Minimum Amount of Time to Fill Cups
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Topics:
Greedy AlgorithmsArrays
You have a water dispenser that can dispense cold, warm, and hot water. Every second, you can either fill up 2 cups with different types of water, or 1 cup of any type of water.
You are given a 0-indexed integer array amount of length 3 where amount[0], amount[1], and amount[2] denote the number of cold, warm, and hot water cups you need to fill respectively. Return the minimum number of seconds needed to fill up all the cups.
Example 1:
Input: amount = [1,4,2] Output: 4 Explanation: One way to fill up the cups is: Second 1: Fill up a cold cup and a warm cup. Second 2: Fill up a warm cup and a hot cup. Second 3: Fill up a warm cup and a hot cup. Second 4: Fill up a warm cup. It can be proven that 4 is the minimum number of seconds needed.
Example 2:
Input: amount = [5,4,4] Output: 7 Explanation: One way to fill up the cups is: Second 1: Fill up a cold cup, and a hot cup. Second 2: Fill up a cold cup, and a warm cup. Second 3: Fill up a cold cup, and a warm cup. Second 4: Fill up a warm cup, and a hot cup. Second 5: Fill up a cold cup, and a hot cup. Second 6: Fill up a cold cup, and a warm cup. Second 7: Fill up a hot cup.
Example 3:
Input: amount = [5,0,0] Output: 5 Explanation: Every second, we fill up a cold cup.
Constraints:
amount.length == 30 <= amount[i] <= 100
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:
- Are the values in the input array `amount` non-negative integers?
- What are the maximum possible values for the numbers in the `amount` array?
- If the input array `amount` contains all zeros, what should the return value be?
- Is the length of the `amount` array always guaranteed to be exactly 3?
- Are we optimizing for space, or is it purely for time?
Brute Force Solution
Approach
The brute force method to minimize the time to fill cups involves simulating every possible action we can take at each step. We explore all combinations of filling the cups until all are empty, and then pick the quickest way.
Here's how the algorithm would work step-by-step:
- Start with the initial number of each type of cup to fill.
- Consider all possible actions: filling two different types of cups at once, or filling one cup of one type.
- For each action, update the number of cups remaining for each type.
- Repeat the process of considering all possible actions with the updated cup counts.
- Keep doing this until all cups of all types are filled (number of cups for each type is zero).
- Record the total time taken for each possible sequence of actions.
- Compare all the recorded times and choose the smallest one. That's the minimum time needed.
Code Implementation
def minimum_time_to_fill_cups_brute_force(amount):
min_time = float('inf')
def solve(current_amount, current_time):
nonlocal min_time
if all(cup_count == 0 for cup_count in current_amount):
min_time = min(min_time, current_time)
return
# Try filling two cups at once
for first_cup_type in range(3):
if current_amount[first_cup_type] > 0:
for second_cup_type in range(first_cup_type, 3):
if current_amount[second_cup_type] > 0:
new_amount = list(current_amount)
new_amount[first_cup_type] -= 1
# Ensure that the same cup type is not reduced twice.
if first_cup_type == second_cup_type:
solve(new_amount, current_time + 1)
else:
new_amount[second_cup_type] -= 1
solve(new_amount, current_time + 1)
# Try filling only one cup
for cup_type in range(3):
if current_amount[cup_type] > 0:
new_amount = list(current_amount)
new_amount[cup_type] -= 1
# If one cup is filled, we need to explore this path.
solve(new_amount, current_time + 1)
solve(amount, 0)
# Return the minimum time found across all paths.
return min_timeBig(O) Analysis
Time Complexity
O(3^n) – The brute force approach explores all possible combinations of filling the cups until all are empty. At each step, we can choose to fill two different types of cups or fill one cup. With three cup types, at each step, we have three choices: filling cup type 1 and 2, filling cup type 1 and 3, or filling cup type 2 and 3, or filling only cup type 1, or filling only cup type 2, or filling only cup type 3. This creates a decision tree where each node branches out into multiple possibilities. Since the maximum number of cups is represented as n (sum of all cup types), and each action may reduce the amount of cups, the height of this decision tree is bounded by n. Therefore, the total number of possible action sequences we explore grows exponentially with n and is approximately 3 raised to the power of n, leading to a time complexity of O(3^n).
