Bloomberg LP Interview Question for Software Engineers


Country: United States
Interview Type: Phone Interview




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1
of 1 vote

// ZoomBA

d_1 = rfold(  str1.toCharArray , dict()  ) as {   [ $.o.key  ,  $.o.i ] }
     found_index = lfold( str2.toCharArray , -1 ) as {  break(  $.o.key @  d_1 ){  $.o.value  }     }

- NoOne September 13, 2016 | Flag Reply
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0
of 0 vote

var Str1 = "ad7f6ysh";
var Str2 = "1234567";

var Str2Arr = Str2.split("");
var minIndex = -1;
var Elem = "";
for(var i=0; i<Str2Arr.length;i++){
currentElemIndex = Str1.indexOf(Str2Arr[i]);
if( currentElemIndex != -1){
if(minIndex == -1){
minIndex = currentElemIndex;
Elem = Str2Arr[i];
}else if(minIndex > currentElemIndex){
minIndex = currentElemIndex;
Elem = Str2Arr[i];
}
}

}
if(minIndex != -1){
console.log(Elem+ " Element is found at index "+ minIndex);
}

- V!R September 13, 2016 | Flag Reply
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0
of 0 vote

One optimization I could think of was to put the shorter string in the HashMap. This could be helpful in the 1st and last cases:

public class findIndexOfChar {
    public static void main(String[] args) {
        System.out.println(findIndexOfChar("adf6ysh","12345678762364798723577857398476089023857435826"));
        System.out.println(findIndexOfChar("adfysh","12345678"));
        System.out.println(findIndexOfChar("adf6as6h","12345678"));
        System.out.println(findIndexOfChar("1234567876236479872357826798347534509809384509384509","adfh6ysh"));
    }

    private static int findIndexOfChar(String str1, String str2) {
        HashMap<Character,Integer> hm = new HashMap<Character, Integer>();
        String longer;
        String shorter;
        boolean str1IsLonger;
        if(str1.length()>str2.length()){
            longer = str1;
            shorter = str2;
            str1IsLonger = true;
        }
        else{longer = str2;
            shorter = str1;
            str1IsLonger = false;
        }

        int index = 0;
        for (Character ch : shorter.toCharArray()) {
            if (!hm.containsKey(ch)) {
                hm.put(ch,index);
            }
            index++;
        }
        for (int i = 0; i < longer.length(); i++) {
            if (hm.containsKey(longer.charAt(i))) {
                if(str1IsLonger) {return i;}
                else {return hm.get(longer.charAt(i));}
            }
        }
        return -1;
    }
}

- ctrlV September 17, 2016 | Flag Reply
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0
of 0 vote

I think this code in C is good for this case assuming ASCII chars - O(n) runtime, O(1) space.

#include <stdlib.h>
#include <stdio.h>
#include <string.h>

// assuming ASCII chars
#define NUM_CHARS 256
// define boolean
typedef enum { false, true } bool;

// find the first occurence of any char from s2 in s1
// in no such occurence found return -1
int firstOccurence(char * s1, char * s2) {
	int index = -1;
	unsigned int i;
	// array to indicate which chars are in s2
	bool charMap[NUM_CHARS];

	// init values with -1
	for (i = 0; i < NUM_CHARS; i++) {
		charMap[i] = false;
	}

	// fill in array where chars show in s2
	for (i = 0; i < strlen(s2); i++) {
		int cVal = s2[i];
		if (charMap[cVal] == false) {
			charMap[cVal] = true;
		}
	}

	// find first char in s1 that appered in s2
	for (i = 0; i < strlen(s1); i++) {
		int cVal = s1[i];
		if (charMap[cVal] == true) {
			index = i;
			break;
		}
	}
	// if no occurence found, index will be -1
	// else index will be the first occurence of any char from s2 in s1
	return index;
}

int main() {
	// strings
	char  s1[] = "abdgilcnhjfcs";
	char  s2[] = "oooooci";

	// expect 4 for this example... as i from 
	// s2 appears first in s1 at index 4 (zero based).
	int first = firstOccurence(s1, s2);
	printf("First occurence is %d\n", first);
}

- ronenmiller1 November 01, 2016 | Flag Reply
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0
of 0 vote

public int findFirstOccurence(String str1, String str2) {
int[] table = new int[128];
char[] charArrayStr1 = str1.toCharArray();
for(char character : charArrayStr1) {
table[character]++;
}
char[] charArrayStr2 = str2.toCharArray();
for(int i = 0; i < charArrayStr2.length; i++) {
if(table[charArrayStr2[i]] > 0) return i;
}
return -1;
}

