Bloomberg LP Interview Question for Interns


Team: London
Country: United States
Interview Type: Phone Interview




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4
of 4 vote

Solution in java

public String charToFrequency(String str) {
        if (str == null || str.isEmpty()) {
            return str;
        }

        char current = str.charAt(0);
        int count = 1;

        StringBuilder output = new StringBuilder();

        for (int i=1; i<str.length(); i++) {
            if (str.charAt(i) == current) {
                count++;
            } else {
                output.append(count).append(current);
                current = str.charAt(i);
                count = 1;
            }
        }

        output.append(count).append(current);
        return output.toString();
    }

Time Complexity: O(n)
Space Complexity: O(n)

- slvrhwk January 25, 2017 | Flag Reply
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1
of 1 vote

In Ruby:

string_in = "aaaggbbbbc"
string_out = ""
while string_in.length > 0
  string_out << (string_in.count(string_in[0]).to_s + string_in[0])
  string_in.delete!(string_in[0])
end
puts string_out

- Tomy8s January 25, 2017 | Flag Reply
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1
of 1 vote

#include <iostream>
#include <string>

using namespace std;

string compressString(string);

int main(){
string str="Ashiq assdewfefkljhdsklfdsklkljlkjklll";
cout<<str<<" : "<<compressString(str)<<endl;
return 0;
}

string compressString(string data){
string result="";
if(data.length()<1) return result;
int count=1;
for(int index = 0; index < data.length()-1; index++){
if(data.at(index) != data.at(index+1)){
result +=to_string(count)+data.at(index);
count=1;
}
else count++;
}
result +=to_string(count)+data.at(data.length()-1);

return result;
}

- Muhammed Ashiq January 25, 2017 | Flag Reply
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0
of 0 vote

public class Solution {
  public static String encodeRepeatedChars(String source) {
      StringBuffer result = new StringBuffer();
      for (int idx = 0; idx < source.length(); idx++) {
          int runLength = 1;
          while (idx + 1 < source.length()
                  && source.charAt(idx) == source.charAt(idx + 1)) {
              runLength++;
              idx++;
          }
          result.append(runLength);
          result.append(source.charAt(idx));
      }
      return result.toString();
  }

  public static void main(String[] args) {
      String example = "aaaggbbbbc";
      System.out.println(encodeRepeatedChars(example));
  }
}

Time Complexity: O(N)

- R@M3$H.N January 24, 2017 | Flag Reply
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0
of 0 vote

char *encode(char *src) {

- Anonymous January 24, 2017 | Flag Reply
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0
of 0 vote

char *encode(char *src) {
{
int length = strlen(src);

if(length == 0) return NULL;

char * first = *src;
char * next = *(++first);
char * result = new char[length];

int occur=1;

while (*next != '\0')
{
if(*first == *next)
{
occur++;
}
else
{
sprintf(result, "%d%c", occur, *first);
occur =1;
}

first++; next++;
}

sprintf(result, "%d%c", occur, *first);
}












}

- AjayL January 24, 2017 | Flag Reply
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0
of 0 vote

class Main {
public static void main(String[] args) {
String str = "aaacccbbfffffff";
int count = 0;
for(int i=0; i< str.length(); i++) {
if(i+1 < str.length() && str.charAt(i) == str.charAt(i+1)){
count++;
} else if(i+1 < str.length() && str.charAt(i) != str.charAt(i+1)){
System.out.print(count+1 + "" + str.charAt(i));
count = 0;
} else {
System.out.print(count+1 + "" + str.charAt(i));
count = 0;
}
}
}
}

- Anonymous January 24, 2017 | Flag Reply
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0
of 0 vote

public static String charFreq(String input){
		StringBuilder output=new StringBuilder();
		Character ch = null;
		int count = 0;
		for(char eachC: input.toCharArray()){
			if(ch != null && eachC != ch){
				output.append(count).append(ch);
				ch = eachC;
				count=1;
			}else{
				ch = eachC;
				count++;
			}
		}	
		//Last set of character not handled in above for loop
		output.append(count).append(ch);
		return output.toString();
	}

