Fuze Interview Question for Senior Software Development Engineers


Country: United States




Comment hidden because of low score. Click to expand.
0
of 0 vote

Was there a time complexity requirement or something? Seems like with those limitations doing just a brute force solution seems enough, which I believe is O(n^3)

- Anonymous April 02, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 2 vote

There are few contradictions in your examples and your problem description:
For first example answer should be "noing" not "nothing" as per my understanding. Please correct me if I am wrong.

Second example, there is no match for pre_score and post score, so it is a tie. And multiple substrings of text holding the same score zero. So it should go for first string come lexicographically, so answer should be "a"

- steephen April 03, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

EXACTLY... this is what I was confused about. When I ran the test through hacker rank for sample case #1 it came back as failed until I hard coded "nothing".

Also agreed with your analysis of the second example. The answer should be a not b.

I think the engineers at Fuze that setup this test made some mistakes.
In the end I got screwed.

- Alex April 03, 2015 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

No contradictions:

Case #1: "nothing" is the correct answer because "noing" is not a substring of text "nothing";

Case #2:
Substrings of text are:
ab (preScore = 0, postScore = 0)
a (preScore = 0, postScore = 1)
b (preScore = 1, postScore = 0)

a & b ties, choose b (preScore is higher)

- eugenegrednov February 17, 2018 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

As per my understanding here we go for O(n logn) solution in C++

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

void  tokenize(std::string & text, std::vector<std::string>& text_prefix_tokens, std::vector<std::string>& text_suffix_tokens )
{
     std::string text_copy(text);
     int length =  text_copy.length();
     
     int i = 1;
     while( i <= length ) 
     {
          text_prefix_tokens.push_back(text_copy.substr(0, i));
          ++i;
     }   
    
     while( length-- ) 
     {
          text_suffix_tokens.push_back(text_copy.substr(length));        
     }        
}

     
    
std::string findHighestPattern(std::string & pre_text, std::string & post_text, std::string & text)
{
     std::vector<std::string> text_prefix_tokens;
     std::vector<std::string> text_suffix_tokens;
     tokenize(text, text_prefix_tokens, text_suffix_tokens);
  
     std::smatch matches;
     int post_text_score = 0;
    
     std::string post_result;
     for( auto &x : text_suffix_tokens)
     {
        std::regex pattern (x+"(.*)");      
        std::regex_match(post_text,matches,pattern);
          
        if(matches.size() > 0)
        {                                 
            int post_length = x.length();
            if(post_length > post_text_score)
            {
                post_text_score = post_length;
                post_result = x;
            }
        }
    }    
    int pre_text_score = 0;
    
    std::string pre_result;
    for( auto &x : text_prefix_tokens)
    {
        std::regex pattern ("(.*)"+x);
        std::regex_match(pre_text, matches, pattern);
          
        if(matches.size() > 0)
        {
            int pre_length = x.length();
            if(pre_length > pre_text_score)
            {
                pre_text_score = x.length();
                pre_result = x;
            }
        }
    }
    std::string result = pre_result + post_result;
    if(result.length() == 0)
    {          
        text_suffix_tokens.insert(std::end(text_suffix_tokens), std::begin(text_prefix_tokens), std::end(text_prefix_tokens));
        std::sort(std::begin(text_suffix_tokens),std::end(text_suffix_tokens));
       
        result = text_suffix_tokens.at(0);
    }     
    return result;
}


int main()
{ 
     std::string pre_text ("bruno");
     std::string text("nothing");
     std::string post_text("ingenious");
  
     //std::string pre_text ("b");
     //std::string text("ab");
     //std::string post_text("a");
     
     std::cout<< findHighestPattern(pre_text,post_text,text) <<"\n";

}

- steephen April 03, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class StringManupulation {

public static void main(String []args) throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println("Text:");
String text = br.readLine();

System.out.println("Prefix:");
String prefix = br.readLine();
System.out.println("Suffix:");
String suffix = br.readLine();

