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Cisco 两轮面经 (全印阵容)

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yi1san3fendi 发表于 2016-7-2 14:46:53 | 显示全部楼层 |阅读模式

2016(1-3月) 码农类 硕士 全职@Cisco - 内推 - 技术电面 |Failfresh grad应届毕业生

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Cisco网上自己投过,也找人内推过,一天突然一个印度recruiter发邮件过来联系电面,也不知是哪个起了作用
两轮面试后再无消息,3个多月过去了,估计是印度人把我当作炮灰做political correctness了

第一轮:一个印度人用webex给我电话面试,开视频,出了道 Phonebook 的 OO design, 并且有一些客户要求如果Phonebook上某些人联系方式改变,会被通知;我意识到用obeserver pattern,写了java code,  通过,说会有下一轮面试

第二轮:印度recruiter联系我,说onsite面试报销太麻烦(现在想起来应该是只不过准备把我当炮灰),说继续用webex video面试,但是需要一个下午时间,面试官全印阵容, 4个印度人
1. 有一堆文档,给一个单词,返回包含这个单词的所有文档路径,并且返回这些文档跟这个单词对应的instances
        word ->  <docPath1, docPath2, ..., docPathn>.1point3acres缃

package cisco;
import java.io.*;
import java.util.*;
. more info on 1point3acres.com
public class Solution1 {//assume the doc path is unique, and we can use it as ID. 鐣欏鐢宠璁哄潧-涓浜╀笁鍒嗗湴
    HashMap<String, HashMap<String, Integer>> wordPathHm;//<word, <docID/path, number of instances>>
    public Solution1(){
        wordPathHm = new HashMap<String, HashMap<String, Integer>>();
    }
   
    public void buildIndex(String path) throws IOException{
        Scanner doc = new Scanner(new FileReader(path));
        while(doc.hasNextLine()){
            String line = doc.nextLine();
            String[] words = line.split("\\s+");
            for(String word : words){
                if(wordPathHm.containsKey(word)){//if the word already exists, update the hashmap
                    HashMap<String, Integer> oldlist = wordPathHm.get(word);
                    Integer currCnt = oldlist.get(path);
                    oldlist.put(path, currCnt+1);
                } else {//if the word does not exist, add a new entry
                    HashMap<String, Integer> newlist = new HashMap<String, Integer>();. 1point3acres.com/bbs
                    newlist.put(path, 1);
                    wordPathHm.put(word, newlist);. from: 1point3acres.com/bbs
                } 鏉ユ簮涓浜.涓夊垎鍦拌鍧.
            }. visit 1point3acres.com for more.
        }
        doc.close();
    }
    .鐣欏璁哄潧-涓浜-涓夊垎鍦
    public List<String> getDocsPaths(String word){
        List<String> paths = new ArrayList<String>();
        if(wordPathHm.containsKey(word)){
            HashMap<String, Integer> hmRecord = wordPathHm.get(word);
            for(String docPath : hmRecord.keySet()){
                paths.add(docPath);
            }
        }
        . visit 1point3acres.com for more.
        return paths;
    }
    . From 1point 3acres bbs
    public int getDocInstanceNum(String word, String docPath){. From 1point 3acres bbs
        if(wordPathHm.containsKey(word)){
            HashMap<String, Integer> hmRecord = wordPathHm.get(word);
            if(hmRecord.containsKey(docPath)){
                return hmRecord.get(docPath);. 1point 3acres 璁哄潧
            } else {
                return -2; //the doc does not exist. 1point 3acres 璁哄潧
            }
        } else {
            return -1; // the word does not exist. 1point 3acres 璁哄潧
        }        
    }
. 1point 3acres 璁哄潧
}

然后又问如果用户很多该怎么办
DHT
hash(docID/path) -> which server should be queried for this doc
    the queried server should return the path of the documents, and how many instances in each document
multiple hashmaps
each hashmap return the number of instances for this key word, and add the count  together
. from: 1point3acres.com/bbs . 1point3acres.com/bbs
. from: 1point3acres.com/bbs

