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import java.util.stream.IntStream;
class CollatzCalculator {
long computeStepCount(int start) {
if (start < 1) {
throw new IllegalArgumentException("Only positive integers are allowed");
}
return IntStream.iterate(start, num -> num != 1, num -> (num & 1) == 1 ? 3 * num + 1 : num >> 1).count();
}
}
This approach starts by importing from packages for what is needed.
The real work begins by calling the IntStream.iterate() method which is passed the input number
as the initial element.
Before each iteration a lambda is used to check if the input number is not the value of 1.
If the value is not 1, the iteration runs.
If the value is 1, then all iteration stops.
Another lambda is used to process the input number.
Bitwise operators are used to check if the number is odd and to divide it in half if it is even.
The bitwise AND operator (&) compares the number with 1 to see if it is odd.
Another way to go about checking if the number is even or odd is to see if it is evenly divided by 2
by using the modulo (also know as the remainder) operator.
So, to see if it is even we could use start % 2 == 0.
That might be slightly less performant but perhaps more readable for intent, especially for those who don't "speak binary".
If the number is even, then the right shift operator (>>) shifts all of the bits once to the right, which is the equivalent
of dividing the number by 2.
Another way would be to use the division operator start \ 2 which might be slighly less performant but more readable
for intent, especially for those who don't "speak binary".
When the input number is the value of 1, then the count() method is called to count the number of iteratons for the
IntStream.
The function returns the result of the count() method.