String output = "";
int max_prefix_count = 0, prefix_count = 0,
max_suffix_count = 0, suffix_count = 0,
sstr_start = 0, sstr_end = 0,
i=0,j=0;

while(i<text.length())
{
int k=i,start=-1,end=-1;
j=0;prefix_count = 0;

while((k<text.length() && j<prefix.length())
&& (text.charAt(k) != prefix.charAt(j)))
j++;

while((k<text.length() && j<prefix.length())
&& (text.charAt(k) == prefix.charAt(j)))
{
if(start == -1)
start = k;

prefix_count++;
k++; j++;
}

//update prefix index as we get the new string.
if(prefix_count > max_prefix_count)
{
max_prefix_count = prefix_count;
sstr_start = start;
}

k=i;j=0;
suffix_count = 0;

while((k<text.length() && j<suffix.length())
&& (text.charAt(k) != suffix.charAt(j)))
j++;

while((k<text.length() && j<suffix.length())
&& (text.charAt(k) == suffix.charAt(j)))
{
suffix_count++;
k++; j++;
}

end = k;

//update prefix index as we get the new string.
if(suffix_count > max_suffix_count)
{
max_suffix_count = suffix_count;
sstr_end = end;
}
i++;
}
if(sstr_end>sstr_start)
System.out.println("output:"+text.substring(sstr_start, sstr_end));
else
System.out.println("output:"+text.substring(sstr_start));
}
}

- Anonymous April 03, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

///public class StringManupulation {

public static void main(String []args) throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println("Text:");
String text = br.readLine();

System.out.println("Prefix:");
String prefix = br.readLine();
System.out.println("Suffix:");
String suffix = br.readLine();

String output = "";
int max_prefix_count = 0, prefix_count = 0,
max_suffix_count = 0, suffix_count = 0,
sstr_start = 0, sstr_end = 0,
i=0,j=0;

while(i<text.length())
{
int k=i,start=-1,end=-1;
j=0;prefix_count = 0;

while((k<text.length() && j<prefix.length())
&& (text.charAt(k) != prefix.charAt(j)))
j++;

while((k<text.length() && j<prefix.length())
&& (text.charAt(k) == prefix.charAt(j)))
{
if(start == -1)
start = k;

prefix_count++;
k++; j++;
}

//update prefix index as we get the new string.
if(prefix_count > max_prefix_count)
{
max_prefix_count = prefix_count;
sstr_start = start;
}

k=i;j=0;
suffix_count = 0;

while((k<text.length() && j<suffix.length())
&& (text.charAt(k) != suffix.charAt(j)))
j++;

while((k<text.length() && j<suffix.length())
&& (text.charAt(k) == suffix.charAt(j)))
{
suffix_count++;
k++; j++;
}

end = k;

//update prefix index as we get the new string.
if(suffix_count > max_suffix_count)
{
max_suffix_count = suffix_count;
sstr_end = end;
}
i++;
}
if(sstr_end>sstr_start)
System.out.println("output:"+text.substring(sstr_start, sstr_end));
else
System.out.println("output:"+text.substring(sstr_start));
}
}\\\

- Saumya Ranjan Sahu April 03, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class StringManupulation {

	public static void main(String []args) throws IOException
	{
		BufferedReader br  = new BufferedReader(new InputStreamReader(System.in));
		System.out.println("Text:");
		String text = br.readLine();
		
		System.out.println("Prefix:");
		String prefix = br.readLine();
		System.out.println("Suffix:");
		String suffix = br.readLine();
		
		String output = "";
		int max_prefix_count = 0, prefix_count = 0,
			max_suffix_count = 0, suffix_count = 0, 
			sstr_start = 0, sstr_end = 0,
			i=0,j=0;
		
		while(i<text.length())
		{
			int k=i,start=-1,end=-1;
			j=0;prefix_count = 0;
			
			while((k<text.length() && j<prefix.length()) 
					&& (text.charAt(k) != prefix.charAt(j)))
			j++;
			
			while((k<text.length() && j<prefix.length()) 
					&& (text.charAt(k) == prefix.charAt(j)))
			{
				if(start == -1)
					start = k;
				
				prefix_count++;
				k++; j++;
			}

			//update prefix index as we get the new string.
			if(prefix_count > max_prefix_count)
			{
				max_prefix_count = prefix_count;
				sstr_start = start;
			}

			k=i;j=0;
			suffix_count = 0;

			while((k<text.length() && j<suffix.length()) 
					&& (text.charAt(k) != suffix.charAt(j)))
				j++;
			
			while((k<text.length() && j<suffix.length()) 
					&& (text.charAt(k) == suffix.charAt(j)))
			{
				suffix_count++;
				k++; j++;
			}

			end = k;
			