- Decent Solution November 04, 2016 | Flag Reply
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0
of 0 vote

public class LPMain001 {

	static int indexOf(String str1, String str2 ) {
		for (int i=0; i<str2.length(); i++ ) {
			char c = str2.charAt(i);
			if ( str1.indexOf(c)>-1 ) {
				return str1.indexOf(c);
			}
		}
		return -1;
	}
	
	public static void main(String[] args) {
		
		String str1 = "adf6ysh";
		String str2 = "123678";
		String str3 = "12378";
		
		assert indexOf(str1,str2) == 3;
		assert indexOf(str1,str3) == -2;
		
	}
	
}

- Anonymous November 21, 2016 | Flag Reply
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0
of 0 vote

public class LPMain001 {

	static int indexOf(String str1, String str2 ) {
		for (int i=0; i<str2.length(); i++ ) {
			char c = str2.charAt(i);
			if ( str1.indexOf(c)>-1 ) {
				return str1.indexOf(c);
			}
		}
		return -1;
	}
	
	public static void main(String[] args) {
		
		String str1 = "adf6ysh";
		String str2 = "123678";
		String str3 = "12378";
		
		assert indexOf(str1,str2) == 3;
		assert indexOf(str1,str3) == -2;
		
	}
	
}

- Perry November 21, 2016 | Flag Reply
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0
of 0 vote

//
// Returns the zero based index of the first occurance of 
// any character of str2 in str1
//
// (c) 2016 Perry Anderson
//  http://perryanderson.com/
//

public class LPMain001 {

	static int indexOf(String str1, String str2 ) {
		for (int i=0; i<str2.length(); i++ ) {
			char c = str2.charAt(i);
			if ( str1.indexOf(c)>-1 ) {
				return str1.indexOf(c);
			}
		}
		return -1;
	}
	
	public static void main(String[] args) {
		
		String str1 = "adf6ysh";
		String str2 = "123678";
		String str3 = "12378";
		
		assert indexOf(str1,str2) == 3;
		assert indexOf(str1,str3) == -1;
		
	}
	
}

- perry.anderson November 21, 2016 | Flag Reply
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0
of 0 vote

public class Driver {

public static void main(String[] args) {
StringMatching sm = new StringMatching();
String str1 = "adfysh26";
String str2 = "123678";
System.out.println(sm.match(str1, str2));
}
}

class StringMatching {

public int match(String str1, String str2) {

for (int i = 0; i < str1.length(); i++) {
char xchar = str1.charAt(i);
if (Character.isDigit(xchar)) {
int idx = str2.indexOf(xchar);

if (idx >= 0) {
return idx;
}
}
}

return -1;
}
}

- Parkash Arjan January 11, 2017 | Flag Reply
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0
of 0 vote

public class Driver {

	public static void main(String[] args) {
		StringMatching sm = new StringMatching();
		String str1 = "adfysh26";
		String str2 = "123678";
		System.out.println(sm.match(str1, str2));
	}
}

class StringMatching {

	public int match(String str1, String str2) {

		for (int i = 0; i < str1.length(); i++) {
			char xchar = str1.charAt(i);
			if (Character.isDigit(xchar)) {
				int idx = str2.indexOf(xchar);

				if (idx >= 0) {
					return idx;
				}
			}
		}

		return -1;
	}
}

- JJ January 11, 2017 | Flag Reply
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0
of 0 vote

import java.util.regex.Matcher;
import java.util.regex.Pattern;
 



 class RegExpr8 {

	public static void main(String[] args) {
		String str1 = "adf6ysh7";
		String str2 = "123678";

		Pattern p = Pattern.compile("[\\d]");
		Matcher m1 = p.matcher(str1);

		boolean isFound = false;
		while (m1.find()) {

			String strTemp1 = m1.group();
			System.out.println("Finding " + "  ---  " + strTemp1);

			Matcher m2 = p.matcher(str2);
			while (m2.find()) {
				String strTemp2 = m2.group();
				if (strTemp1.equals(strTemp2)) {
					System.out.println("Found " + " --- " + strTemp2 + "  at  " + m2.start());
					isFound = true;
					break;
				}

			}
			if (isFound)
				break;
		}
	}
}

- Parkash Arjan January 11, 2017 | Flag Reply
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0
of 0 vote

public static void main(String args[]) {
String s1 = "adf6ysh";
String s2 = "123678";

int length;
if (s1.length() > s2.length()) {
length = s1.length();
}
else {
length = s2.length();
}

for (int i = 0; i < length; i++) {
char c1 = s1.charAt(i);
char c2 = s2.charAt(i);

if (c1 == c2) {
System.out.println(c1);
break;
}
}
}

- Anonymous March 29, 2017 | Flag Reply
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0
of 0 vote

Check this out

public static int FindFirstOccurrence(string str1, string str2)
{
	// error condition checks
	var charcount = Dictionary<char,int>();
	for(int i = 0 ; i < str2.Length;i++)
	{
		if(!charcount.Contains(str2[i]))
		{
			charcount[str2[i]] = 1;
		}
		else
		{
			charcount[str2[i]]++;
		}
	}
	for(int i = 0 ; i < str1.length;i++)
	{
		if(charcount.Contains(str1[i])
		{
			return i;
		}
	}
	return -1;
}

Complexity of this is O(n) + O(1), as the char count is constant.