2. 问了操作系统调度问题,我刚好看了一点
Completely Fair Scheduler   Linux 2.6
we have different priority weights for different tasks. Waral 鍗氬鏈夋洿澶氭枃绔,
it is difficult for us to predict how much time.鐣欏璁哄潧-涓浜-涓夊垎鍦

previous exectuion time . 鐣欏鐢宠璁哄潧-涓浜╀笁鍒嗗湴
virtual running time(vrt) = already execution time/ priority weights

build red-black tree -> an approximately balanced binary search tree
if vrt is smaller , it means it should be assigned to the processor earlier
start from leftmost nodes

               18. 1point 3acres 璁哄潧
        12           26
    4        16     20    28

又问top K words. 1point3acres.com/bbs
package cisco;
import java.io.FileReader;
import java.util.*;. more info on 1point3acres.com

public class Solution2 {
    public List<String> getTopKWords(String path, int k) throws Exception{
        List<String> result = new ArrayList<String>();
        if(k <= 0)
            return result;
        Scanner doc = new Scanner(new FileReader(path));
        HashMap<String, Integer> wordCnt = new HashMap<String, Integer>();
        while(doc.hasNextLine()){ // find the total number of occurrences for each word
            String line = doc.nextLine();
            String[] words = line.split("\\s+");
            for(String word : words){
                   if(wordCnt.containsKey(word)){
                       wordCnt.put(word, wordCnt.get(word)+1);
                   } else {
                       wordCnt.put(word, 1);
                   }
                }
        }
        doc.close();

        Set<Map.Entry<String, Integer>> set = wordCnt.entrySet();
        List<Map.Entry<String, Integer>> list = new ArrayList<Map.Entry<String, Integer>>(set);
        Collections.sort(list, new Comparator<Map.Entry<String, Integer>>(){
            @Override
            public int compare(Map.Entry<String, Integer> o1, Map.Entry<String, Integer> o2){
                return o2.getValue()-o1.getValue(); // from the highest occurrences to the lowest occurrences
            }
        });
        
        int upperbound = Math.min(k, list.size());
        for(int i=0; i<upperbound; i++){// if k is larger than the number of words, then return all words
            result.add(list.get(i).getKey());
        }
        
        return result;
    }
. 鐗涗汉浜戦泦,涓浜╀笁鍒嗗湴
}. visit 1point3acres.com for more.


最后还剩8分钟,拷贝粘贴扔给我一道大题,关于字符串编码解码的,题目文字描述其中还有一堆乱码;他说这题你还能用java来做?我跟他确认题目意思都用了8分钟;最后说我可以面试完后做了发给他(好吧,他拿到了java版本的答案,真精明) 为了方便大家看,我剔除乱码,整理如下:
Where Did Those Numbers Come From?.鏈枃鍘熷垱鑷1point3acres璁哄潧
In both modes, the program maintains a linked list of words that have appeared in the input. In compression mode, the program reads the next word from the input file and searches for it in the list. If it is not found, the program writes the word to the output file and inserts it at the front of the list. If the word is found in the i-th position in the list, the program writes out i, removes the word from the list, then reinserts it at the front (this is the move-to-front feature). For example, let us consider the text:
LOVE,   LOVE   ME   DO!      YOU   KNOW   I   LOVE   YOU,   YOU,   YOU.
which after compression becomes     
LOVE,   1   ME   DO!      YOU   KNOW   I   6   4,   1,   1.
Notice that LOVE is replaced by both 1 and 6, and that 1 replaces both LOVE and YOU. This is due to word insertions in the list and the effect of move-to-front. After the second LOVE is processed, the word list is simply
(LOVE). From 1point 3acres bbs
hence the second instance of LOVE is replaced by the number 1. After the I is processed, the word list is
(I, KNOW, YOU, DO, ME, LOVE). From 1point 3acres bbs
Next, the third instance of LOVE is processed, the word is found in the 6th position (thus it is
replaced by number 6), and the list becomes:
(LOVE, I, KNOW, YOU, DO, ME)
When the second YOU is processed, YOU is found in the 4th position, and the list becomes:
(YOU, LOVE, I, KNOW, DO, ME). 鐣欏鐢宠璁哄潧-涓浜╀笁鍒嗗湴
Finally, when the last two instances of YOU are processed, the word is found in the first position.
In decompression mode, a word read from input is written out again and inserted at the front of the list. When the number i is read, the word in the i-th position is written out, then the word is deleted from the list and reinserted at the front. Decompression ends when the number 0 is read. The decompressed version should be identical to the original in every respect! (In Unix, you can use the utility diff to check this.)


