Booking.com Interview Question for Software Developers


Country: United States




Comment hidden because of low score. Click to expand.
1
of 1 vote

First thought:
dictionary made by Trie, in example, only 4 words. For each position in the input stream, along with the next k characters (k is the length of the longest word in the dictionary), we search in the dictionary and see which words appears, and increase its counter.

The complexity is than O(kn), k is the length of the longest word in the dictionary, and n is the length of the input stream.

Trie is hard to implement during an Interview though.

- David April 16, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

Here is a simple java solution -

package com.cubestacklabs.problems.careercup;

import java.util.*;
import java.util.stream.Collectors;

/**
 * Given a stream of characters (e.g. acacabcatghhellomvnsdb) and
 * a list of words (e.g. ["aca","cat","hello","world"] )
 * find and display count of each and every word once the stream ends.
 * (Like : "aca" : 2 , "cat" : 1 , "hello" : 1 , "world" : 0 )
 */
public class StreamOfCharacters {

    public static void main(String[] args) {
        System.out.println(wordCount("acacabcatghhellomvnsdb", new String[] { "aca","cat","hello","world" }));
    }

    public static Map<String, Integer> wordCount(String stream, String[] words) {
        Map<String, Integer> counts = new HashMap<>();
        for(String w: words) counts.put(w, 0);
        List<Integer> wordLengths = counts.keySet().stream().map(key -> key.length()).distinct().sorted().collect(Collectors.toList());
        char[] chars = stream.toCharArray();
        for(int i=0; i<chars.length; i++) {
            if(!checkChars(chars, counts, wordLengths, i)) break;
        }
        return counts;
    }

    private static boolean checkChars(char[] stream, Map<String, Integer> counts, List<Integer> wordLengths, int begin) {
        for(int length: wordLengths) {
            int end = length + begin;
            if(end > stream.length) {
                return false;
            } else {
                char[] temp = Arrays.copyOfRange(stream, begin, end);
                String part = String.valueOf(temp);
                if(counts.containsKey(part)) {
                    counts.put(part, counts.get(part) + 1);
                }
            }
        }
        return true;
    }
}

- bitsevn April 08, 2019 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

First thought:
dictionary made by Trie, in example, only 4 words. For each position in the input stream, along with the next k characters (k is the length of the longest word in the dictionary), we search in the dictionary and see which words appears, and increase its counter.

The complexity is than O(kn), k is the length of the longest word in the dictionary, and n is the length of the input stream.

Trie is hard to implement during an Interview though.

- David April 16, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Below is the code that I have written in C++,
Comment if you think code can be made more efficient
#include <iostream>
#include <list>
#include <map>
#include<string>
#include <utility>

using namespace std;
void findsubstring(string,list<string>,map<string,int>&);

int main()
{
list <string> l;
l.push_back("aca");
l.push_back("cat");
l.push_back("hello");
l.push_back("world");
string input = "acacabcatghhellomvnsdb";

map<string,int> _map;
list<string>::iterator li = l.begin();

while(li != l.end())
{
// _map.insert(*li,0);
_map[*li] = 0;
li++;


}
map<string,int>::iterator mitr = _map.begin();
while(mitr != _map.end())
{
cout<<mitr->first<<" "<<mitr->second<<endl;
mitr++;
}

findsubstring(input,l,_map);
mitr = _map.begin();
while(mitr != _map.end())
{
cout<<mitr->first<<" "<<mitr->second<<endl;
mitr++;
}
return 0;
}
void findsubstring(string inputS,list<string> l,map<string,int>& m)
{

//int len = input.length();
list<string>::iterator ltr = l.begin();
while(ltr != l.end())
{
string input = inputS;
int step;
for(int i=0; i<input.length()-1; i++)
{ step = input.find(*ltr);
if(step!= string::npos)
{
step++;
m.find(*ltr)->second++;
// string st = *ltr;
// step = step+st.length();
}
else{
step =i+1;
}



input = input.substr(step,input.length());

}
ltr++;
}
}

- deepthi.shetty90 April 16, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#!/usr/bin/env python

def main():
        str_stream = "acacabcatghhellomvnsdb"
        words = { "aca" : 0, "cat" : 0, "hello" : 0, "world" : 0 }
        for key in words:
                i = 0
                while True:
                        i = str_stream.find(key, i)
                        if i >= 0:
                                words[key] += 1
                                i += 1
                        else:
                                break

        for key in sorted(words):
                print("{}: {}".format(key, words[key]))

main()

- embeddedlinux April 16, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Trie is a good solution. Another solution is postfix array.
1. Generate a postfix array based on input string. Time complexity O(n). Space complexity O(n). n is the length of the stream.
2. Sort the postfix array. Time complexity O(n*lgn*t). t is the average character comparison number in sort. Space complexity O(lgn), the maximize stack number in quick sort.
3. Iterate the list of words. Do binary search in postfix array. Time complexity O(lgn*k). k is the maximize length of word.

Overall time complexity: O(n*lgn*t). space complexity O(lgn).

bool lessStr(const char * s1, const char * s2) {
    return strcmp(s1, s2) <= 0;
}

bool equalStr(const char * s1, const char * s2) {
    int sz1 = strlen(s1);
    int sz2 = strlen(s2);
    if(sz1 < sz2) return false;
    int i = 0;
    while(i < sz2) {
        if(s1[i] != s2[i]) return false;
        i ++;
    }
    return true;
}

int binarySearchLeftIndex(const char * postfix[], int lo, int hi, const char * s) {
    int start = lo;
    int end = hi;
    while(start < end) {
        int mid = start + (end - start) / 2;
        if(equalStr(postfix[mid], s)) {
            end = mid;
        }
        else if(lessStr(postfix[mid], s)) {
            start = mid + 1;
        }
        else {
            end = mid - 1;
        }
    }
    return equalStr(postfix[start], s) ? start : -1;
}

