for (int i = 1; i < a; i++){
    for(int j = 1; j < b; j++){
        Function() <-- O(1)
    }
}

In this case, the outer loop will be executed 'a'times(O(a)), and
the inner loop will be executed 'b/a' times(O(b/a)).
Then the total time complexity will be O(a * b/a ) = O(b)?