package cisco;

import java.util.*;. Waral 鍗氬鏈夋洿澶氭枃绔,

//the requirement does not specify "LOVE" or "love" is the same word, so I assume words are all in uppercase
public class MoveToFrontCoding {
    public String compression(String str) {
        LinkedList<String> list = new LinkedList<>();.1point3acres缃
        StringBuilder ret = new StringBuilder();
        int last = 0;
        for (int i = 0; i < str.length(); i++) {
            if (Character.toUpperCase(str.charAt(i)) >= 'A' && Character.toUpperCase(str.charAt(i)) <= 'Z') {. 鐗涗汉浜戦泦,涓浜╀笁鍒嗗湴
                continue;
            } else {. visit 1point3acres.com for more.
                if (last != i) {
                    String s = str.substring(last, i);
                    int idx = list.indexOf(s);
                    if (idx >= 0) {// if the word already exists, replace it with a number
                        list.remove(idx);
                        ret.append(idx + 1);. 1point 3acres 璁哄潧
                    } else {// if the word does not exist, keep it
                        ret.append(s);
                    }
                    list.offerFirst(s);//insert the word at the beginning
                }
                last = i + 1;
                ret.append(str.charAt(i));//insert the word at the beginning. from: 1point3acres.com/bbs
            }
        }. 鐣欏鐢宠璁哄潧-涓浜╀笁鍒嗗湴
        if (last != str.length()) {
            String s = str.substring(last, str.length());. visit 1point3acres.com for more.
            int idx = list.indexOf(s);.1point3acres缃
            if (idx >= 0) {// if the word already exists, replace it with a number
                list.remove(idx);
                ret.append(idx + 1);
            } else {// if the word does not exist, keep it
                ret.append(s);
            }
            list.offerFirst(s);//insert the word at the beginning
        }
        return ret.toString();
    }

    public String decompression(String str) {
        LinkedList<String> list = new LinkedList<>();
        StringBuilder ret = new StringBuilder();
        int last = 0;
        for (int i = 0; i < str.length(); i++) {. 鍥磋鎴戜滑@1point 3 acres
            if ((str.charAt(i) >= '0' && str.charAt(i) <= '9') || (Character.toUpperCase(str.charAt(i)) >= 'A' && Character.toUpperCase(str.charAt(i)) <= 'Z')) {
                continue;
            } else {
                if (last != i) {
                    String ss = str.substring(last, i);
                    if (ss.matches("\\d+")) {//  the word already exists, replace the index with the word
                        int idx = Integer.parseInt(ss) - 1;
                        ret.append(list.get(idx));
                        String s = list.remove(idx);
                        list.offerFirst(s);//insert the word at the beginning
                    } else {
                        ret.append(ss);. 涓浜-涓夊垎-鍦帮紝鐙鍙戝竷
                        list.offerFirst(ss);//insert the word at the beginning
                    }
                }
                last = i + 1;.鏈枃鍘熷垱鑷1point3acres璁哄潧
                ret.append(str.charAt(i));. 涓浜-涓夊垎-鍦帮紝鐙鍙戝竷
            }
        }. Waral 鍗氬鏈夋洿澶氭枃绔,
        if (last != str.length()) {-google 1point3acres
            String ss = str.substring(last, str.length());
            if (ss.matches("\\d+")) {//  the word already exists, replace the index with the word
                int idx = Integer.parseInt(ss) - 1;
                ret.append(list.get(idx));
                String s = list.remove(idx);
                list.offerFirst(s);//insert the word at the beginning
            } else {
                ret.append(ss);. 鐗涗汉浜戦泦,涓浜╀笁鍒嗗湴
                list.offerFirst(ss);//insert the word at the beginning
            }
        }
        return ret.toString();
    }. more info on 1point3acres.com

    public static void main(String[] args) {
        MoveToFrontCoding s = new MoveToFrontCoding();
        String originalStr = "LOVE, LOVE ME DO!!  YOU KNOW I LOVE YOU, YOU, YOU.";
        System.out.println("original string: ");
        System.out.println(originalStr);
        System.out.println("Compressing...");
        String s1 = s.compression(originalStr);
        System.out.println(s1);-google 1point3acres
        System.out.println("Decompressing...");
        String s2 = s.decompression(s1);
        System.out.println(s2);
    }. 1point3acres.com/bbs
}
鏉ユ簮涓浜.涓夊垎鍦拌鍧.

