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public static int getTotalX(List<int> a, List<int> b)
{
int total = 0;
int number = a.Max();
Enumerable.Range(number, b.Min())
.ToList()
.ForEach(n =>
{
if (a.All(e => number % e == 0 || e % number == 0)
&& b.All(e => number % e == 0 || e %number == 0))
total++;
number++;
});
return total;
}
I noticed that Your answer goes through way too many iterations and doesn't take large enough steps, there can be no correct answer between a.Max() and 2a.Max(), therefore skipping those is a good idea.
Looping through by generating an amount of integers also isnt really necessary, just start from lowest possible correct answer ( a.Max() ) and make sure youre not over the largest possible ( b.Min() ) correct answer.
Works like a charm :).
Here is the resulting code with comments.
public static int getTotalX(List<int> a, List<int> b)
{
/*Starting from 0 found, storing largest value in A and smallest value in b. Setting starting point to largest value in A.*/
int foundCount = 0, maxA = a.Max(), minB = b.Min(), current = maxA;
while (current <= minB)
{
/*If the current value is divisible by all members of both arrays, then it's the one we want.*/
if (a.All(e => current % e == 0 || e % current == 0) && b.All(e => current % e == 0 || e % current == 0))
foundCount++;
/*Iterate the value by largest member of divisors, no reason to take smaller steps.*/
current+= maxA;
};
return foundCount;
}
" C# " for everyone CTRL+F-ing.
//Stored a.Max() and b.Min() as dsgarg71 suggested below :).
Between Two Sets
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Small
i am not getting the logic can you please explain?
It's pretty easy
range(1,int(nmin/nmax)+1) -
Loops through the possibilites between nmax and nmin
sum((inmax)%n for n in a) == 0
Finds whether chosen n is multiple of nmax by using modulus function
sum(n%(inmax) for n in b) == 0
Finds whether chosen n is a factor of bmin
I hope you got it..
We need to find numbers b/w two arrays, we can take max from a and min from b, then we can do calculation to find numbers.
This is what i'm doing in m beloved JS.
function getTotalX(a, b) { let total = 0; let maxA = Math.max(...a); let minB = Math.min(...b); let number = maxA;
}
That makes it very easy for me.
@gkjha009 -- This is really good.
i think u shouldn't compute numberIsMultipleOfAll if allElementsAreMultiple is false
This is a great solution. I implemented it in Swift
Nice logic!
The same logic in C#:
Hello!
I noticed that Your answer goes through way too many iterations and doesn't take large enough steps, there can be no correct answer between a.Max() and 2a.Max(), therefore skipping those is a good idea.
Looping through by generating an amount of integers also isnt really necessary, just start from lowest possible correct answer ( a.Max() ) and make sure youre not over the largest possible ( b.Min() ) correct answer.
Works like a charm :).
Here is the resulting code with comments.
" C# " for everyone CTRL+F-ing.
//Stored a.Max() and b.Min() as dsgarg71 suggested below :).
Very nice optimization! I really was not realized it. Thanks. :)
Thanks :). dsarg71 found another neat optimization. I've added that as well.
Wow It did the magic. Thanks
Thats n^3, Why not to store a.Max/b.Min and other such stuff once got so that their compute does not repeat.
You're right, the computation does indeed get repeated, I've changed the code accordingly. Good eye!
This is a fucking work of art.
nice job
why "i x nmax"?