We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
I know it's been a long time since you asked this question, and I apologize for not being active here for a while. But if you are still interested, Here is my answer:
The for loop before this statement finds that the Element I is divisible by all the elements in array A, a[j]. Thus at any given point if any element a[j] cannot divide I, it should shut off and move on to the next element I + 1.
How do we know if all the elements in array A actually were able to divide I? - If you are still in the loop, and your j is n-1, meaning, you've reached the last element in array A (As the n is the length of A), all the elements at position (J==n-1) satisfied the division (otherwise, the loop would've exited - See j = n in else condition - that is exit logic).
Now if all a[j] were satisfying, we need to check if the element I can divide all elements in B, b[k]. The same logic used in checking if all elements in B are divisible by I - meaning if (K==m-1) where m is length of B - All elements at position m-1 were divisible by I.
At this point we found our element - So answer = answer + 1.
Between Two Sets
You are viewing a single comment's thread. Return to all comments →
can somebody please Explain this part to me?
if(j == n-1){ for(int k = 0; k < m; k++){ if(b[k]%i == 0){ if(k == m-1){ answr++; } }
I know it's been a long time since you asked this question, and I apologize for not being active here for a while. But if you are still interested, Here is my answer:
The for loop before this statement finds that the Element I is divisible by all the elements in array A, a[j]. Thus at any given point if any element a[j] cannot divide I, it should shut off and move on to the next element I + 1.
How do we know if all the elements in array A actually were able to divide I? - If you are still in the loop, and your j is n-1, meaning, you've reached the last element in array A (As the n is the length of A), all the elements at position (J==n-1) satisfied the division (otherwise, the loop would've exited - See j = n in else condition - that is exit logic).
Now if all a[j] were satisfying, we need to check if the element I can divide all elements in B, b[k]. The same logic used in checking if all elements in B are divisible by I - meaning if (K==m-1) where m is length of B - All elements at position m-1 were divisible by I.
At this point we found our element - So answer = answer + 1.