price.cpp

#include <cassert>
#include <vector>

// Given some price information at discrete time intervals, returns a suitable
// value for the price at a given time interval. You may assume that times is
// sorted in ascending order, and that there are at least two datapoints (i.e.
// prices for at least two times).
auto stock_price(int time, std::vector<int> const &times,
                 std::vector<float> const &prices) -> float {
    return 0;
}

auto main() -> int {
    // Some sample time and price data. You might want to call stock_price for
    // different time arguments to test your function.
    auto const times = std::vector<int>{1, 3, 6, 11, 17, 20};
    auto const prices =
        std::vector<float>{100.00, 106.00, 109.00, 104.00, 107.00, 113.00};
}

price.py

from typing import List


def stock_price(time: int, times: List[int], prices: List[float]) -> float:
    """
    Given some price information at discrete time intervals, returns a suitable
    value for the price at a given time interval. You may assume that times is
    sorted in ascending order, and that there are at least two datapoints (i.e.
    prices for at least two times).
    """
    return 0


if __name__ == "__main__":
    # Some sample time and price data. You might want to call stock_price for
    # different time arguments to test your function.
    times  = [     1,      3,      6,     11,     17,     20]
    prices = [100.00, 106.00, 109.00, 104.00, 107.00, 113.00]

Sample solution using the interp function from NumPy:

from typing import List
import numpy as np


def stock_price(t: int, times: List[int], prices: List[float]) -> float:
    """
    Given some price information at discrete time intervals, return a suitable
    value for the price at a given time interval. You may assume that times is
    sorted in ascending order, and that there are at least two datapoints (i.e.
    a price at at least two times).
    """
    return np.interp(t, times, prices)


if __name__ == "__main__":
    times  = [     1,      3,      6,     11,     17,     20]
    prices = [100.00, 106.00, 109.00, 104.00, 107.00, 113.00]

    for time, price in zip(times, prices):
        assert stock_price(time, times, prices) == price

    assert stock_price(2, times, prices) == 103.00
    assert stock_price(4, times, prices) == 107.00
    assert stock_price(5, times, prices) == 108.00

One note about this solution compared to the last one is that np.interp will simply return the first/last price for times before/after the range of given information. For example, at t = 21 the price returned will be 113.00 instead of 115.00.

Other nice financial-themed problems:

• Number of orders in the backlog
• Stock Price Fluctuation