- sonesh April 01, 2017 | Flag Reply
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0
of 0 vote

public static int FindFirstOccurance(string str1, string str2)
        {
            int retVal = -1;
            for (int i=0; i<str1.Length - 1; i++ )
            {
                if (str2.IndexOf(str1[i]) > -1)
                    return i;
            }

            return retVal;

}

- Anonymous February 27, 2018 | Flag Reply
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-1
of 1 vote

import java.util.*;

public class FirstOccurrences{
	public static void main(String[] args){
		String str1 = "adf6ysh";
		String str2 = "1234567";
		
		HashSet<Character> s2 = new HashSet<Character>();
		for(int i = 0; i < str2.length(); i++){
			s2.add(str2.charAt(i));
		}
		
		HashMap<Character, Integer> map = new HashMap<Character, Integer>();
		for(int i = 0; i < str1.length(); i++){
			char c = str1.charAt(i);
			if(s2.contains(c)){
				map.put(c, i);
				s2.remove(c);
			}
		}
		
		for(Character c : map.keySet()){
			System.out.print(c + ": " + map.get(c));
		}
	}

}

- Anonymous September 07, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

import java.util.*;

public class FirstOccurences{
	public static void main(String[] args){
		String str1 = "adf6ysh";
		String str2 = "1234567";
		
		HashSet<Character> s2 = new HashSet<Character>();
		for(int i = 0; i < str2.length(); i++){
			s2.add(str2.charAt(i));
		}
		
		HashMap<Character, Integer> map = new HashMap<Character, Integer>();
		for(int i = 0; i < str1.length(); i++){
			char c = str1.charAt(i);
			if(s2.contains(c)){
				map.put(c, i);
				s2.remove(c);
			}
		}
		
		for(Character c : map.keySet()){
			System.out.print(c + ": " + map.get(c));
		}
	}

}

- TC September 07, 2016 | Flag Reply
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-1
of 1 vote

#include <iostream>
using namespace std;

int main()
{
string first,second;
cin>>first>>second;
int index=0;
int i=0,j=first.length()-1;
int p=0,q=second.length()-1;
while(i<j&&p<q)
{
if(first[i]==second[p])
{
index=i;
break;
}
i++;
p++;
}
if(index==0&&first[i]==second[p])
{
index=i;
}
cout<<index;
return 0;
}

- rbihemu September 07, 2016 | Flag Reply
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-1
of 1 vote

public class FirstOccurence {
    public static void main(String[] args){
        String s1 = "adf6ysh";
        String s2 = "1234567";
        boolean found = false;
        
        int[] s2Count = new int[96]; // Assuming alphabet, number and special characters only in s1 and s2 ASCII(127-31)
        
        for(char c:s2.toCharArray()){
            s2Count[(c-' ')]++;
        }
        
        char[] s1Array = s1.toCharArray();
        
        for(int i=0;i<s1Array.length;i++){
            if(s2Count[s1Array[i]-' ']>0) 
            {
                System.out.println("Character at "+i+"+1 Matches");
                found = true;
            }
        }
        if(!found){
            System.out.println("No Match");
        }
    }
}

- kamaldheeraj September 08, 2016 | Flag Reply
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-1
of 1 vote

#include<iostream>
using namespace std;
int main()
{
int t,n,i;
cin>>t;
while(t--){
string s1,s2;
cin>>s1;
cin>>s2;
int a[127]={0};
//for(i=0;i<127;i++)
// a[i]=0;
for(i=0;i<s1.length();i++)
a[s1[i]]++;
for(i=0;i<s2.length();i++)
if(a[s2[i]]!=0){
cout<<i<<endl;
break;
}
}
}

- SAHA HINLEY SHRUNG September 10, 2016 | Flag Reply
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-1
of 1 vote

static int compareTwostrings(char[] str1,char[] str2)
{
int nCommon=0;
Arrays.sort(str1, 0, str1.length);
for(int i=0;i<str1.length;++i)
{
nCommon=Arrays.binarySearch(str1, str2[i]);

if(nCommon>=0)
{
return i;
}

}

System.out.println("Element not found");
return -1;
}

- Anonymous September 26, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

static int compareTwostrings(char[] str1,char[] str2)
{
int nCommon=0;
Arrays.sort(str1, 0, str1.length);
for(int i=0;i<str1.length;++i)
{
nCommon=Arrays.binarySearch(str1, str2[i]);

if(nCommon>=0)
{
return i;
}

}

System.out.println("Element not found");
return -1;
}

- sai September 26, 2016 | Flag Reply


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