- Sachin.jha1 January 25, 2017 | Flag Reply
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0
of 0 vote

inputstr = 'aaaggbbbbc'
outputstr = '3a2g4b1c'
from collections import OrderedDict

def encode(string):
    assert (len(string) > 0), 'Must pass a valid string'
    i = 0
#    charmap = {}
    charmap = OrderedDict()
    while i < len(string):
        x = string[i]
#        assert (x.isspace() == False), 'Contiguous letters expected'
        if x.isalpha():
            v = charmap.get(x)
            if v is not None:
                charmap[x] += 1
            else:
                charmap.setdefault(x, 1)
        i += 1
    strbuilder = ''.join(['%d%s' % (v, k) for k,v in charmap.items()])
    return strbuilder

result = encode(inputstr)
print('Decoded string: %s' % (result))
assert (result == outputstr) , 'Input and Output must match'

- Joy January 25, 2017 | Flag Reply
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0
of 0 vote

inputstr = 'aaaggbbbbc'
outputstr = '3a2g4b1c'
from collections import OrderedDict

def encode(string):
    assert (len(string) > 0), 'Must pass a valid string'
    i = 0
#    charmap = {}
    charmap = OrderedDict()
    while i < len(string):
        x = string[i]
#        assert (x.isspace() == False), 'Contiguous letters expected'
        if x.isalpha():
            v = charmap.get(x)
            if v is not None:
                charmap[x] += 1
            else:
                charmap.setdefault(x, 1)
        i += 1
    strbuilder = ''.join(['%d%s' % (v, k) for k,v in charmap.items()])
    return strbuilder

result = encode(inputstr)
print('Decoded string: %s' % (result))
assert (result == outputstr) , 'Input and Output must match'

- joy January 25, 2017 | Flag Reply
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0
of 0 vote

Here is my Java Solution using LinkedHashMap...

package com.careercup;

/**
 * Created by ejangpa on 1/25/2017.
 */
import java.util.*;

public class RepeatedCharacters {
    public static void main(String[] args) {
        String str = "aaaggbbbbc"; // we can take input from scanner also
        char[] strArray = str.toCharArray();
        Map<Character,Integer> map = new LinkedHashMap<>();         // taking LinkedHashMap to
                                                                    // maintain the insetion order
        for (char ch : strArray ) {
            if (map.containsKey(ch)) {
                int value = map.get(ch);
                map.put(ch, value + 1);
            } else {
                map.put(ch, 1);
            }
        }
        Set set = map.entrySet();
        Iterator iterator = set.iterator();

        while (iterator.hasNext()) {
           Map.Entry entry = (Map.Entry) iterator.next();
            System.out.print(entry.getValue() +""+ entry.getKey());
        }
    }
}

- Jabongg January 25, 2017 | Flag Reply
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0
of 0 vote

#include <iostream>
#include <string>

using namespace std;

string compressString(string);

int main(){
	string str="Ashiq assdewfefkljhdsklfdsklkljlkjklll";
	cout<<str<<" : "<<compressString(str)<<endl;
	return 0;
}

string compressString(string data){
	string result="";
	if(data.length()<1) return result;
	int count=1;
	for(int index = 0; index < data.length()-1; index++){
		if(data.at(index) != data.at(index+1)){
			result +=to_string(count)+data.at(index);
			count=1;
		}
		else count++;
	}
	result +=to_string(count)+data.at(data.length()-1);

	return result;
}

- Muhammed Ashiq January 25, 2017 | Flag Reply
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0
of 0 vote

My Solution In C#

string S = "aaaggbbbbcccf";
int Cnt = 1;
StringBuilder str = new StringBuilder();

for (int i=0;i<S.Length-1;i++)
{ 
     if (S[i] != S[i+1])
      {
        str.Append(Cnt);
		str.Append(S[i]);
		   Cnt=1;
      }
     else
      {
        Cnt = Cnt+ 1;
      }
	  
	  if (i == S.Length -2)
	  {
	     if (S[i] != S[i+1])
		 {
		    str.Append(Cnt);
		    str.Append(S[i+1]);
		 }
		 else
		 {
		    str.Append(Cnt);
		    str.Append(S[i]);
			}
	  }
}

Console.WriteLine("Output: " + str);

