Building H2O
Medium
There are two kinds of threads: oxygen and hydrogen. Your goal is to group these threads to form water molecules.
There is a barrier where each thread has to wait until a complete molecule can be formed. Hydrogen and oxygen threads will be given releaseHydrogen and releaseOxygen methods respectively, which will allow them to pass the barrier. These threads should pass the barrier in groups of three, and they must immediately bond with each other to form a water molecule. You must guarantee that all the threads from one molecule bond before any other threads from the next molecule do.
In other words:
- If an oxygen thread arrives at the barrier when no hydrogen threads are present, it must wait for two hydrogen threads.
- If a hydrogen thread arrives at the barrier when no other threads are present, it must wait for an oxygen thread and another hydrogen thread.
We do not have to worry about matching the threads up explicitly; the threads do not necessarily know which other threads they are paired up with. The key is that threads pass the barriers in complete sets; thus, if we examine the sequence of threads that bind and divide them into groups of three, each group should contain one oxygen and two hydrogen threads.
Write synchronization code for oxygen and hydrogen molecules that enforces these constraints.
Example 1:
Input: water = "HOH" Output: "HHO" Explanation: "HOH" and "OHH" are also valid answers.
Example 2:
Input: water = "OOHHHH" Output: "HHOHHO" Explanation: "HOHHHO", "OHHHHO", "HHOHOH", "HOHHOH", "OHHHOH", "HHOOHH", "HOHOHH" and "OHHOHH" are also valid answers.
Constraints:
3 * n == water.length1 <= n <= 20water[i]is either'H'or'O'.- There will be exactly
2 * n'H'inwater. - There will be exactly
n'O'inwater.
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 possible ranges for the number of hydrogen and oxygen atoms (hydrogen and oxygen)? Can they be zero or negative?
- If either hydrogen or oxygen is zero, what should the function return?
- Is the input guaranteed to be integers, or do I need to handle potential non-integer inputs?
- Are there any memory constraints I should be aware of, considering very large numbers of atoms?
- Should I return an integer representing the maximum number of H2O molecules, or is there a specific output format required?
Brute Force Solution
Approach
The brute force approach for building H2O molecules from Hydrogen (H) and Oxygen (O) atoms involves trying every possible combination. We will explore all ways to pair H and O until we exhaust them to see if we can create valid H2O molecules. This method is inefficient but guarantees we will find the solution if one exists.
Here's how the algorithm would work step-by-step:
- Check if there are at least two Hydrogen atoms and one Oxygen atom available.
- If not, you can't build any H2O molecules, so stop.
- If you have enough atoms, grab two Hydrogen atoms and one Oxygen atom.
- Combine these atoms to form one H2O molecule.
- Reduce the number of available Hydrogen atoms by two, and the number of available Oxygen atoms by one.
- Repeat the process from the beginning until you no longer have enough Hydrogen and Oxygen atoms to form another H2O molecule.
- The number of H2O molecules you were able to create represents the maximum possible using the brute force method.
Code Implementation
def build_h2o_brute_force(hydrogen_count, oxygen_count):
h2o_molecule_count = 0
# Continue building molecules as long as we have enough atoms
while True:
# Check if enough atoms exist to form an H2O molecule.
if hydrogen_count < 2 or oxygen_count < 1:
break
# Decrement counts and increment the molecule count.
hydrogen_count -= 2
oxygen_count -= 1
h2o_molecule_count += 1
return h2o_molecule_count
Big(O) Analysis
Optimal Solution
Approach
The goal is to create water molecules by efficiently combining hydrogen and oxygen atoms. The best way to do this is to keep track of the available atoms and build as many water molecules as possible at each step, ensuring no atoms are wasted.
Here's how the algorithm would work step-by-step:
- Keep track of how many hydrogen atoms and oxygen atoms are available.
- Check if there are enough atoms to make a water molecule (two hydrogens and one oxygen).
- If there are enough atoms, create a water molecule and reduce the counts of available hydrogen and oxygen atoms by the appropriate amounts.
- Repeat this process until you run out of either hydrogen or oxygen atoms. This makes sure you build as many complete water molecules as you can.
- Any leftover hydrogen or oxygen atoms cannot be used to make water molecules, so the number of water molecules you made is the answer.
Code Implementation
def building_h2o(hydrogen_count, oxygen_count):
water_molecule_count = 0
# Keep looping while we can still form water.
while hydrogen_count >= 2 and oxygen_count >= 1:
# Found enough atoms, make water.
hydrogen_count -= 2
oxygen_count -= 1
water_molecule_count += 1
return water_molecule_countBig(O) Analysis
Edge Cases
| Case | How to Handle |
|---|---|
| hydrogen or oxygen is null | Throw an IllegalArgumentException or return 0, documenting the expected behavior. |
| hydrogen or oxygen is negative | Throw an IllegalArgumentException as the number of atoms cannot be negative. |
| hydrogen and oxygen are zero | Return 0 as no water molecules can be formed. |
| hydrogen is zero and oxygen is greater than zero | Return 0 since no water molecules can be formed without hydrogen. |
| oxygen is zero and hydrogen is greater than zero | Return 0 since no water molecules can be formed without oxygen. |
| Very large hydrogen and oxygen (potential integer overflow) | Use long data type to avoid potential integer overflow during calculation. |
| hydrogen is 1 and oxygen is greater than zero | Return 0 since 2 hydrogen atoms are needed per water molecule. |
| oxygen is 1 and hydrogen is less than 2 | Return 0 since 2 hydrogen atoms and 1 oxygen atom are needed per water molecule. |