Space Complexity
O(3^max(cups[0], cups[1], cups[2])) – The brute force approach explores all possible combinations of filling cups. Each decision (filling two different types or filling one) branches the search. The depth of this recursion tree is bounded by the maximum number of cups of any single type, max(cups[0], cups[1], cups[2]). Since each node in the recursion tree can potentially branch into three possibilities, the maximum number of nodes in the call stack could grow exponentially with a base of 3 and an exponent based on the max of the cups, leading to O(3^max(cups[0], cups[1], cups[2])) space due to the recursion stack.
Optimal Solution
Approach
To minimize the time needed to fill the cups, we focus on pouring from the two fullest cup types in each step. This ensures that we are always making progress on the biggest amounts. By repeating this until all cups are empty, we efficiently determine the minimum time.
Here's how the algorithm would work step-by-step:
- Find the counts for the three types of cups and identify the two with the highest counts.
- Pour one cup from each of the two fullest types. This takes one unit of time.
- Reduce the counts of the two types you poured from.
- Repeat the process of finding the two highest counts, pouring, and reducing the counts until all cup counts are zero.
- The total number of pour actions is the minimum time required to fill all cups.
Code Implementation
def minimum_time_to_fill_cups(cups):
total_seconds = 0
cups.sort()
while cups[1] > 0:
# Handle the case where the two largest are zero
if cups[2] == 0:
total_seconds += cups[1]
break
# Pour from the two fullest cups.
cups[1] -= 1
cups[2] -= 1
total_seconds += 1
cups.sort()
# Handle the remaining cups if any
total_seconds += cups[2]
return total_secondsBig(O) Analysis
Time Complexity
O(max(cups)) – The algorithm's runtime is determined by how many times we need to pour cups until all counts are zero. In each iteration, we identify the two largest cup counts, decrement them, and increment the time. The number of iterations is directly related to the initial maximum value among the 'cups' array because the two largest cup counts are reduced by 1 in each iteration. Therefore, the time complexity is O(max(cups)).
Space Complexity
O(1) – The provided solution focuses on repeatedly finding the two largest values and decrementing them until all values are zero. It primarily involves in-place modifications and comparisons of a fixed number of cup counts. The algorithm does not create any auxiliary data structures that scale with the input size. Therefore, the auxiliary space used is constant, regardless of the initial number of cups.
Edge Cases
| Case | How to Handle |
|---|---|
| Cups array is null or empty | Return 0 immediately as no cups need filling. |
| Cups array contains only one cup amount (e.g., [5]) | Return the single cup amount as that is the minimum time. |
| Cups array contains extremely large values causing integer overflow | Use long integer types if language supports or handle with appropriate data structures like BigInteger, throw an exception if overflow is not avoidable. |
| Two cups have very large amounts, one has a very small amount (e.g., [100000, 100000, 1]) | The greedy strategy of picking the two largest amounts will still produce the optimal solution. |
| Cups array contains zeros (e.g., [5, 0, 2]) | The zeros have no impact on time, and the algorithm will correctly handle the other values. |
| Cups array has all values equal (e.g., [5, 5, 5]) | The algorithm should still arrive at the correct solution (in this case, 5). |
| Input cups array is very large (performance considerations). | The constant time operations (max heap operations, sorting small array) should scale well, but the algorithm should be reviewed for potential performance bottlenecks if the cup array is astronomically large. |
| Cups array contains maximum allowed integer values. | Ensure that intermediate calculations do not cause overflow, potentially by using larger data types, and ensure comparison operators handle maximum values correctly. |
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