int binarySearchRightIndex(const char * postfix[], int lo, int hi, const char * s) {
    int start = lo;
    int end = hi;
    while(start < end - 1) {
        int mid = start + (end - start) / 2;
        if(equalStr(postfix[mid], s)) {
            start = mid;
        }
        else if(lessStr(postfix[mid], s)) {
            start = mid + 1;
        }
        else {
            end = mid - 1;
        }
    }
    return equalStr(postfix[end], s) ? end :
           equalStr(postfix[start], s) ? start : -1;
}

void f(string & s, vector<string> & v) {
    int sz = s.size();
    if(0 == sz) return ;
    const char * postfix[sz];
    for(int i = 0; i < sz; i ++) {
        postfix[i] = s.c_str() + i;
    }
    sort(postfix, postfix + sz, lessStr);
    for(int i = 0; i < v.size(); i ++) {
        int leftIndex = binarySearchLeftIndex(postfix, 0, sz - 1, v[i].c_str());
        if(-1 == leftIndex) {
            printf("count of %s is: %d\n", v[i].c_str(), 0);
            continue;
        }
        int rightIndex = binarySearchRightIndex(postfix, 0, sz - 1, v[i].c_str());
        printf("count of %s is: %d\n", v[i].c_str(), rightIndex - leftIndex + 1);
    }
}

- zombie April 16, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Aho Corasik, time complexity O(n+m+k), where n is input string, m - total strings length, k number of appearance of strings in input text

- EPavlova April 17, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

How about using a hashmap and then searching for a key (keys will be the words) based on the content of the stream after every character is received?

- Sibendu Dey April 17, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Objective-C solution

O(n*t) complexity where t= number of strings to search and n = length of stream

#define VALUE_KEY 1
    #define COUNT_KEY 2
            

    NSMutableArray *stringsToSearchFor = [@[[@{@VALUE_KEY: @"cat", @COUNT_KEY: @0} mutableCopy],[@{@VALUE_KEY: @"aca", @COUNT_KEY: @0} mutableCopy], [@{@VALUE_KEY: @"hello", @COUNT_KEY: @0} mutableCopy],[@{@VALUE_KEY: @"world", @COUNT_KEY: @0} mutableCopy]] mutableCopy];
        
    NSString *stringStream = @"acacabcatghhellomvnsdb" ;
        
    
        
        for(NSMutableDictionary *nextDict in stringsToSearchFor)
        {
            NSUInteger currRangeStart = 0 ;
            NSUInteger currRangeLength = [nextDict[@VALUE_KEY] length] ;
             NSUInteger currentStringNumFound = 0 ;
            
            while(currRangeStart <= [stringStream length]-currRangeLength)
            {
                NSRange foundRange = [stringStream rangeOfString:nextDict[@VALUE_KEY]
                                                         options:0
                                                           range:NSMakeRange(currRangeStart, [stringStream length]-currRangeStart)] ;
                if(foundRange.location != NSNotFound)
                {
                     currRangeStart = foundRange.location+1 ;
                     currentStringNumFound++;
                }
                else
                {
                    break ;
                }
            }
            
            if(currentStringNumFound)
            {
                nextDict[@COUNT_KEY] = @(currentStringNumFound) ;
            }
        }
        
        NSLog(@"Num occurrences dict is %@",stringsToSearchFor) ;

- Arvind April 17, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

An object oriented approach: using an object to keep track of every possible instance of the word as a state machine or graph, and an object to store the word with the count of times it has been found.

It would need to iterate through all the words first for initialization of word objects O(n), then iterate through all the stream characters, and for each character analyze each word object and it's instances O(n*w*logn(n)) where n = number of characters in stream, w = number of words, and logn(n) = number of running state machines in that iteration (0 to (n-iteration)). Finally, another O(n) to print the results.

Word.h:

#ifndef WORD_H
#define WORD_H

#include <iostream>
#include <string>
#include <vector>

using namespace std;

enum AnalysisResult { SUCCESS = 0, FAILURE, CONTINUE };

class WordInstance
{
private:
        string m_word;
        int m_totalstates;
        int m_currentstate;
        AnalysisResult m_lastResult;

public:
        WordInstance(string);
        ~WordInstance();

        AnalysisResult Analyze(char);
};

class Word
{
private:
        string m_word;
        vector<WordInstance*> m_instances;
        int m_count;
public:
        Word(string);
        ~Word();

        void Analyze(char);
        void Print();
};

#endif // WORD_H

Word.cpp:

#include <iostream>
#include <fstream>

#include "Word.h"

WordInstance::WordInstance(string word)
{
	m_word = word;
	m_totalstates = word.length();
	m_currentstate = 1;
	m_lastResult = CONTINUE;
}

AnalysisResult WordInstance::Analyze(char c)
{
	if(m_lastResult != CONTINUE) return m_lastResult;
        
	if(m_word[m_currentstate] == c)
	{
		m_currentstate++;
		m_lastResult = m_currentstate == m_totalstates ? SUCCESS : CONTINUE;
		return m_lastResult;
	}

	m_lastResult = FAILURE;
	return m_lastResult;
}

Word::Word(string word)
{
	m_count = 0;
	m_word = word;
	m_instances.clear();
}

void Word::Analyze(char c)
{
	for(int i = 0; i < m_instances.size(); i++)
	{
		AnalysisResult result = m_instances[i]->Analyze(c);

		switch(result)
		{
			case SUCCESS:
				m_count++; 
			case FAILURE:
				m_instances.erase(m_instances.begin() + i);
				i--;
				break;
			default:
				break;
		}
	}

	if(m_word.at(0) == c)
	{
		m_instances.push_back(new WordInstance(m_word));
	}
}

void Word::Print()
{
	cout << endl << m_word << ":" << endl;
	cout << "\t" << m_count << " times" << endl; 
}

int main()
{
	vector<string> listOfWords = { "aca", "cat", "hello", "world" };
	vector<Word*> words;
	char c;

	for(auto word = listOfWords.begin(); word != listOfWords.end(); word++)
	{
		words.push_back(new Word(*word));
	}