. 鐣欏鐢宠璁哄潧-涓浜╀笁鍒嗗湴
Result:


original string:
LOVE, LOVE ME DO!!  YOU KNOW I LOVE YOU, YOU, YOU.
Compressing...
LOVE, 1 ME DO!!  YOU KNOW I 6 4, 1, 1.
Decompressing...
LOVE, LOVE ME DO!!  YOU KNOW I LOVE YOU, YOU, YOU.

3  这个印度人完全问我的项目经历
. Waral 鍗氬鏈夋洿澶氭枃绔,
4. 这个印度人问完我的项目经历,让我用两种方法做string reverse,并且分析每一步的复杂度

A fox jumped over the fence
. 1point3acres.com/bbs
package cisco;

public class Solution3 {. more info on 1point3acres.com
       public String reverseStr1(String input){
            String s = input.trim();
            int len = s.length();. 1point 3acres 璁哄潧
            if(len == 0). 鐣欏鐢宠璁哄潧-涓浜╀笁鍒嗗湴
                return input;//the input only contains space, return it
            String[] arrStr = input.split("\\s+");  // compute O(n)  memory read O(n)  write O(n)
            StringBuilder sb = new StringBuilder();. 1point3acres.com/bbs
            for(int i=arrStr.length-1; i>=0; i--){  // compute O(n)  memory read O(n)  write O(n)
                sb.append(arrStr[i]+" ");        
            }
            String result = new String(sb); //compute O(n)  read O(n)  write O(n)
            result = result.trim();// remove the last empty space  compute O(n) read O(n)  write O(1)
            return result;
       }.1point3acres缃
      
        public String reverseStr2(String input) {
            String s = input.trim();
            int len = s.length(); 鏉ユ簮涓浜.涓夊垎鍦拌鍧.
            if(len == 0)
                return input;//the input only contains space, return it
            char[] cArr = s.toCharArray();
            for(int i=0; i<len/2; i++){// reverse the whole string  compute O(n)  memory read O(n)  write O(n)
                char tmp = cArr[i];
                cArr[i] = cArr[len-1-i];
                cArr[len-1-i] = tmp;
            }
            
            int m=0;
            for(int i=0; i<=len; i++){. Waral 鍗氬鏈夋洿澶氭枃绔,
                /* reverse each substring delimited by space
                 * compute O(n)  memory read O(n)  write O(n)*/
                if(i==len || cArr[i]==' '){
                    int len1 = i-m;
                    for(int j=0; j<len1/2; j++){
                        char tmp1 = cArr[m+j];
                        cArr[m+j] = cArr[len1+m-1-j];
                        cArr[len1+m-1-j] = tmp1;. visit 1point3acres.com for more.
                    }
                    m = i+1;
                }
            }
            . 鐣欏鐢宠璁哄潧-涓浜╀笁鍒嗗湴
            return String.valueOf(cArr);
        }      
}
. 鐣欏鐢宠璁哄潧-涓浜╀笁鍒嗗湴

现在感觉我完全在给印度人提供面试题答案了;还是分享给大家比较好-google 1point3acres

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八和九生 发表于 2016-7-2 14:55:55 | 显示全部楼层
我勒个去,这是new grad吗?感觉好难啊。。。
楼主好厉害= =
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八和九生 发表于 2016-7-2 15:02:04 | 显示全部楼层
八和九生 发表于 2016-7-2 14:55
我勒个去,这是new grad吗?感觉好难啊。。。
楼主好厉害= =

楼主的top k words写的好棒啊= = 学习了。。。
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lajiwushi 发表于 2016-7-6 07:04:21 | 显示全部楼层
谢谢楼主的分享!!!
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yingchal 发表于 2016-9-30 12:20:46 | 显示全部楼层
谢谢楼主分享,第一题题干真的好长啊
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