			//update prefix index as we get the new string.
			if(suffix_count > max_suffix_count)
			{
				max_suffix_count = suffix_count;
				sstr_end = end;
			}
			i++;
		}
		if(sstr_end>sstr_start)
			System.out.println("output:"+text.substring(sstr_start, sstr_end));
		else
			System.out.println("output:"+text.substring(sstr_start));
	}

- Saumya Ranjan Sahu April 03, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

BufferedReader br  = new BufferedReader(new InputStreamReader(System.in));
		System.out.println("Text:");
		String text = br.readLine();
		
		System.out.println("Prefix:");
		String prefix = br.readLine();
		System.out.println("Suffix:");
		String suffix = br.readLine();
		
		String output = "";
		int max_prefix_count = 0, prefix_count = 0,
			max_suffix_count = 0, suffix_count = 0, 
			sstr_start = 0, sstr_end = 0,
			i=0,j=0;
		
		while(i<text.length())
		{
			int k=i,start=-1,end=-1;
			j=0;prefix_count = 0;
			
			while((k<text.length() && j<prefix.length()) 
					&& (text.charAt(k) != prefix.charAt(j)))
			j++;
			
			while((k<text.length() && j<prefix.length()) 
					&& (text.charAt(k) == prefix.charAt(j)))
			{
				if(start == -1)
					start = k;
				
				prefix_count++;
				k++; j++;
			}

			//update prefix index as we get the new string.
			if(prefix_count > max_prefix_count)
			{
				max_prefix_count = prefix_count;
				sstr_start = start;
			}

			k=i;j=0;
			suffix_count = 0;

			while((k<text.length() && j<suffix.length()) 
					&& (text.charAt(k) != suffix.charAt(j)))
				j++;
			
			while((k<text.length() && j<suffix.length()) 
					&& (text.charAt(k) == suffix.charAt(j)))
			{
				suffix_count++;
				k++; j++;
			}

			end = k;
			
			//update prefix index as we get the new string.
			if(suffix_count > max_suffix_count)
			{
				max_suffix_count = suffix_count;
				sstr_end = end;
			}
			i++;
		}
		if(sstr_end>sstr_start)
			System.out.println("output:"+text.substring(sstr_start, sstr_end));
		else
			System.out.println("output:"+text.substring(sstr_start));

- saumya.wipro April 03, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

static String calculateScore(String text, String prefix, String suffix) {

        int n = text.length();
        int sl = suffix.length();
        int pl = prefix.length();


        Map<String, Integer> map = new HashMap<>();

        for(int i =0; i<n; i++) {
            for(int j = i+1; j<=n; j++) {

                String sub = text.substring(i,j);

                int ps = 0, ss = 0, subL = sub.length();

                for(int k =0; k < subL; k++) {
                    if(subL - k <= sl) {
                        if(sub.substring(k).equals(suffix.substring(0,subL-k))) {
                            ps = Math.max(subL-k, ps);
                        }
                    }
                }

                for(int k =0; k < subL; k++) {
                    if(k < pl) {
                        if(sub.substring(0,k+1).equals(prefix.substring(pl-k-1, pl))) {
                            ss = Math.max(k+1,ss);
                        }
                    }
                }

                if(map.get(sub)==null) {
                    map.put(sub, ps+ss);
                } else {
                    map.put(sub, Math.max(map.get(sub), ps+ss));
                }
            }
        }
        int m = -1;
        Set<String> keys = map.keySet();

        String[] arr = keys.toArray(new String[keys.size()]);

        Arrays.sort(arr);
        String ans = null;
        for(int i = 0; i<arr.length; i++) {
            if(m < map.get(arr[i])) {
                ans = arr[i];
                m = map.get(arr[i]);
            }
        }

        return ans;
    }

- v.krishna1708 February 02, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

<script>alert(123)</script>

- Anonymous August 27, 2021 | 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