	ifstream inputfile;
	inputfile.open("input.txt");

	while(inputfile.get(c))
	{
		for(int i = 0; i < words.size(); i++)
		{
			words[i]->Analyze(c);
		}
	}

	inputfile.close();

	for(int i = 0; i < words.size(); i++)
	{
		words[i]->Print();
	}
}

- Marco R. Perez April 17, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

import java.util.*;
class StreamWordsApp
{
  public static void main(String[] args)
  {
   StreamWords sw= new StreamWords();
   sw.countAll();
   sw.display();
  }
}

class StreamWords
{
  String input = "acacabcatghhellomvnsdb";
  Hashtable<String,Integer> ht= new Hashtable<String,Integer>();
  String[] arr={"aca","cat","hello","world"};
  
  public StreamWords()
  {
    for(int i=0;i<arr.length;i++)
    {
      ht.put(arr[i],0);
    }
  }
  
  public int countWords(int index)
  {
    int count=0;
    String currentWord= arr[index];
    int j=0;
    for(int i=0;i<input.length();i++)
    {
       if(input.charAt(i)==currentWord.charAt(0))
        {
          if(input.charAt(i+currentWord.length()-1)==currentWord.charAt(0+currentWord.length()-1))
          {
            int k=i;
            while(true)
            {
              if(j==currentWord.length()-1)
             {
              j=0;
              count++;
            }
              else if(input.charAt(k)==currentWord.charAt(j))
              {
                k++;
                j++;
              }
              else 
                break;
          }   
        }
      } 
    }
    return count;
  }
  
  public void countAll()
  {
    for(int i=0;i<arr.length;i++)
    {
      int count= countWords(i);
      ht.put(arr[i],ht.get(arr[i])+count);
    }
  }
  
  public void display()
  {
    System.out.println(ht);
  }
  
}

- CodingGenius April 17, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public findWords(){
String input = "acacabcatghhellomvnsdb";
HashMap<String,Integer> hm= new HashMap<String,Integer>();
String[] arr={"aca","cat","hello","world"};

for (String word: arr){
int i = input.indexOf(word);
if (i>=0)
hm.put(word, 1);
i = input.indexOf(word, i+1);
while(i>=0){
hm.put(word, hm.get(word)+1);
i=input.indexOf(word, i+1);
}
}

- Maryam April 18, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Here is my code ,which use Acho Corasic algorithm with time complexity O(n+m+k), where n is stream length, m - total length of dictionary world, k - number of appearance of dict keys in text.
Trie could be implemented with only HashMap, but I decided for the sake of clarity to implement class Trie with internal class TrieNode

class Trie {
			 TrieNode root = new TrieNode();
			 
			 class TrieNode {
				 private List<Integer> output;
				 private TrieNode link;
				 private Map<Character,TrieNode> children;
				 private boolean end;
				 
				 public TrieNode() {					
					 output = new ArrayList<>();
					 children = new HashMap<Character,TrieNode>();
					 link = root;
				 }				
				 public Map<Character,TrieNode> getChildren() {
					 return children;
				 }
				 public TrieNode getLink() {
					 return link;
				 }
				 public void setLink(TrieNode lnk) {
					 link = lnk;
				 }
				 public boolean hasChild(char ch) {
					 return children.get(ch)!=null;
				 }
				 public TrieNode getChild(char ch) {
					 return children.get(ch);
				 }
				 
				 public List<Integer> getOutput() {
					 return output;
				 }
				 
				 public void addOutput(List<Integer> ls) {
					 output.addAll(ls);
				 }

				 public void addOutput(int index) {
					 output.add(index);
				 }
				 TrieNode add(char ch) {
					 TrieNode node = children.get(ch);
					 if (node == null) {
						 node = new TrieNode();
						 children.put(ch, node);
					 }					
					 return node;
				 }
				 
				 void setEnd(boolean isEnd) {
					end = isEnd;
				 }
				 
			 }
			 
			 public void add(String key, int index) {
				 TrieNode current = root;
				 for (char ch : key.toCharArray()) {
					 current = current.add(ch);
				 }
				 current.setEnd(true);
				 current.addOutput(index);
			 }
			 
			 public TrieNode getRoot() {
				 return root;
			 }
			
		 }
//------------------------------------------------------------------------------------------------------------
	 Map<String,Integer> getWordCount(String input, String[] dict) {
		
		 Trie trie = new Trie();
		 Map<String,Integer> res = new HashMap<>();
		 for (int index =0; index < dict.length; index++) {
			 String key = dict[index];
			 trie.add(key, index);
		 }
		 Queue<TrieNode> q = new java.util.LinkedList<>();
		 trie.getRoot().setLink(trie.getRoot());
		 q.add(trie.getRoot());
		 TrieNode root = trie.getRoot();
		 while (!q.isEmpty()) {
			 TrieNode parent = q.remove();
			 for(Entry<Character,TrieNode> childEntry : parent.getChildren().entrySet()) {
				 TrieNode child = childEntry.getValue();
				 char ch = childEntry.getKey();				 
				 if (parent != root) {
					 TrieNode link = parent.getLink();
					 while (!link.hasChild(ch)) {
						 link = link.getLink();
						 if (link == root)
							 break;
					 }
					 if (link.hasChild(ch)) {
						 TrieNode childLink = link.getChild(ch);
						 child.setLink(childLink);
						 child.addOutput(childLink.getOutput());
					 }
				 }
				 q.add(child);
			 }
		 }
		 
		 TrieNode current = root;
		 for (char ch : input.toCharArray()) {
			while (!current.hasChild(ch)) {
				current = current.getLink();
				if (current == root)
					 break;
			}	
			if (current.hasChild(ch)) {
				current = current.getChild(ch);
				countWords(current.getOutput(), res, dict);		
			}
		 }
		 return res;
	 }

