Minimum Cost to Set Cooking Time

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Greedy Algorithms

A generic microwave supports cooking times for:

at least 1 second.
at most 99 minutes and 99 seconds.

To set the cooking time, you push at most four digits. The microwave normalizes what you push as four digits by prepending zeroes. It interprets the first two digits as the minutes and the last two digits as the seconds. It then adds them up as the cooking time. For example,

You push 9 5 4 (three digits). It is normalized as 0954 and interpreted as 9 minutes and 54 seconds.
You push 0 0 0 8 (four digits). It is interpreted as 0 minutes and 8 seconds.
You push 8 0 9 0. It is interpreted as 80 minutes and 90 seconds.
You push 8 1 3 0. It is interpreted as 81 minutes and 30 seconds.

You are given integers startAt, moveCost, pushCost, and targetSeconds. Initially, your finger is on the digit startAt. Moving the finger above any specific digit costs moveCost units of fatigue. Pushing the digit below the finger once costs pushCost units of fatigue.

There can be multiple ways to set the microwave to cook for targetSeconds seconds but you are interested in the way with the minimum cost.

Return the minimum cost to set targetSeconds seconds of cooking time.

Remember that one minute consists of 60 seconds.

Example 1:

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.

Example 2:

Input: startAt = 0, moveCost = 1, pushCost = 2, targetSeconds = 76
Output: 6
Explanation: The optimal way is to push two digits: 7 6, interpreted as 76 seconds.
The finger moves to 7 (with cost 1), pushes 7 (with cost 2), moves to 6 (with cost 1), and pushes 6 (with cost 2). The total cost is: 1 + 2 + 1 + 2 = 6
Note other possible ways are 0076, 076, 0116, and 116, but none of them produces the minimum cost.

Constraints:

0 <= startAt <= 9
1 <= moveCost, pushCost <= 10⁵
1 <= targetSeconds <= 6039

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the input parameters `startAt`, `moveCost`, `pushCost`, and `targetSeconds`? Are they non-negative integers?
What is the maximum value of `targetSeconds`? Is there a limit on the number of digits?
If `targetSeconds` is zero, what should the function return?
If there are multiple ways to achieve the target time with the same minimum cost, is any valid solution acceptable?
Can the digits of the keypad (0-9) be assumed to be in a standard numerical order or is there a defined order? Is the '0' button always in the same row/column as assumed?

Brute Force Solution

Approach

The brute force strategy explores every possible combination of digits to see which one results in the minimum cost. We essentially try out all valid times to cook the food and then choose the cheapest option. It is an exhaustive search of all possibilities.

Here's how the algorithm would work step-by-step:

First, list every single combination of digits you could possibly press to set the cooking time.
Then, remove any combination that isn't a valid time. For example, if the cooking time requested is 50 seconds, 99 seconds isn't valid, nor is a number greater than 99 minutes and 99 seconds.
For each of the remaining valid combinations, figure out exactly which buttons need to be pressed on the cooking device.
Calculate the cost of pressing those buttons. Remember to add the cost to press the 'start' button at the end.
Keep track of the minimum cost you've found so far and which button combination gave you that cost.
After you've checked all the valid combinations, the button combination that resulted in the minimum cost is the answer.

Code Implementation

def min_cost_to_set_time(start_at, move_cost, push_cost, target_seconds):
    minimum_cost = float('inf')

    for minutes in range(100):
        for seconds in range(100):
            total_seconds = minutes * 60 + seconds
            if total_seconds != target_seconds:
                continue

            # Build all possible button sequences
            time_string = str(minutes).zfill(2) + str(seconds).zfill(2)
            button_presses = []
            leading_zero = True

            for digit_char in time_string:
                digit = int(digit_char)

                if leading_zero and digit == 0:
                    continue
                else:
                    leading_zero = False
                    button_presses.append(digit)

            if not button_presses:
                button_presses = [0] # handle zero case

            current_cost = 0
            current_position = start_at

            for button in button_presses:
                if button != current_position:

                    current_cost += move_cost

                current_cost += push_cost

                current_position = button

            minimum_cost = min(minimum_cost, current_cost)

    return minimum_cost

Big(O) Analysis

Time Complexity

O(1) – The algorithm iterates through all possible digit combinations for minutes and seconds, which is limited to 99 minutes and 99 seconds. This means the number of iterations is constant (100 * 100 = 10000) regardless of the target cooking time. Therefore, the time complexity is O(1) because the number of operations does not scale with any input variable. Effectively, we have a fixed number of combinations to explore.

Space Complexity

O(1) – The brute force strategy explores combinations and keeps track of the minimum cost found so far. It primarily uses variables to store the current time combination being evaluated, the minimum cost, and the corresponding button sequence. The space needed for these variables remains constant regardless of the requested cooking time. Therefore, the auxiliary space used is constant, resulting in O(1) space complexity.

Optimal Solution

Approach

The most efficient way to find the minimum cost is to carefully consider two possibilities: directly entering the time and converting the time into digit presses. We avoid exploring every single combination and instead focus on these two dominant strategies to determine the cheapest way.

