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Dynamic Holding, LLC interview – Dynamic Holding, LLC VO 面试真题 面经 – 一亩三分地

Dynamic Holding, LLC is a quantitative proprietary trading company specializing in systematic trading using quantitative models derived from mathematical and statistical analysis with a focus on Index futures, Commodity futures and equities. Our team collaborates to solve tough problems in the most complex and dynamic markets and focuses on understanding how the markets dynamically change by building quantitative models. We convert our intellectual outcomes into sophisticated portfolios with sustainable returns and strategic partnerships with financial institutions. We establish and engineer models to dissect the financial markets, and we strive to build a fundamental, cause-and-effect understanding of the markets.

近期Dynamic Holding, LLC 公司火热招聘中,它能够支持签证等事项,我们一起来看看它的其中一道面试真题吧。

You are given an array 𝐴A of 𝑁N positive integers. In one move, you can pick a segment (a contiguous fragment) of 𝐴A and a positive integer 𝑋X, and then increase all elements within that segment by 𝑋X.

An array is strictly increasing if each element (except for the last one) is smaller than the next element.

Write a function def solution(A) that, given an array 𝐴A of 𝑁N integers, returns the minimum number of moves needed to make the array strictly increasing.

Example:

  1. Given 𝐴=[4,2,4,1,3,5]A=[4,2,4,1,3,5], the function should return 2. One possible solution is to add 𝑋=3X=3 to the segment [2,4][2,4], and then add 𝑋=8X=8 to the segment [1,3,5][1,3,5]. As a result of these two moves, 𝐴A is now strictly increasing: [4,2,4,1,3,5]→[4,5,7,1,3,5]→[4,5,7,9,11,13][4,2,4,1,3,5]→[4,5,7,1,3,5]→[4,5,7,9,11,13]
  2. Given 𝐴=[3,5,7,7]A=[3,5,7,7], the function should return 1.
  3. Given 𝐴=[1,5,6,10]A=[1,5,6,10], the function should return 0.

为了将给定的数组 𝐴A 转变成严格递增序列,并求出需要的最少操作次数,我们可以采取以下简单的思路:

  1. 识别非递增点:遍历数组,识别所有那些当前元素大于或等于下一个元素的位置。这些位置表明需要进行操作以确保整个数组是严格递增的。
  2. 确定操作的边界:每次识别到非递增的点,意味着我们需要从这个点开始或之前的某个位置开始操作,直到数组的某个后续位置,使得操作后的这一部分满足严格递增的条件。
  3. 最小化操作次数:尽可能地将多个需要操作的点合并到一次操作中。如果一个操作可以覆盖多个非递增的点,那么总的操作次数可以减少。
  4. 优化选择:对于每个需要增加的段,选择最小的增量 𝑋X,这个 𝑋X 必须使得整个段在操作后成为严格递增的。

这个问题可以通过动态规划或贪心算法解决,核心在于识别需要操作的关键点,并计算涵盖这些点的最少操作次数。每个操作都应当尽可能长时间地推迟数组中后续的非递增问题。

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