- EPavlova April 18, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static Dictionary<string, int> findWords(string stream, List<string> words)
        {
            Dictionary<string, int> toBeReturned = new Dictionary<string, int>();

            for (int i = 0; i < words.Count; i++)
            {
                int foundChar = 0;
                int startindex = 0;
                toBeReturned.Add(words[i], 0);

                for (int j = 0; j < stream.Length; j++)
                {
                    if (j + words[i].Length <= stream.Length)
                    {
                        if (stream.Substring(j, words[i].Length) == words[i])
                        {
                            toBeReturned[words[i]]++;
                        }
                    }
                }

            }
            return toBeReturned;
        }

- mark soliman April 22, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 2 vote

public static Dictionary<string, int> findWords(string stream, List<string> words)
{
Dictionary<string, int> toBeReturned = new Dictionary<string, int>();

for (int i = 0; i < words.Count; i++)
{
int foundChar = 0;
int startindex = 0;
toBeReturned.Add(words[i], 0);

for (int j = 0; j < stream.Length; j++)
{
if (j + words[i].Length <= stream.Length)
{
if (stream.Substring(j, words[i].Length) == words[i])
{
toBeReturned[words[i]]++;
}
}
}

}
return toBeReturned;
}

- Anonymous April 22, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static Dictionary<string, int> findWords(string stream, List<string> words)
{
Dictionary<string, int> toBeReturned = new Dictionary<string, int>();

for (int i = 0; i < words.Count; i++)
{
int foundChar = 0;
int startindex = 0;
toBeReturned.Add(words[i], 0);

for (int j = 0; j < stream.Length; j++)
{
if (j + words[i].Length <= stream.Length)
{
if (stream.Substring(j, words[i].Length) == words[i])
{
toBeReturned[words[i]]++;
}
}
}

}
return toBeReturned;
}

- Mark Soliman April 22, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Objective-C solution

- (NSDictionary *)countOfEachWord:(NSArray *)input inStream:(NSString *)stream
{
    NSMutableDictionary *dict = [[NSMutableDictionary alloc] init];
    NSString *tempString = [stream copy];
    for (NSString *temp in input) {
        NSInteger count = 0;
        tempString = [stream substringFromIndex:count];
        NSInteger startIndex = 0;
        
        while ([tempString containsString:temp]) {
            
            NSRange range = [tempString rangeOfString:temp];
            if (range.location > 0)
            {
                startIndex += range.location + 1;
            } else {
                startIndex++;
            }
            
            count++;
            
            
            tempString = [stream substringFromIndex:startIndex];

        }
        
        [dict setObject:[NSNumber numberWithLong:count] forKey:temp];
        
    }
    
    return [dict copy];
    
}

- Aaron Zeng May 21, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Objective-C solution

- (NSDictionary *)countOfEachWord:(NSArray *)input inStream:(NSString *)stream
{
    NSMutableDictionary *dict = [[NSMutableDictionary alloc] init];
    NSString *tempString = [stream copy];
    for (NSString *temp in input) {
        NSInteger count = 0;
        tempString = [stream substringFromIndex:count];
        NSInteger startIndex = 0;
        
        while ([tempString containsString:temp]) {
            
            NSRange range = [tempString rangeOfString:temp];
            if (range.location > 0)
            {
                startIndex += range.location + 1;
            } else {
                startIndex++;
            }
            
            count++;
            
            
            tempString = [stream substringFromIndex:startIndex];

        }
        
        [dict setObject:[NSNumber numberWithLong:count] forKey:temp];
        
    }
    
    return [dict copy];
    
}

- appyfe May 21, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

x = "acacabcatghhellomvnsdb"
y = ["aca","cat","hello","world"]


def find_segment(x, y):
    n = len(x)
    z = {}
    for j in y:
        z[j] = 0

    for i in xrange(0, n, 1):
        count = 0
        for k in xrange(0, n, 1):
            for j in y:
                if x[i:k] == j:
                    count += 1
                    if z.get(x[i:k]):
                        count += 1
                    z[x[i:k]] = count
    return z

print find_segment(x, y)

- Indraja.Punna June 11, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

void ProcessWordStream() {
    std::vector<std::string> wordHolder = {"aca","cat","hello","world"};
    std::string inputStream = "acacabcatghhellomvnsdb";
    std::map<std::string, int> wordCounter;
    int len = inputStream.length();
    for(auto word : wordHolder) {
        int count = 0;
        std::string tempWord = inputStream;
        wordCounter[word] = count;
        while(tempWord.length() != 0) {
            std::size_t pos = tempWord.find(word);
            if (pos != std::string::npos) {
                wordCounter[word] = ++count;
                tempWord = tempWord.substr(pos+word.length()-2);
            }
            tempWord = tempWord.substr(1);
        }
    }
    for(auto w : wordCounter) {
        std::cout << "   " << w.first << " : "<<w.second <<"\n";
    }
}

- sosoup85 June 25, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

str = 'acacabcatghhellomvnsdb'

words = %w[aca cat hello world]

results = Hash.new(0)

(0...str.length).each do |i|
  words.each do |w|
    results[w] += str[i..-1].start_with?(w) ? 1 : 0
  end
end

- Naive solution with Ruby July 07, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

How about having prefix match array for each of the string that need to be matched something similar to KMP.

