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每日一练 | Data Scientist & Business Analyst & Leetcode 面试题 306

大数据应用  · 公众号  · 大数据  · 2018-03-07 10:00

正文

自2017年6月15日起,数据应用学院与你一起温习数据科学(DS)和商业分析(BA)领域常见的面试问题。 从2017年10月4号起,每天再为大家分享一道Leetcode算法题。

希望积极寻求相关领域工作的你每天关注我们的问题并且与我们一起思考,我们将会在第二天给出答案。

Day 206

DS Interview Questions

You are given a data set on cancer detection. You’ve build a classification model and achieved an accuracy of 96%. Why shouldn’t you be happy with your model performance? What can you do about it?

BA Interview Questions

R language:

Using the following variable:

x=as.Date("10/11/2017","%d/%m/%Y")

# type a repeat () loop that increment x until x is equal to 31/12/2017.

LeetCode Questions

  • Description:

    • Given two binary trees, write a function to check if they are equal or not.

    • Two binary trees are considered equal if they are structurally identical and the nodes have the same value.

  • Input: 两颗树相同

  • Output: true

欲知答案如何?请见下期分解!

Day 205 答案揭晓

DS Interview Questions

You are given a data set. The data set has missing values which spread along 1 standard deviation from the median. What percentage of data would remain unaffected? Why?

Since, the data is spread across median, let’s assume it’s a normal distribution. We know, in a normal distribution, ~68% of the data lies in 1 standard deviation from mean (or mode, median), which leaves ~32% of the data unaffected. Therefore, ~32% of the data would remain unaffected by missing values.

BA Interview Questions

R language


Using the following variable:

x=100

y=50

i=1

type a while() or repeat () loop that incrementing i computes x=x-i and y=y+i until x


while(x>y){

x=x-i

y=y+i

i=i+1

print(c(x,y))

}


repeat{

x=x-i

y=y+i

i=i+1

print(c(x,y))

if(x

}


Leetcode Questions

  • Description:

    • Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

    • You should preserve the original relative order of the nodes in each of the two partitions.

  • Input: 1->4->3->2->5->2 and x = 3

  • Output: 1->2->2->4->3->5







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