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Decode the Coding Interview in Rust: Real-World Examples

Feature #8: Divide in Power Save Mode

Explore how to implement the divide function in Rust optimized for power save mode. Understand techniques to handle integer division adaptively depending on device battery conditions. Learn an efficient algorithm using doubling and bitwise operations to achieve logarithmic time complexity with constant space, preparing you to handle compiler optimization challenges effectively.

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Description

When people want to use mobile devices for a long duration and require basic functionality like audio calls and text messages, then the power-save mode is of great help. A drawback of power-save mode is that it disables power-hungry features, which the user may sometimes need. Therefore, we want our compiler to generate an adaptive code that behaves differently under different battery conditions. When the device is in power-save mode, it generates code that will adapt to conditions. An example of an operation that is disabled in power-save mode is integer division.

Let’s see the following illustration to get an idea of the divide function:

Solution

We will implement the divide() function, which takes two inputs: the dividend and the divisor. Over here, we will assume the constraint that the divisor will not be 0.

Let’s discuss how we will implement this method:

  1. First, the function will check if the given dividend equals the minimum possible number and that the divisor equals -1. Then, it will return the maximum possible number, using the i32::MAX function.

  2. If either value (dividend or divisor) is positive, we’ll convert it to negative by multiplying it with -1. This will help us avoid negative number overflow. In the end, we can adjust the sign of the result.

  3. Now, we can pick increasing multiples of the divisor until we get the smallest integer that is greater than the dividend. However, this is a slow process. Instead of this, we will keep doubling the divisor until we find that number. Effectively, we will be looking for the highest value of i for which 2 ...