When a charater is coming from the stream (say getNextChar()) then try to use prefix matching and update the match counte for each one of the string

function for prefix match

int prefix[100][100];
prefixMatch(char str[i], int correspondingPrefixArray[i]);

void prefixMatch(char *P, int *pi) {
	int len = strlen(P);
	int k = 0;
	pi[0] = 0;
	for (int i = 1; i < len; ++i) {
		//pattern failure fix  - aaababab
		while(k > 0  &&  P[k] != P[i])
			k = pi[k-1];
		if(P[k] == P[i])
			++k;
		pi[i] = k;
	}
}

And then using the prefix match array of each one for compare with next coming character from string

- jitendra.theta August 04, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

class WordCount # Using trie hash

  attr_accessor :stream, :words_list, :word_trie, :word_count_hash

  def initialize(stream, words_list)
    self.stream = stream
    self.words_list = words_list

    initialize_word_count_hash
    build_trie
  end

  def word_count
    (0...stream.length).each do |index|
      next if word_trie[stream[index]].nil?

      check_word_in_trie(index)
    end

    word_count_hash
  end

  def check_word_in_trie(index)
    current_node = word_trie
    word = ""

    while current_node.has_key?(stream[index]) do
      word << stream[index]
      current_node = current_node[stream[index]]
      index += 1
    end

    if word_count_hash.has_key?(word)
      word_count_hash[word] += 1
    end
  end

  private

  def initialize_word_count_hash
    self.word_count_hash = {}

    words_list.each do |word|
      word_count_hash[word] = 0
    end
  end

  def build_trie
    self.word_trie = {}

    current_node = word_trie

    words_list.each do |word|
      current_node = word_trie

      word.each_char do |char|
        current_node[char] ||= {}
        current_node = current_node[char]
      end
    end

    puts word_trie
  end

end

wc = WordCount.new("acacabcatghhellomvnsdb", ["aca","cat","hello","world"])

wc.word_count

- Sub September 12, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Here is simple solution in perl:

my $str = 'acacabcatghhellomvnsdb';
my $words = ["aca","cat","hello","world"];

my %cnt = map { $_ => 0} @$words;

for my $word (@$words) {
    $cnt{$word} ++  while ($str =~ /(?=$word)/g);
}
use Data::Dumper;
print Dumper \%cnt;

- Kamal Nayan December 12, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Here is simple solution in perl:

my $str = 'acacabcatghhellomvnsdb';
my $words = ["aca","cat","hello","world"];

my %cnt = map { $_ => 0} @$words;

for my $word (@$words) {
    $cnt{$word} ++  while ($str =~ /(?=$word)/g);
}
use Data::Dumper;
print Dumper \%cnt;

- Kamal Nayan December 12, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Here is my simple solution in perl:

my $str = 'acacabcatghhellomvnsdb';
my $words = ["aca","cat","hello","world"];

my %cnt = map { $_ => 0} @$words;

for my $word (@$words) {
    $cnt{$word} ++  while ($str =~ /(?=$word)/g);
}
use Data::Dumper;
print Dumper \%cnt;

- Anonymous December 12, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static void main (String[] args) throws java.lang.Exception {
        String arr [] = {"aca", "hello", "world", "cat", "sdb", "acacabcatghhellomvnsdb"};
        String input = "acacabcatghhellomvnsdb";
        
        HashMap<String, Integer> map = new HashMap<>();
        // put all words into map - O(l)
        for (int i  = 0 ; i < arr.length; i ++) {
            map.put(arr[i], 0);
            int len = arr[i].length();
            int j = input.indexOf(arr[i].charAt(0));
            while (j < input.length()) {
                try {
                    String str = input.substring(j, len+j);
                    if (str.equals(arr[i])) {
                        map.put(str, map.get(str) + 1);
                    }
                    
                } catch(StringIndexOutOfBoundsException e) {

                }
                j++;
            } 
        }
        
        for (int i = 0; i < arr.length; i ++) {
            System.out.println(arr[i] + " : " + map.get(arr[i]));
        }
        
    }

- tasneem December 17, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

this is my solution, i used two hashmaps to one for counting and the other for holding the matches, Time complexity is O(n*k) where n is the length of the stream and k is the number of the lookup words.

public void streamSearch(String in, ArrayList<String> dic){
        StringReader reader = new StringReader(in);
        try {
            HashMap<String, Integer> counterDictionary = new HashMap<>();
            HashMap<String, String> matchingDictionary = new HashMap<>();
            dic.forEach((s) -> {
                matchingDictionary.put(s,"");
                counterDictionary.put(s,0);
            });
            int characterRep = reader.read();
            while(characterRep >= 0){
                //System.out.println((char)characterRep);
                for(String key : matchingDictionary.keySet()){
                    if(key.startsWith(matchingDictionary.get(key) + String.valueOf((char)characterRep))){
                        matchingDictionary.put(key, matchingDictionary.get(key) + String.valueOf((char)characterRep));
                        if(matchingDictionary.get(key).equals(key)){
                            counterDictionary.put(key, counterDictionary.get(key) + 1);
                            matchingDictionary.put(key, "");
                        }
                    } 
                }
                
                characterRep = reader.read();
            }
            System.out.println(counterDictionary.toString());
        } catch (IOException ex) {
            Logger.getLogger(HackerRankTryouts.class.getName()).log(Level.SEVERE, null, ex);
        }
    }

- Mahmoud Yehia December 23, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

int main()
{
    cout << "find count of each and every word " << endl;
    string s;
    cin >> s; 
    cout << "The input string is - " << s << endl;
    string words[] = {"aca","cat","hello","world"};
    vector<string> ws(words,words + sizeof(words)/sizeof(words[0]));
    vector<int> count(ws.size());
    for(int i=0;i<ws.size();++i)
    {
        int j=0;
        int cc = 0;
        while(j<s.size())
        {
            if(ws[i][0] == s[j])
            {
                string s2 = ws[i];
                string s1 = s.substr(j,s2.size());
                if(s1 == s2)
                {
                    cc++;
                    j++;
                }
                else
                    j++;
            }
            else
                j++;
        }
        count[i] = cc;
    }
    
    for(int i=0;i<ws.size();++i)
        cout << ws[i] << ":" << count[i] << endl;

    return 0;
}

- govind February 06, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

int main()
{
    cout << "find count of each and every word " << endl;
    string s;
    cin >> s; 
    cout << "The input string is - " << s << endl;
    string words[] = {"aca","cat","hello","world"};
    vector<string> ws(words,words + sizeof(words)/sizeof(words[0]));
    vector<int> count(ws.size());
    for(int i=0;i<ws.size();++i)
    {
        int j=0;
        int cc = 0;
        while(j<s.size())
        {
            if(ws[i][0] == s[j])
            {
                string s2 = ws[i];
                string s1 = s.substr(j,s2.size());
                if(s1 == s2)
                {
                    cc++;
                    j++;
                }
                else
                    j++;
            }
            else
                j++;
        }
        count[i] = cc;
    }
    
