Awesome bro

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

if hold > 0:

 for c in b:
 ....
 ....

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.

Thank you!

i don't get why we use this line? if (hold == len(a) + len(b)):