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Small
def getTotalX(a, b): nmax,nmin,count = max(a),min(b),0 for i in range(1,int(nmin/nmax)+1): if(sum((i*nmax)%n for n in a)+sum(n%(i*nmax) for n in b))==0: count+=1 return count
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;
let allElementsAreMultiple = false; let numberIsMultipleOfAll = false; while(number <= minB){ console.log('aMax ', maxA,'bMin ', minB, 'number ', number); // Every element of array must be a multiple of considerd number allElementsAreMultiple = a.every(ele => number%ele === 0 ); numberIsMultipleOfAll = b.every(ele => ele%number === 0 ); if( allElementsAreMultiple && numberIsMultipleOfAll ) total++; number++; } return total;
}
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
func getTotalX(a: [Int], b: [Int]) -> Int { var allElementsAreMultiple = false var numberIsMultipleOfAll = false let minB = b.reduce(b[0], min) var total = 0 for i in 1...minB { allElementsAreMultiple = a.map { return i % $0 == 0 }.reduce(true, {$0 && $1}) numberIsMultipleOfAll = b.map { return $0 % i == 0 }.reduce(true, {$0 && $1}) if (allElementsAreMultiple && numberIsMultipleOfAll) { total += 1 } } return total }
This is a fucking work of art.
nice job
why "i x nmax"?
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
This is a fucking work of art.
nice job
why "i x nmax"?