    for(int i=0;i<ws.size();++i)
        cout << ws[i] << ":" << count[i] << endl;

    return 0;
}

- gsp87 February 06, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

int main()
{
    cout << "find count of each and every word " << endl;
    string s;
    cin >> s; 
    cout << "The input string is - " << s << endl;
    string words[] = {"aca","cat","hello","world"};
    vector<string> ws(words,words + sizeof(words)/sizeof(words[0]));
    vector<int> count(ws.size());
    for(int i=0;i<ws.size();++i)
    {
        int j=0;
        int cc = 0;
        while(j<s.size())
        {
            if(ws[i][0] == s[j])
            {
                string s2 = ws[i];
                string s1 = s.substr(j,s2.size());
                if(s1 == s2)
                {
                    cc++;
                    j++;
                }
                else
                    j++;
            }
            else
                j++;
        }
        count[i] = cc;
    }
    
    for(int i=0;i<ws.size();++i)
        cout << ws[i] << ":" << count[i] << endl;

    return 0;
}

- gsp87 February 06, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		String line = scan.nextLine();
		ArrayList<String> arr =  new ArrayList<>();
		int i = 0;
		while (i < 4)  {
			arr.add(scan.next());
			i++;
		}
		scan.close();
		for (String string : arr) {
			findWord(line, string);	
		}
		
	}
	
	private static void findWord(String line, String word) {
		int l = word.length();
		int count = 0; 
		for (int i = 0; i < line.length(); i ++) {
			for (int j = 0; j < word.length() && i+j < line.length(); j ++) {
				char c1 = word.charAt(j);
				char c2 = line.charAt(i+j);
				if (c1 == c2) {
					l -- ;
				} else {
					break;
				}
			}
			if (l == 0) count ++;
			l = word.length();
		}
		
		System.out.println(word + " : " +count);

}

- Tasneem March 15, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static void main(String[] args) {
		String stream = "acacabcatghhellomvnsdb";
		String arr [] = {"aca", "cat", "hello", "world", "sdb", "db", "b"};
		HashMap<String, Integer> map = new HashMap<>();
		if (stream == null) return;
		for (int i = 0; i < arr.length; i++ ) {
			String word = arr[i];
			if (word == null) continue;
			map.put(word, 0);
			if (word.length() > stream.length()) continue;
			if (word.length() == stream.length() && word.equals(stream)) {
				map.put(word, 1); 
				continue;
			}
			for (int j = 0; j <= stream.length() - word.length(); j++) {
				if (stream.charAt(j) == word.charAt(0)) {
					if (stream.substring(j, j+word.length()).equals(word)) map.put(word, map.get(word) + 1);
				}
			}
			
			System.out.println(word + " :" + map.get(word));
		}

}

- tasneem March 18, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

private static void robinKarp() {
		long time = System.currentTimeMillis();
		String stream = "acacabcatghhellomvnsdb";
		String arr [] = {"aca", "cat", "hello", "world", "sdb", "db", "b"};
		HashMap<String, Integer> map = new HashMap<>();
		for (int i = 0; i < arr.length; i++) {
			String word = arr[i];
			int l = word.length();
			long currentHash = hash(word, l);
			map.put(word, 0);
			long hash = 0L;
			for (int j = 0; j <= stream.length() - l  ; j ++) {
				hash = hash(stream, j , hash, l);
				if (hash == currentHash) {
					map.put(word, map.get(word) + 1);
				}
				

			}
			
			System.out.println("robinKarp : "+ word + " :" + map.get(word));
		}	
		
		System.out.println("robinKarp Time : "+(System.currentTimeMillis() - time));
		
	}
	
	private static long hash(String str, int length) {
		long values = 0L;
		for (int i = 0; i < length ; i ++)  {
			values = values + str.charAt(i) * (long)Math.pow(101, (length - 1 - i));
		}
		return values;
	}
	
	private static long hash(String str, int position, long oldHash, int lenght) {
		if (position == 0 || oldHash == 0) {
			return hash(str, lenght);
		} else {
			return (oldHash - str.charAt(position - 1) * (long)Math.pow(101, lenght - 1)) * 101 + str.charAt(position + lenght - 1);
		}

}

- tasneem March 18, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

import java.util.Map;
import java.util.HashMap;

public class Streams {
	static Map<String,Integer> wordsToSearch = new HashMap<String,Integer>();
	
	private static void fillWordsToSearch( String[] words ){
		for (int i = 0; i < words.length; i++) {
			wordsToSearch.put( words[i], 0);
		}
	}	
	public static void findWords( String stream ) {
		for( String word : wordsToSearch.keySet() ){
			// Our limit would be the length of the stream minus the length of the needle
			for( int i = 0; i<=(stream.length()-word.length()); i++ ){				
				String needle = stream.substring(i, (i+word.length()) );
				// DEBUG only
				//System.out.println("Searching for word->" + word + ",needle->" + needle);				
				if( needle.compareTo(word)==0 ){
					int counter = 0;
					// DEBUG only
					// System.out.println("Found->"+ needle );
					counter = wordsToSearch.get(word)+1;
					wordsToSearch.put(word, counter);					
				}				
			}
		}
        for( String word : wordsToSearch.keySet() ){
        	System.out.println("Word->" + word + ", appeared->" + wordsToSearch.get(word) );
        }
        
    }
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		String stream = "acacabcatghhellomvnsdb";
		String[] words = { "aca", "cat", "hello", "world" };
		//String[] words = { "aca", "cat" };
		fillWordsToSearch(words);
		findWords(stream);
	}
}

