Quantitative strategist Interview Questions

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Goldman Sachs
Quantitative Strategist was asked...December 17, 2010

Insert a new data point at the head of a singly linked circular list in constant time. Given a pointer to the current head of the circular list.

3 Answers

Insert the new element after the current head, and then swap the new value with the current head. Less

For a singly linked circular list, assuming you have a pointer to the tail of the list (rather than the head) to make insertion at head and tail constant time, here's how to insert at the beginning (head) in constant time: temp = Node(val2BeInserted) temp.setNext(self.tail.getNext()) self.tail = temp (given the classes Node and SinglyLinkedCircularList) with (value,next=None) and (tail) attributes, respectively. Less

For a singly linked circular list, assuming you have a pointer to the tail of the list (rather than the head) to make insertion at head and tail constant time, here's how to insert at the beginning (head) in constant time: temp = Node(val2BeInserted) temp.setNext(self.tail.getNext()) self.tail = temp (given the classes Node and SinglyLinkedCircularList) with (value,next=None) and (tail) attributes, respectively. Less

CV questions

1 Answers

Just answer it clearly, HR is a very cute person.

Knight Capital

Suppose you have a one-dimensional array A with length n that contains both positive and negative numbers. Design an algorithm to find the maximum sum of any contiguous subarray.

1 Answers

Make two arrays: accumulated sum and maximum sum. Start from index 0 and move to end. maximum sum would be max of accumulated sum till current position. Refresh accumulated sum once its smaller than the current number. Less

Morgan Stanley

Given a stick that is randombly broken in 2 places, what is the probability that the pieces can form a triangle?

1 Answers

The (formal) solution lies in the distribution of order statistics for a sample of 2 from the standard uniform. It is not too hard to show that the solution is 1-P(Max 1/2)-P(Range>1/2), and with a combinatorics/conditional probability argument, it is equivalent to 1-3*P(Max <1/2). P(Max< 1/2) = 1/4, so the probability of that the pieces can form a triangle is 1/4. Less

Title Trading

A few poker game questions I didn't really anticipate.

1 Answers

Intermediate probability solutions

Citadel

How would you measure the risk of an equity portfolio.

1 Answers

Open ended question, think they are looking for someone with the "right experience" rather than textbook answers. Less

Virtu Financial

20 points are randomly distributed on a circle. You pick 2 pairs of points at random, what is the probability for the two lines connecting each pair to intersect

1 Answers

The locations of the 4 picked points on the circle don't matter, you will always have three possibilities to divide them into two pairs, of which one yields intersecting lines. Therefore, no matter if you distribute 20 or 10000 points on the circle, the probability of intersecting lines of two pairs will always be 1/3 Less

Credit Suisse

basic math, programming skills, problem solving, and mathematical finance

1 Answers

I wasn't prepared

Virtu Financial

Return the integer that corresponds to the minimum number of steps required to change X to a Fibonacci number.

1 Answers

What is the step size?

Knight Capital

A random walk inside some boundaries. Ask to calculate the probability as time goes to infinite. Not fully understand the question.

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