Here's how the algorithm would work step-by-step:

First, analyze the target cooking time to determine how many digits it has (e.g., 600 seconds has 3 digits).
Consider directly entering the cooking time, which is simply the number of digits times the push cost. This is our initial minimum cost.
Next, intelligently convert the cooking time. Start by identifying valid two-digit numbers (between 0 and 99), one-digit numbers (0 to 9) or three or four digit numbers based on what's present in the target cooking time. Also check the extreme case of 600 seconds being equivalent to cooking for 10 minutes (600/60) to determine the optimal way to represent your time
For each valid number, calculate the cost of entering each digit of the number and any leading zeros needed.
Compare the cost of this conversion with our current minimum cost and update it if the conversion is cheaper.
Return the overall minimum cost of the push button presses between direct entry or the optimal conversion approach.

Code Implementation

def minimum_cost_to_set_cooking_time(start_at, move_cost, push_cost, target_seconds):

    def calculate_cost(digits):
        cost = 0
        current_position = start_at
        for digit in digits:
            digit = int(digit)
            if digit != current_position:
                cost += move_cost
            cost += push_cost
            current_position = digit
        return cost

    # Direct entry: cost is based on the number of digits.
    direct_entry_cost = len(str(target_seconds)) * push_cost
    minimum_cost = direct_entry_cost

    # Explore alternative time representations.
    minutes = target_seconds // 60
    seconds = target_seconds % 60

    # Limit the valid time to avoid scenarios > 99.
    if minutes > 99:
        return minimum_cost

    time_string = str(minutes * 100 + seconds)

    # Only consider times with a valid number of digits.
    if len(time_string) <= 4:
        while len(time_string) < 4 and time_string[0] == '0' and len(time_string) > 1:
            time_string = time_string[1:]
        cost = calculate_cost(time_string)
        minimum_cost = min(minimum_cost, cost)

    # Check cost from seconds directly to minutes.
    if target_seconds >= 60:
        minutes = target_seconds // 60
        seconds = target_seconds % 60

        time_string = str(minutes).zfill(2) + str(seconds).zfill(2)

        if int(str(minutes)) <= 99:
            cost = calculate_cost(time_string)
            minimum_cost = min(minimum_cost, cost)

    return minimum_cost

Big(O) Analysis

Time Complexity

O(1) – The algorithm's runtime is dominated by a fixed number of arithmetic calculations and comparisons. It analyzes the cooking time to extract digits and convert it, but the number of digits is bounded by the maximum possible cooking time (e.g., 9999 seconds). The operations to check for direct entry and optimal conversions are all performed on a small, fixed number of possibilities. Therefore, the time complexity does not depend on the size of the input and remains constant, or O(1).

Space Complexity

O(1) – The algorithm primarily uses a fixed number of integer variables to store the target cooking time, minimum cost, and intermediate calculations for digit conversions. No auxiliary data structures like arrays, lists, or hash maps are created that scale with the input target cooking time (N). Therefore, the auxiliary space used remains constant regardless of the input size, resulting in a space complexity of O(1).

Edge Cases

Case	How to Handle
Target time is zero	Return the cost of pressing the button representing '0' directly, if allowed, or the cost to wait (if waiting is cheaper and allowed).
Button press costs are all zero	Find the shortest sequence regardless of the number of button presses, likely just pressing the digits of the target time.
Only one button is available	If it's '0', return cost * target time; otherwise, see if target time is divisible by button value, returning the corresponding cost or infinity.
Target time is extremely large	The solution should avoid integer overflow by checking against a reasonable maximum value for time, returning infinity or appropriate error if exceeded.
Target time requires more than 4 digits.	Return infinity or a similar sentinel value indicating that the time is unreachable, as it goes beyond the standard clock format.
Waiting is not allowed and some digits of the target time are unavailable as buttons	Return infinity indicating that the target time is unreachable because you cannot wait or form the required digits.
The '00' case, when the target time has leading zeros.	Calculate the cost of either waiting, using the direct digits or by using the special double zero feature if available; return the minimum cost
Integer overflow when calculating cost for a very long sequence of button presses	Use a data type large enough to hold the maximum possible cost (e.g., long long in C++, long in Java) to avoid overflow.

Target time is zero

How to Handle:

Return the cost of pressing the button representing '0' directly, if allowed, or the cost to wait (if waiting is cheaper and allowed).

Button press costs are all zero

How to Handle:

Find the shortest sequence regardless of the number of button presses, likely just pressing the digits of the target time.

Only one button is available

How to Handle:

If it's '0', return cost * target time; otherwise, see if target time is divisible by button value, returning the corresponding cost or infinity.

Target time is extremely large

How to Handle:

The solution should avoid integer overflow by checking against a reasonable maximum value for time, returning infinity or appropriate error if exceeded.

Target time requires more than 4 digits.

How to Handle:

Return infinity or a similar sentinel value indicating that the time is unreachable, as it goes beyond the standard clock format.

Waiting is not allowed and some digits of the target time are unavailable as buttons

How to Handle:

Return infinity indicating that the target time is unreachable because you cannot wait or form the required digits.

The '00' case, when the target time has leading zeros.

How to Handle:

Calculate the cost of either waiting, using the direct digits or by using the special double zero feature if available; return the minimum cost

Integer overflow when calculating cost for a very long sequence of button presses

How to Handle:

Use a data type large enough to hold the maximum possible cost (e.g., long long in C++, long in Java) to avoid overflow.