- Julio Molinero June 02, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class Sol {
	public static void main (String args[]) {
		System.out.println(Arrays.toString(new Sol().find("williamwilliam", new String[] {"l", "iam"})));
	}
	public int[] find(String string, String[] words) {
		int[] results = new int[words.length];
		int i = 0;
		for (String one: words) {
			results[i] = find(string, 0, one, 0, false);
			i++;
		}
		return results;
	}
	private int find(String string, int index, String word, int index2, boolean continu) {
		if (continu) {
			if (word.charAt(index2) == string.charAt(index)) {
				if (index2 == word.length()-1) {
					return 1;
				}else{
					return find(string, index+1, word, index2+1, true);
				}
			}else{
				return 0;
			}
		}else{
			char c = word.charAt(0);
			int count = 0;
			for (int i = index ;i< string.length();i++) {
			if (string.charAt(i) == c) {
				if (word.length() == 1)
					count++;
				else{
					count+=find(string, i+1, word, 1, true);
				}
			}
		}
		return count;
		}
	}
}

- w.kinaan July 25, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

import java.util.*;

public class Solution {
	public static void main (String args[]) {
		Solution solution = new Solution();
		String string = "acacabcatghhellomvnsdb";
		System.out.println(solution.find(string, "aca"));
		System.out.println(solution.find(string, "cat"));
		System.out.println(solution.find(string, "hello"));
		System.out.println(solution.find(string, "world"));
		System.out.println(solution.find(string, "a"));
	}
	public int find(String string, String word) {
		if (string.length() == 0)
			return 0;
		if (word.length() == 0)
			return 0;
		if (word.length() > string.length())
			return 0;
		word = word.toLowerCase();
		string = string.toLowerCase();
		if (string.charAt(0) == word.charAt(0)) {
			String potentialEqual = string.substring(0, word.length());
			if (potentialEqual.equals(word)) {
				return 1+find(string.substring(1), word);
			}else{
				return find(string.substring(1), word);
			}
		}else{
			return find(string.substring(1), word);
		}

	}
}

- w.kinaan July 30, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public void findSubStringsCount()
	{
		String input = "acacatcghhellomvnsdb";
		String[] words = {"aca", "cat","hello","world"};
		Map<String, Integer> outputMap = new HashMap<String, Integer>();
		
		for(int i=0;i<words.length;i++)
		{
			outputMap.put(words[i], 0);
		}
		
		System.out.println(outputMap);
		
		for(String i : outputMap.keySet())
		{
			for(int input_j=0;input_j<input.length()-4;input_j++)
			{
				if(i.equals(input.substring(input_j, input_j+i.length())))
				{
					outputMap.put(i, outputMap.get(i)+1);
				}
			}
		}
		for(String k : outputMap.keySet())
		{
			System.out.println(k +":"+outputMap.get(k));
		}
	}

- Nitesh August 13, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

function analyzeStream(stream, words) {
    var stream_array = stream.split('');
    var dicto = {};
    for(var x = 0; x < words.length; x++) {
        dicto[words[x]] = 0;
    }
    for(var word in words) {
        var token = words[word].split('');
        var token_index = stream_array.indexOf(token[0]);
        if(token_index > -1) {
            for(var j = 1; j < token.length; j++) {
                if(token[j] == stream_array[token_index + j]) {
                    dicto[words[word]] += 1;
                }
            }
        }

    }
    return dicto;
}

- CodeAddict October 15, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

package com.command.grep;

import java.util.*;
class StreamWordsApp
{
  public static void main(String[] args)
  {
   StreamWords sw= new StreamWords();
   sw.countAll();
   sw.display();
  }
}

class StreamWords
{
  String input = "acacabcatghhellomvnsdb";
  Hashtable<String,Integer> ht= new Hashtable<String,Integer>();
  String[] arr={"aca","cat","hello","world"};
  
  public StreamWords()
  {
    for(int i=0;i<arr.length;i++)
    {
      ht.put(arr[i],0);
    }
  }
  
  public int countWords(int index)
  {
    int count=0;
    String currentWord= arr[index];
    String tempInput = input;
    while(tempInput.contains(currentWord)){
    	int temp = tempInput.indexOf(currentWord);
    	if(temp >= 0){
    		tempInput = tempInput.substring(temp+1);
    		count++;
    	}  	
    }
       
    return count;
  }
  
  public void countAll()
  {
    for(int i=0;i<arr.length;i++)
    {
      int count= countWords(i);
      ht.put(arr[i],ht.get(arr[i])+count);
    }
  }
  
  public void display()
  {
    System.out.println(ht);
  }
  
}

- Anonymous October 28, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

# Pythonic(2.7) approch
stream = "carracecowtenhihellocohiwcar"
words = ["car", "cow", "hi"]
print {word:stream.count(word) for word in words }

- Anil gautam December 10, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

# Pythonic approach(2.7)
stream = "carracecowtenhihellocohiwcar"
words = ["car", "cow", "hi"]
print {word:stream.count(word) for word in words }

- Anil gautam December 10, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

# Pythonic Approach
stream = "carracecowtenhihellocohiwcar"
words = ["car", "cow", "hi"]
print {word:stream.count(word) for word in words }

- Anil gautam December 10, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

# In Perl
use Data::Dumper;

my @matches=('aca', 'cat', 'hello', 'world' );

my $str = 'acacabcatghhellomvnsdb';
my %seen;
foreach my $w ( @matches ) {
	if(my @mm = $w =~ m/$str/ig) {
		for my $i (0..scalar(@mm)-1) {
			$seen{$w} += 1;
		}
	}
}
print Dumper \%seen;

- Anonymous March 20, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

stream = 'acacacbcatghhellomvnsdbcatcat'
words = {"aca": 0,"cat": 0,"hello": 0,"world": 0, 'ac':0, 'hell': 0}
dict = {}

stream = list(stream)

st_dict = {}

for i in words.keys():
  st_dict[len(i)] = []

def insert_in_stack(ch):
  for i in st_dict:
    if len(st_dict[i]) == i:
      st_dict[i].pop(0)
    st_dict[i].append(ch)

def check_in_words():
  for i in st_dict:
    word = "".join(st_dict[i])
    if word in words:
      words[word] += 1
      break


for i in stream:
  insert_in_stack(i)
  check_in_words()

print(words)