- PremDiva January 25, 2017 | Flag Reply
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0
of 0 vote

public static void calculateAndReplaceRepeatedChars()
        {
            string text = "aaaggbbbbc";
            char[] chars = text.ToCharArray();
            short counter = 0;
            string current = "a";
            string output = "";

            for (int i = 0; i < chars.Length; i++)
            {
                if (current == chars[i].ToString())
                {
                    counter++;
                }
                else
                {
                    output += (counter.ToString() + current);

                    current = chars[i].ToString();
                    counter = 1;
                }     
            }

            output += (counter.ToString() + current);

            Console.WriteLine(output);

}

- Jean Brand January 25, 2017 | Flag Reply
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0
of 0 vote

public static void calculateAndReplaceRepeatedChars()
        {
            string text = "aaaggbbbbc";
            char[] chars = text.ToCharArray();
            short counter = 0;
            string current = "a";
            string output = "";

            for (int i = 0; i < chars.Length; i++)
            {
                if (current == chars[i].ToString())
                {
                    counter++;
                }
                else
                {
                    output += (counter.ToString() + current);

                    current = chars[i].ToString();
                    counter = 1;
                }     
            }

            output += (counter.ToString() + current);

            Console.WriteLine(output);
        }

- Jean Brand January 25, 2017 | Flag Reply
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0
of 0 vote

from itertools import groupby
#Solution in Python
def CountOccurences(mystring):
    for val,occur in groupby(mystring):
        char,num=val,len(list(occur))
        mystring=mystring.replace(char*(num),str((num))+char,1)
    return mystring
if __name__ == "__main__":
    print CountOccurences("aaaggbbbbcaaaa") #Outputs 3a2g4b1c4a

- ankitpatel2100 January 29, 2017 | Flag Reply
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0
of 0 vote

def CountOccurences(mystring):
    mynewstring=""
    prev=mystring[0]
    count=0
    for char in mystring:
        if char != prev :
            mynewstring += str(count)+prev
            count=0
        count+=1
        prev=char
    mynewstring += str(count)+prev
    return mynewstring
if __name__ == "__main__":
    print CountOccurences("aaaggbbbbcaaaa") #Outputs 3a2g4b1c4a

- ankitpatel2100 January 30, 2017 | Flag Reply
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0
of 0 vote

public static void main(String args[])
	{
		StringBuffer str = new StringBuffer("aaabbbgggcc");
		
		int curStartPos = 0;
		int curEndPos = 0;
		int count = 1;
		
		for(int i=0; i<str.length()-1; i++)
		{
			if(str.charAt(i) == str.charAt(i+1))
			{
				count++;
				curEndPos++;
			}
			else
			{
				String replaceMentStr = Integer.toString(count)+str.charAt(i);
				str.replace(curStartPos, curEndPos+1, replaceMentStr);
				curStartPos+= 2;
				curEndPos = curStartPos;
				i = curStartPos-1;
				count = 1;
			}
		}
		//Update for Last character
		if(count != 0)
		{
			String replaceMentStr = Integer.toString(count)+str.charAt(str.length()-1);
			str.replace(curStartPos, curEndPos+1, replaceMentStr);
		}
		
		System.out.println(str.toString());
	}

Time Complexity = O(n)
Space Complexity = O(1)

- Naveen January 31, 2017 | Flag Reply
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0
of 0 vote

#include <iostream>
#include <string>
#include <map>
#include <sstream>
using namespace std;

int main(int argc, char* argv[]) {

	if(argc != 2 ) {
		cerr<<"Error: Usage: "<<endl;
		return -1;
	}


	string input = string(argv[1]);
	map<char, int> char_counting;
	int times = 1;
	for(int i = 0; i < input.length(); i++) {
		if(char_counting.find(input[i]) !=  char_counting.end() ) {
			times = char_counting[input[i]];
			char_counting[input[i]] = ++times;
		}
		char_counting[input[i]] = times;
	}
	string output;
	for(map<char, int>::iterator itr = char_counting.begin(); itr != char_counting.end(); itr++)
	{
		 stringstream ss;

		 ss << itr->second;
		 ss << itr->first;	
		 output += ss.str();	
	}	
	cout<<output<<endl;
}

