How do you solve Minimum Cost to Set Cooking Time on LeetCode?

Minimum Cost to Set Cooking Time (LeetCode #2162) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The range of minutes is small (0-99), allowing for enumeration.

What is the brute force solution for Minimum Cost to Set Cooking Time?

The brute force approach for Minimum Cost to Set Cooking Time is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible combinations of minutes and seconds that can be represented with four digits. For each combination, we calculate the cost of moving to and pushing the digits to set the cooking time. This approach is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Cost to Set Cooking Time?

Minimum Cost to Set Cooking Time uses the following concepts: Math, Enumeration. The recommended patterns to study are: Enumeration, Dynamic Programming.

What are the interview tips for Minimum Cost to Set Cooking Time?

Always clarify the problem constraints before jumping into coding. Think about edge cases, such as the minimum and maximum values for inputs. Explain your thought process clearly while coding.

What are common mistakes when solving Minimum Cost to Set Cooking Time?

Not normalizing the seconds correctly when calculating costs. Forgetting to check valid ranges for minutes and seconds.

#2162

Minimum Cost to Set Cooking Time

Medium

MathEnumerationEnumerationDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Instead of brute-forcing through all combinations, we can directly calculate the cost for each valid minute and second combination that sums to targetSeconds. This reduces unnecessary calculations and speeds up the process.

⚙️

Algorithm

5 steps

1Step 1: Calculate target minutes and seconds from targetSeconds.
2Step 2: Loop through all valid minutes from 0 to 99.
3Step 3: For each minute, calculate the corresponding seconds needed to reach targetSeconds.
4Step 4: Use the Cost function to compute the cost for valid combinations.
5Step 5: Track the minimum cost.

solution.py18 lines

1def minCost(startAt, moveCost, pushCost, targetSeconds):
2    minCost = float('inf')
3    for mm in range(100):
4        ss = targetSeconds - mm * 60
5        if 0 <= ss < 60:
6            cost = Cost(mm, ss, startAt, moveCost, pushCost)
7            minCost = min(minCost, cost)
8    return minCost
9
10def Cost(mm, ss, startAt, moveCost, pushCost):
11    digits = f'{mm:02}{ss:02}'
12    cost = 0
13    for digit in digits:
14        if int(digit) != startAt:
15            cost += moveCost
16        cost += pushCost
17        startAt = int(digit)
18    return cost

ℹ

Complexity note: The time complexity is O(n) because we iterate through a maximum of 100 minutes, and for each valid minute, we perform a constant amount of work. The space complexity is O(1) since we only use a fixed amount of space.

1The range of minutes is small (0-99), allowing for enumeration.
2Cost calculation is dependent on the sequence of digits pushed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.