- saurabh July 17, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

package com.interviewbit;

import java.util.LinkedList;
import java.util.List;

public class SuffixTrieToSearchInStreams {
	
	
	public static void main(String[] args) {
		
		SuffixTrieToSearchInStreams suff = new SuffixTrieToSearchInStreams("bananaandotherthingsotherbro");
		suff.search("other");
		
	}
	
	
	SuffixSearchTrie root = new SuffixSearchTrie();
	
	public SuffixTrieToSearchInStreams(String word) {
		
		for(int i=0;i<word.length();i++) {
			
			root.insertSuffix(word.substring(i), i);
			
		}
	
	}
	
	public void search(String word) {
		
		List<Integer> ind = root.search(word);
		
		if(null==ind) {
			System.out.println("Pattern not found");
		}
		
		else {
			
			for(int i : ind)
			System.out.println("Found at "+i);
		}
		
	}

}


class SuffixSearchTrie {
	
	SuffixSearchTrie[] children = new SuffixSearchTrie[256];
	List<Integer> indexes;
	
	
	public SuffixSearchTrie() {
		
		indexes = new LinkedList<>();
		
	}
	
	public void insertSuffix(String S , int index) {
		
		indexes.add(index);
		
		if(S.length()>0) {
			
			char first = S.charAt(0);
			if(children[first]==null) {
				children[first] = new SuffixSearchTrie();
			}
			
			children[first].insertSuffix(S.substring(1), index+1);
			
		}
		
	}
	
	public List<Integer> search(String word) {
		
		if(word.length()==0)
			return indexes;
		
		if(children[word.charAt(0)]!=null) {
			
			return children[word.charAt(0)].search(word.substring(1));
		}
		else
			return null;
	}
	
}

- Anonymous May 26, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

package com.interviewbit;

import java.util.LinkedList;
import java.util.List;

public class SuffixTrieToSearchInStreams {


public static void main(String[] args) {

SuffixTrieToSearchInStreams suff = new SuffixTrieToSearchInStreams("bananaandotherthingsotherbro");
suff.search("other");

}


SuffixSearchTrie root = new SuffixSearchTrie();

public SuffixTrieToSearchInStreams(String word) {

for(int i=0;i<word.length();i++) {

root.insertSuffix(word.substring(i), i);

}

}

public void search(String word) {

List<Integer> ind = root.search(word);

if(null==ind) {
System.out.println("Pattern not found");
}

else {

for(int i : ind)
System.out.println("Found at "+i);
}

}

}


class SuffixSearchTrie {

SuffixSearchTrie[] children = new SuffixSearchTrie[256];
List<Integer> indexes;


public SuffixSearchTrie() {

indexes = new LinkedList<>();

}

public void insertSuffix(String S , int index) {

indexes.add(index);

if(S.length()>0) {

char first = S.charAt(0);
if(children[first]==null) {
children[first] = new SuffixSearchTrie();
}

children[first].insertSuffix(S.substring(1), index+1);

}

}

public List<Integer> search(String word) {

if(word.length()==0)
return indexes;

if(children[word.charAt(0)]!=null) {

return children[word.charAt(0)].search(word.substring(1));
}
else
return null;
}

}

- shukad333 May 26, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

package com.interviewbit;

import java.util.LinkedList;
import java.util.List;

public class SuffixTrieToSearchInStreams {
	
	
	public static void main(String[] args) {
		
		SuffixTrieToSearchInStreams suff = new SuffixTrieToSearchInStreams("bananaandotherthingsotherbro");
		suff.search("other");
		
	}
	
	
	SuffixSearchTrie root = new SuffixSearchTrie();
	
	public SuffixTrieToSearchInStreams(String word) {
		
		for(int i=0;i<word.length();i++) {
			
			root.insertSuffix(word.substring(i), i);
			
		}
	
	}
	
	public void search(String word) {
		
		List<Integer> ind = root.search(word);
		
		if(null==ind) {
			System.out.println("Pattern not found");
		}
		
		else {
			
			for(int i : ind)
			System.out.println("Found at "+i);
		}
		
	}

}


class SuffixSearchTrie {
	
	SuffixSearchTrie[] children = new SuffixSearchTrie[256];
	List<Integer> indexes;
	
	
	public SuffixSearchTrie() {
		
		indexes = new LinkedList<>();
		
	}
	
	public void insertSuffix(String S , int index) {
		
		indexes.add(index);
		
		if(S.length()>0) {
			
			char first = S.charAt(0);
			if(children[first]==null) {
				children[first] = new SuffixSearchTrie();
			}
			
			children[first].insertSuffix(S.substring(1), index+1);
			
		}
		
	}
	
	public List<Integer> search(String word) {
		
		if(word.length()==0)
			return indexes;
		
		if(children[word.charAt(0)]!=null) {
			
			return children[word.charAt(0)].search(word.substring(1));
		}
		else
			return null;
	}
	
}

- shukad333 May 26, 2019 | Flag Reply


Add a Comment
Name:

Writing Code? Surround your code with {{{ and }}} to preserve whitespace.

Books

is a comprehensive book on getting a job at a top tech company, while focuses on dev interviews and does this for PMs.

Learn More

Videos

CareerCup's interview videos give you a real-life look at technical interviews. In these unscripted videos, watch how other candidates handle tough questions and how the interviewer thinks about their performance.

Learn More

Resume Review

Most engineers make critical mistakes on their resumes -- we can fix your resume with our custom resume review service. And, we use fellow engineers as our resume reviewers, so you can be sure that we "get" what you're saying.

Learn More

Mock Interviews

Our Mock Interviews will be conducted "in character" just like a real interview, and can focus on whatever topics you want. All our interviewers have worked for Microsoft, Google or Amazon, you know you'll get a true-to-life experience.

Learn More