- Anonymous March 08, 2017 | Flag Reply
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0
of 0 vote

#include <iostream>
#include <string>
#include <map>
#include <sstream>
using namespace std;

int main(int argc, char* argv[]) {

	if(argc != 2 ) {
		cerr<<"Error: Usage: "<<endl;
		return -1;
	}


	string input = string(argv[1]);
	map<char, int> char_counting;
	int times = 1;
	for(int i = 0; i < input.length(); i++) {
		if(char_counting.find(input[i]) !=  char_counting.end() ) {
			times = char_counting[input[i]];
			char_counting[input[i]] = ++times;
		}
		char_counting[input[i]] = times;
	}
	string output;
	for(map<char, int>::iterator itr = char_counting.begin(); itr != char_counting.end(); itr++)
	{
		 stringstream ss;

		 ss << itr->second;
		 ss << itr->first;	
		 output += ss.str();	
	}	
	cout<<output<<endl;
}

- C++ geek March 08, 2017 | Flag Reply
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0
of 0 vote

Here you go

public string compressString(string input)
{
	if(input == null)
	{
		throw new ArgumentException();
	}
	var output = new StringBuilder();
	char currentchar;
	int occurance = 0;
	for(int i = 0;i<input.Length;i++)
	{
		occurance = 1;
		if(input[i] < 48 && input[i] > 57)
		{
			currentchar = input[i];
			for(int j = i; j < input.Length;j++)
			{
				if(input[j] == currentchar)
				{
					occurance++;
				}
				else
				{
					break;
				}
			}
			output.Append(occurance);
			output.Append(currentchar);
			i = j-1;
		}
		output.Append(input[i]);
	}
	return output.ToString();
}

The complexity of this algorithm is O(N) + O(N).
Still looking for O(N) inplace for this problem, general input are all having worst case O(N) space complexity

- sonesh March 31, 2017 | Flag Reply
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0
of 0 vote

public class Exm{
public static void main(String[]args)
{
String data="aaaabbggggcccccyyy";
String now="";
String charr="";
int c=0;
for(int k=0;k<data.length();k++)
{
if(k==0)
{
c++;
charr=data.charAt(k)+"";
now=data.charAt(k)+"";
}else
{
if(charr.equals(data.charAt(k)+""))
{
c++;

}else{
now+=c+"";
now+=data.charAt(k);
charr=data.charAt(k)+"";
c=1;
}
}

}
now+=c;
System.out.print(now);
}
}

- Sangee April 26, 2017 | Flag Reply
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0
of 0 vote

public String charToFrequency(String input) {
int size = input.length();
StringBuilder sb = new StringBuilder();
for(int i=0;i<size;i++){
char current = input.charAt(i);
sb.append(Math.abs(i - (input.lastIndexOf(current)+1))).append(current);
i = input.lastIndexOf(current);
}
return sb.toString();
}

- programmer April 28, 2017 | Flag Reply
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0
of 0 vote

public String charToFrequency(String input) {
                int size = input.length();
		StringBuilder sb = new StringBuilder();
		for(int i=0;i<size;i++){
			char current = input.charAt(i);
			sb.append(Math.abs(i - (input.lastIndexOf(current)+1))).append(current);
			i = input.lastIndexOf(current);
		}
		return sb.toString();

- programmer April 28, 2017 | Flag Reply
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-1
of 1 vote

public class ParentClassStatic {

	public static String encodeRepeatedChars(String source) {
	    StringBuilder sb = new StringBuilder();
	    Map<Character, Integer> map = new HashMap<Character,Integer>();
	    for(int i= 0; i<source.length(); i++){
	    	map.put(source.charAt(i), map.containsKey(source.charAt(i))? map.get(source.charAt(i))+1 : 1 );
	    }
		
	    Set<Entry<Character, Integer>> set = map.entrySet();
		for(Entry<Character,Integer> e : set){
			sb.append(e.getValue()+""+e.getKey());
		}
		
	      return sb.toString();
	  }

	  public static void main(String[] args) {
	      String example = "aaaggbbbbc";
	      System.out.println(encodeRepeatedChars(example));
	  }
}

- surendrapandey3788 January 25, 2017 | Flag Reply


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