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

Between Two Sets

  • [deleted] + 3 comments

    Here is smaller code bro,

    def getTotalX(a, b):
    total = 0
    for x in list(range(a[-1], b[0] + 1)):
        hold = 0
        for v in a:
            if (x % v == 0): hold += 1
            else: hold = 0; break    
        
        for c in b:
            if (c % x == 0): hold += 1
            else: hold = 0; break       
        if (hold == len(a) + len(b)):
            total += 1   
    return total
    • anurag12345 + 0 comments

      Awesome bro

    • chandrakanthshy + 0 comments

      add a if condition above the 3rd for loop like:

      if hold > 0:

       for c in b:
 ....
 ....
    • gkjha009 + 0 comments

      Great,

      We not only need to solve this puzzle but at the same time think of time complexity :).

      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.