# Between Two Sets

# Between Two Sets

+ 0 comments O(n log(n)) solution.

1. find the LCM of all the integers of array**A**.

2. find the GCD of all the integers of array**B**.

3. Count the number of multiples of LCM that evenly divides the GCD.

+ 0 comments I found this outrageously difficult for being "easy" and I would have not even begun to understand what I was supposed to do without reading the discussions. "Factor" ? "Between two sets"? Why not call things by their name : GCD, LCM ?

Congratulations to those who solved this problem without help, this was way over my head.

+ 0 comments For all those guys like me who found this question difficult to understand, here's the simple explanation i got after searching for decades.

**1. Find LCM of the first array a. 2.Find GCD / HCF of the second array b. 3.Find all the multiples of LCM up to GCD, which divides the GCD evenly.**For Example: Here, In the given sample Input, The LCM of array a would be

**4**and the GCD of the array b would be**16**. Now, Find all Multiples of 4,(like 4,8,12,16,...) upto 16, such that,**(16%multiple_of_4_here) should be 0.**Here, 16%4=0 -----> count=1 (suppose this variable.) 16%8=0 -----> count=2 16%12!=0 ---> count=2 16%16=0 ---> count=3.Thus, The answer is 3. Hope this helped you.

+ 0 comments **My javascript solution:**function getTotalX(a, b) { let validCount = 0; for (let x = 1; x <= 100; x++) { if (a.every(int => (x % int == 0))) { if (b.every(int => (int % x == 0))) { validCount++; } } } return validCount; }

+ 0 comments Python with comprehension lists, for each number between max(setA) and min(setB), it will create a list that will hold boolean values, and 'all' checks that all the boolean values in a list are true.

lenA, lenB = map(int, raw_input().split()) setA = map(int, raw_input().split()) setB = map(int, raw_input().split()) maxA = max(setA) minB = min(setB) count = 0 for num in range(maxA, minB + 1): left = all([num % numA == 0 for numA in setA]) right = all([numB % num == 0 for numB in setB]) count += left*right print count

Sort 1945 Discussions, By:

Please Login in order to post a comment