每日一练 | Data Scientist & Business Analyst & Leetcode 面试题 305

大数据应用 · 公众号 · 大数据 · 2018-03-06 09:59

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 204

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?

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

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

欲知答案如何？请见下期分解！

Day 203 答案揭晓

DS Interview Questions

Is rotation necessary in PCA? If yes, Why? What will happen if you don't rotate the components?

Yes, rotation (orthogonal) is necessary because it maximizes the difference between variance captured by the component. This makes the components easier to interpret. Not to forget, that’s the motive of doing PCA where, we aim to select fewer components (than features) which can explain the maximum variance in the data set. By doing rotation, the relative location of the components doesn’t change, it only changes the actual coordinates of the points.

If we don’t rotate the components, the effect of PCA will diminish and we’ll have to select more number of components to explain variance in the data set.

BA Interview Questions

R language:

Using the following variable:

i=10

x=10

# type a while() or repeat () loop that decreasing i computes x=x/i until i=0.

j=0

while(j<10){

x=x/i

i=i-1

j=j+1

print(x)

}

repeat{

x=x/i

i=i-1

print(x)

if(i==0) break;

}

Leetcode Questions

Description:

Given a 2D board and a word, find if the word exists in the grid.
The word can be constructed from lettersofsequentiallyadjacentcell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

Input: board = [ ['A','B','C','E'], ['S','F','C','S'], ['A','D','E','E'] ] word = "ABCCED"
Output: true