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  1. Practice
  2. Algorithms
  3. Implementation
  4. Between Two Sets
  5. Discussions

Between Two Sets

  • adityashaw2 + 1 comment

    This is a very lengthy approach to the problem.

    • rajat_yadav + 4 comments

      small

      static int getTotalX(int[] a, int[] b) {
        int x=1,r=0,j=0,count=0;
        int[] d = new int[101];
        for(x=1;x<101;x++){
        	int c=0;
        		for(int i=0;i<a.length;i++) {
        			if(x%a[i]==0&&x>=a[i]){
        				c++;
        			}
        			}
        		if(c==a.length){
        			d[j]=x;
    				r++;
    				j++;
        		}
        		}
        for(j=0;j<r;j++){
        	int c=0;
            for(int i=0;i<b.length;i++){
         	   if(b[i]%d[j]==0){
         		   c++;
         	   }
            }
            if(c==b.length){
         	   count++;
            }
        }
        return count;
    }
      • NSterling + 1 comment

        Could you explain what exactly is going on with this?

        • vgbutterfly6104 + 0 comments

          There, all the multiples common to first array(a) elements is found upto 100 and stored in 'd'. and then the elements that are in second array(b) that becomes the multiples of any of the elements is found and count is incremented accordingly

      • snkt_pat2 + 6 comments

        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
        • vaishnavipariti + 1 comment

          i am not getting the logic can you please explain?

          • parikshitsharma2 + 0 comments

            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..

        • gkjha009 + 3 comments

          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.

          • hacker117 + 0 comments

            @gkjha009 -- This is really good.

          • aheza007 + 0 comments

            i think u shouldn't compute numberIsMultipleOfAll if allElementsAreMultiple is false

          • boertjie_seun + 0 comments

            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
}
        • andy89 + 0 comments
          [deleted]
        • joshumcode + 0 comments

          This is a fucking work of art.

        • iamkdm + 0 comments

          nice job

        • max_aka21 + 0 comments

          why "i x nmax"?

      • evan51 + 0 comments

        Real programmers can program fortran in any language. Go back to CS101 and learn to use variable names that have meaning. a,b,x,r,j,d,c provide no context. This code would be unacceptable in a commericial application.

      • vivek4296 + 0 comments

        static int gcd(int l,int e) { if(e==0) { return l; } else { return gcd(e,l%e); } } static int lcm(int[] e) {
        int ans=1; for(int i:e) { ans=(ans*i)/gcd(ans,i); } return ans; } static int getTotalX(int[] a, int[] b) { /* * Write your code here. */

             int count=0;
     int g=b[0];
     int l=lcm(a);
     for(int i=1;i<b.length;i++)
     {
         g=gcd(g,b[i]);
     }
     int m=l;
     int i=1;
     while(m<=g)
     {

         m=l*i;
         if(g%m==0)
         {
             count++;
         }
         i++;

     }

return count;

}