Asked in: Amazon
Image of the Question
All Testcases Passed ✔
Solution
def findOptimalLevel(inventoryLevels, x, y):
# Write your code here
inventoryLevels.sort()
def solve(mid):
// ... rest of solution available after purchaseExplanation
```
This problem revolves around performing a set of defined operations to maximize the minimum value in an array after transformations. The allowed operation lets you increase one element by `x` while decreasing another by `y`, and you can repeat this operation as many times as you'd like. The objective is to find the **highest possible value that the smallest element in the array can be made**, under the given rules, while ensuring all final values remain positive.
To approach this, start by understanding the mechanics of the operation. You can pick two distinct indices `i` and `j`. When you do, you increase the value at index `i` by `x`, and decrease the value at index `j` by `y`. This means that the operation is redistributive — you’re moving some “inventory” from one location to another, but with a conversion rate: for every `y` you remove from one element, you only gain `x` in another. Since `x <= y`, this redistribution comes with a net cost — some amount of total inventory is effectively lost with each operation.
This loss implies that you can't indefinitely increase the minimum value of the array by redistributing from high to low values. Eventually, you’ll run out of resources because the more you apply the operation, the more total inventory decreases. Hence, you must aim for an optimal stopping point that results in the highest achievable minimum value.
Think about how you'd evaluate if a certain target value `T` can be achieved as the minimum for all elements in the array. For a given `T`, you can check whether it's feasible to raise all values in the array to at least `T`, by using the allowed operation to redistribute values from higher-than-`T` elements to those lower than `T`.
Let’s formalize this into a thinking strategy. For a specific value `T`, iterate through the array and separate the elements into two categories:
1. **Needy elements**: These are elements currently less than `T`. You want to raise them to `T`, which requires increasing their value. You must calculate how much total "x-increment" is needed to lift them to `T`.
2. **Donor elements**: These are elements currently greater than `T`. You may be able to reduce these values using the operation to supply the x-increments needed elsewhere. For every reduction of `y` from such an element, you gain `x` which can be used to boost a needy element.
So, for a fixed `T`, the idea is to compare:
- The **total x-increments required** to lift all needy elements to at least `T`
- The **total x-increments available**, computed by reducing donor elements (via repeated `y` subtractions) until they reach `T`, and collecting the corresponding x-increments from those reductions
The core idea is that if the available x-increments (from donors) are at least equal to the required x-increments (for needy elements), then it's possible to achieve a minimum value of `T`. Otherwise, it's not.
Once you have a way to evaluate whether a given `T` is achievable, the next step is to find the **maximum such T**. This is where binary search becomes a useful tool. The possible values for `T` lie between 1 and the maximum value in the original array. You can apply binary search within this range:
- For a midpoint `T` in your current range, check if achieving `T` as the minimum value is feasible using the logic described above.
- If it is feasible, try to push the lower bound of the search higher — maybe an even larger minimum is possible.
- If it’s not feasible, reduce the upper bound — `T` is too ambitious given the constraints.
This binary search over feasible minimum values ensures that you converge on the optimal result efficiently, even for large arrays.
Finally, remember to handle edge cases:
- Arrays where all elements are already equal or higher than the result — no operations needed.
- When the difference between `x` and `y` is large, the redistribution can be more effective (though that doesn’t change the fundamental logic).
- Large values of `y` with small `x` result in greater losses, limiting how far you can push the minimum.
To summarize:
- The operation lets you redistribute value at a cost.
- You want to maximize the minimum element in the final array.
- Use a greedy feasibility check for a given target minimum value.
- Apply binary search over the possible values of the minimum to find the maximum achievable one.
- Evaluate feasibility based on required and available x-increments from the defined operation constraints.
This approach ensures you efficiently find the optimal result without exhaustively simulating every possible sequence of operations.
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