Leetcode #2162: Minimum Cost to Set Cooking Time
In this guide, we solve Leetcode #2162 Minimum Cost to Set Cooking Time 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
A generic microwave supports cooking times for: at least 1 second. at most 99 minutes and 99 seconds.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Math, Enumeration
Intuition
There is a mathematical invariant or formula that directly leads to the result.
Using math avoids unnecessary loops and reduces complexity.
Approach
Derive the formula or update rule, then compute the answer directly.
Handle edge cases like overflow or zero carefully.
Steps:
- Identify the math relationship.
- Compute the result with a loop or formula.
- Handle edge cases.
Example
Input: startAt = 1, moveCost = 2, pushCost = 1, targetSeconds = 600
Output: 6
Explanation: The following are the possible ways to set the cooking time.
- 1 0 0 0, interpreted as 10 minutes and 0 seconds.
The finger is already on digit 1, pushes 1 (with cost 1), moves to 0 (with cost 2), pushes 0 (with cost 1), pushes 0 (with cost 1), and pushes 0 (with cost 1).
The cost is: 1 + 2 + 1 + 1 + 1 = 6. This is the minimum cost.
- 0 9 6 0, interpreted as 9 minutes and 60 seconds. That is also 600 seconds.
The finger moves to 0 (with cost 2), pushes 0 (with cost 1), moves to 9 (with cost 2), pushes 9 (with cost 1), moves to 6 (with cost 2), pushes 6 (with cost 1), moves to 0 (with cost 2), and pushes 0 (with cost 1).
The cost is: 2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 = 12.
- 9 6 0, normalized as 0960 and interpreted as 9 minutes and 60 seconds.
The finger moves to 9 (with cost 2), pushes 9 (with cost 1), moves to 6 (with cost 2), pushes 6 (with cost 1), moves to 0 (with cost 2), and pushes 0 (with cost 1).
The cost is: 2 + 1 + 2 + 1 + 2 + 1 = 9.
Python Solution
class Solution:
def minCostSetTime(
self, startAt: int, moveCost: int, pushCost: int, targetSeconds: int
) -> int:
def f(m, s):
if not 0 <= m < 100 or not 0 <= s < 100:
return inf
arr = [m // 10, m % 10, s // 10, s % 10]
i = 0
while i < 4 and arr[i] == 0:
i += 1
t = 0
prev = startAt
for v in arr[i:]:
if v != prev:
t += moveCost
t += pushCost
prev = v
return t
m, s = divmod(targetSeconds, 60)
ans = min(f(m, s), f(m - 1, s + 60))
return ans
Complexity
The time complexity is O(n) or O(1). The space complexity is O(1).
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.