Leetcode #2735: Collecting Chocolates
In this guide, we solve Leetcode #2735 Collecting Chocolates in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a 0-indexed integer array nums of size n representing the cost of collecting different chocolates. The cost of collecting the chocolate at the index i is nums[i].
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Enumeration
Intuition
The constraints allow a direct scan that keeps only the essential state.
By translating the requirements into a clean loop, the logic stays easy to reason about.
Approach
Iterate through the data once, updating the state needed to compute the answer.
Return the final state after the traversal is complete.
Steps:
- Parse the input.
- Iterate and update state.
- Return the computed answer.
Example
Input: nums = [20,1,15], x = 5
Output: 13
Explanation: Initially, the chocolate types are [0,1,2]. We will buy the 1st type of chocolate at a cost of 1.
Now, we will perform the operation at a cost of 5, and the types of chocolates will become [1,2,0]. We will buy the 2nd type of chocolate at a cost of 1.
Now, we will again perform the operation at a cost of 5, and the chocolate types will become [2,0,1]. We will buy the 0th type of chocolate at a cost of 1.
Thus, the total cost will become (1 + 5 + 1 + 5 + 1) = 13. We can prove that this is optimal.
Python Solution
class Solution:
def minCost(self, nums: List[int], x: int) -> int:
n = len(nums)
f = [[0] * n for _ in range(n)]
for i, v in enumerate(nums):
f[i][0] = v
for j in range(1, n):
f[i][j] = min(f[i][j - 1], nums[(i - j) % n])
return min(sum(f[i][j] for i in range(n)) + x * j for j in range(n))
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.