10) == 5  # single element array
assert solution.minCost([1, 1, 1], 100) == 3  # all equal values
assert solution.minCost([100, 1, 100], 1) == 5  # rotations become very valuable
assert solution.minCost([10, 10, 1], 2) == 7  # cheap value spreads through rotations
assert solution.minCost([7, 6, 5, 4], 1000) == 22  # rotations too expensive
assert solution.minCost([1000000000, 1000000000], 1000000000) == 2000000000  # large values
assert solution.minCost([8, 3, 5, 2], 3) == 14  # mixed values with beneficial rotations

Test Summary

Test Why
[20,1,15], x=5 Validates the main rotating optimization
[1,2,3], x=4 Confirms zero rotations can be optimal
[5], x=10 Tests minimum array size
[1,1,1], x=100 Ensures unnecessary rotations are avoided
[100,1,100], x=1 Checks repeated use of a very cheap chocolate
[10,10,1], x=2 Validates cumulative minimum updates
[7,6,5,4], x=1000 Ensures high rotation cost is handled correctly
Large 10^9 values Verifies overflow safety
[8,3,5,2], x=3 Tests a general mixed scenario

Edge Cases

Single Element Array

If the array contains only one chocolate, there is only one type to collect. Rotations do nothing because the type maps back to itself immediately.

A buggy implementation might still attempt unnecessary loops or incorrectly apply rotation costs. This implementation handles the case naturally because the rotation loop never executes when n = 1.

Extremely Large Rotation Cost

When x is very large, performing rotations is usually worse than buying chocolates directly.

For example:

nums = [7,6,5,4]
x = 1000

Even though rotations could reduce some purchase costs, the operation fee dominates the total cost. The implementation correctly compares every rotation scenario against the zero rotation baseline.

Very Cheap Chocolate Dominates

Sometimes one chocolate is dramatically cheaper than all others:

nums = [100,1,100]
x = 1

Rotations allow this cheap chocolate to eventually represent every type. A naive greedy solution might fail to recognize that paying several operation costs is worthwhile.

This implementation correctly tracks the minimum reachable cost for every type across increasing rotations, allowing the cheap chocolate to propagate through all positions.