Leetcode 2735: Collecting Chocolates
You are given an array
nums where each element represents the cost of collecting a chocolate of a specific type. You can perform an operation to change the types of chocolates, and each operation incurs a cost. Your task is to determine the minimum total cost to collect all types of chocolates.Problem
Approach
Steps
Complexity
Input: The input consists of an integer array `nums` and an integer `x`, where `nums[i]` represents the cost of collecting chocolate type `i` and `x` is the cost of performing one operation.
Example: nums = [10, 5, 20], x = 3
Constraints:
• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 10^9
• 1 <= x <= 10^9
Output: Return the minimum total cost to collect chocolates of all types, given that you can perform the operation any number of times.
Example: Output: 11
Constraints:
Goal: To minimize the total cost by strategically performing the operation while collecting chocolates.
Steps:
• Loop over all types of chocolates and calculate the minimum cost for collecting them with the operations applied.
• For each type, calculate the cost of the operation and the cost of collecting the chocolate.
• Return the minimum total cost obtained.
Goal: The array `nums` will always contain integers representing the cost of collecting chocolates, and the length of the array will be between 1 and 1000. The operation cost `x` is within the given bounds.
Steps:
• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 10^9
• 1 <= x <= 10^9
Assumptions:
• You can perform the operation any number of times, but the total cost of performing operations should be minimized.
• The array `nums` represents the cost of chocolates of distinct types.
• Input: nums = [10, 5, 20], x = 3
• Explanation: The initial cost is 35 (10 + 5 + 20). Performing the operation results in the following costs for collecting chocolates, ultimately yielding the minimum cost of 11.
• Input: nums = [5, 1, 5], x = 2
• Explanation: Here, the best strategy is to perform one operation to reduce the total cost, resulting in a minimum total of 6.
Approach: The problem can be solved by calculating the minimum cost by performing operations optimally and collecting chocolates based on the incurred costs.
Observations:
• The operation can be repeated, so we need to find the optimal number of times to perform it.
• It is beneficial to minimize the total cost of collecting chocolates.
• We need to consider all possible shifts of the chocolate types and calculate the total cost for each shift.
Steps:
• For each chocolate type, calculate the total cost for every possible shift.
• Add the cost of performing the operation and collecting chocolates.
• Return the minimum total cost from all possible shifts.
Empty Inputs:
• The array `nums` contains only one type of chocolate.
Large Inputs:
• The array `nums` has the maximum allowed length, and `x` is very large.
Special Values:
• The array `nums` contains chocolates with the same cost.
Constraints:
• The array `nums` is always non-empty.
long long minCost(vector<int>& A, int x) {
int n = A.size();
vector<long long> res(n);
for (int i = 0; i < n; i++) {
res[i] += 1L * i * x;
int cur = A[i];
for (int k = 0; k < n; k++) {
cur = min(cur, A[(i - k + n) % n]);
res[k] += cur;
}
}
return *std::min_element(res.begin(), res.end());
}
1 : Function Definition
long long minCost(vector<int>& A, int x) {
The function `minCost` is defined, which takes a vector `A` and an integer `x` as input and returns a `long long` result.
2 : Variable Initialization
int n = A.size();
The size of the input vector `A` is stored in the variable `n`.
3 : Variable Initialization
vector<long long> res(n);
A vector `res` of size `n` is initialized to store the accumulated results.
4 : Loop - Outer
for (int i = 0; i < n; i++) {
An outer loop iterates over each index `i` in the range [0, n-1].
5 : Cost Calculation
res[i] += 1L * i * x;
For each index `i`, the value `1L * i * x` is added to the corresponding element in the `res` vector.
6 : Variable Initialization
int cur = A[i];
A variable `cur` is initialized to the current element `A[i]`.
7 : Loop - Inner
for (int k = 0; k < n; k++) {
An inner loop iterates over each index `k` in the range [0, n-1].
8 : Min Calculation
cur = min(cur, A[(i - k + n) % n]);
In each iteration of the inner loop, the variable `cur` is updated to the minimum of its current value and the element `A[(i - k + n) % n]` (ensuring circular indexing).
9 : Cost Update
res[k] += cur;
The updated value of `cur` is added to `res[k]` in the `res` vector.
10 : Return Statement
return *std::min_element(res.begin(), res.end());
The function returns the smallest value from the `res` vector, which is found using `std::min_element`.
Best Case: O(n^2) where n is the length of the array.
Average Case: O(n^2) due to iterating over all possible shifts.
Worst Case: O(n^2) since we need to check every shift and calculate the associated cost.
Description: The time complexity is quadratic because for each chocolate type, we are calculating the cost for all possible shifts.
Best Case: O(n) if minimal space is used for calculations.
Worst Case: O(n) due to the additional storage for the result array.
Description: The space complexity is linear since we use a result array to store the computed costs for each shift.
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