Bloomberg LP Interview Question for Developer Program Engineers


Country: United States
Interview Type: Phone Interview




Comment hidden because of low score. Click to expand.
13
of 13 vote

public static void moveZeroes(int[] a){
  	int pos = 0;
  	for(int i = 0; i < a.length; i++)
  		if(a[i] != 0)
  			a[pos++] = a[i];
  	for(int i = pos; i < a.length; i++)
  		a[i] = 0;
  }

- Roman August 13, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
1
of 1 vote

Iterate through the array, keeping track of how many zeroes you've seen. For each value, set arr[i-count] to the current value a[i]. If i + count is equal to the length, set a[i] to 0 (i.e. If you're within "count" of the end, start filling in zeroes.

Time: O(n)
Space: O(1)

Something like this:

public void MoveZeroes(int[] arr) {
   int zeroCount = 0;
   for (int i=0; i<arr.Length; i++) {
      if (arr[i] == 0)
         zeroCount++;
      else
         a[i - zeroCount] = a[i];
      if (arr.Length < i + zeroCount)
         arr[i] = 0;
    }
}

- Andy August 12, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
1
of 1 vote

moveZeroes(std::vector<int>& v)
    {
         // remove zero and get the iterator past the last nonzero element
         auto it = std::remove(v.begin(), v.end(), 0);
         std::fill(it, v.end(), 0);
    }

- jpan215 August 16, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static void placeZeroAtEnd1(int arr[])
	{
		int i = 0;
		while (true)
		{
			while (i < arr.length && arr[i] != 0)
			{
				i++;
			}
			int j = i + 1;
			while (j < arr.length && arr[j] == 0)
			{
				j++;
			}
			if (i >= arr.length || j >= arr.length)
			{
				break;
			}
			if (j > i)
			{
				int temp = arr[j];
				arr[j] = arr[i];
				arr[i] = temp;
				i++;
				j++;

			}

		}
		System.out.println(Arrays.toString(arr));
	}

- koustav.adorable August 12, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

void ReorganizeArray(int arr[], int n)
{
	int write_indx = 0, read_indx = 0;
	while (read_indx < n)
	{
		if (arr[read_indx] != 0)
		{
			if (write_indx != read_indx)
				arr[write_indx++] = arr[read_indx];
			else
				write_indx++;
		}
		read_indx++;
	}
	while (write_indx < n)
		arr[write_indx++] = 0;
}

- LANorth August 13, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static void moveZeroes(int[] a){
  	int pos = 0;
  	for(int i = 0; i < a.length; i++)
  		if(a[i] != 0)
  			a[pos++] = a[i];
  	for(int i = pos; i < a.length; i++)
  		a[i] = 0;
  }

- Anonymous August 13, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static void moveZeroes(int[] arr){
for (int i =0 ;i <arr.length;i++)
{
if (arr[i]==0 ){
for (int j=i+1;j<arr.length;j++){
if ( arr[j]!=0 ) {
arr[i]=arr[j];
arr[j]=0;
break;
}
}

}

}

- Anonymous August 13, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I would suggest two approaches for linear time.
1. Use first loop to places all non-zeros at position counter <index> and keep count for zeros also. In second loop, iterate over only zeros to places zeros at the end. The time complexity will be O(n) still in this case.
2. Do only in one loop -

void placeZeros(int a[],int n) {
	int j = 0;
	int i = 0;

	while(j<n) {
		if(a[i] != 0)	i++, j++;
		else  {
			if(a[j] == 0)
				j++;
			else {
				swap(a[i++],a[j++]);
			} 
		}
	}
}

- amit August 13, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I hope, my interview question is this easy..
Python 3

def moveZerosToTheRight(arglist):
    
    for i in range(len(arglist)):
        
        if arglist[i] == 0:
            del arglist[i]
            arglist.append(0)
            
    return arglist

- nil12285 August 14, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Straightforward. Iterate over the array keeping track of all numbers except zeros. This will overwrite the zeroes in between with the non-zeroes. Fill the rest with Zeroes.

public class Test
{
 public static void main(String ...args)
 {
  int arr[] = {1,3,0,8,1,2,0,4,0,7};

  moveZeroes(arr);
  for(int i =0;i<arr.length;i++)
  {
   System.out.print(arr[i]);
  }


 }




 private static void moveZeroes(int[] arr)
 {
  int pos=0;

  for(int i =0;i<arr.length;i++)
  {
   if(arr[i]!=0)
   {
    arr[pos++] = arr[i];
   }
  }

  for(int i=pos;i<arr.length;i++)
  {
   arr[i]=0;

  }
 }

- labscst August 14, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

First question is: Do i have to make the transformation in place or can i use extra space and other data structures?

- Roberto B Beltran August 15, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def move_zeros(input_list) :
    place_holder = len(list) - 1
    i = 0
    while i <= place_holder :
        if list[i] == 0 :
            list.append(0)
            del list[i]
            place_holder -= 1
        else :
            i += 1
    return list

#test
list = [0,1,0,0,20,0,0,4,5]
print(move_zeros(list))

- Luke Rhinehart August 15, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

func moveZeros() {
arr := []int{1,3,0,8,1,2,0,0,4,0,7}
fmt.Println(arr)
index := -1
for i := 0; i < len(arr); i++ {
if arr[i] == 0 {
if index == -1 {
index = i
}
} else if index > -1 {
arr[index] = arr[i]
arr[i] = 0
index++
}
}
fmt.Println(arr)
}

- Anonymous August 15, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

func moveZeros() {
arr := []int{1,3,0,8,1,2,0,0,4,0,7}
fmt.Println(arr)
index := -1
for i := 0; i < len(arr); i++ {
if arr[i] == 0 {
if index == -1 {
index = i
}
} else if index > -1 {
arr[index] = arr[i]
arr[i] = 0
index++
}
}
fmt.Println(arr)
}

- Vivek August 15, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

func moveZeros() {
	arr := []int{1,3,0,8,1,2,0,0,4,0,7}
	fmt.Println(arr)
	index := -1
	for i := 0; i < len(arr); i++ {
		if arr[i] == 0 {
			if index == -1 {
				index = i
			}
		} else if index > -1 {
			arr[index] = arr[i]
			arr[i] = 0
			index++
		}
	}
	fmt.Println(arr)
}

- Vivek August 15, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Just a little improvement:

public static void moveZeroes(int[] a){
  	int pos = 0;
  	for(int i = 0; i < a.length; i++)
  		if(a[i] != 0)
		{
  			a[pos] = a[i];
			if(pos != i)
				a[i] = 0;
			pos++; 
		}
  	
  }

- Roberto B Beltran August 15, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

import java.util.Scanner;


public class goog_1 {

/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc=new Scanner(System.in);
System.out.println("enter the number of elements :");
int n=sc.nextInt();
int[] a=new int[n];
int c=0;
for(int i=0;i<n;i++)
a[i]=sc.nextInt();
for(int i=0;i<n;i++)
{
if(a[i]!=0)
{
System.out.print(a[i]+" ");
c++;
}
}
while(c!=n)
{
System.out.print("0 ");
c++;
}
}

}

- Sindhujadurairaj August 15, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

import java.util.Scanner;
public class goog_1 {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc=new Scanner(System.in);
System.out.println("enter the number of elements :");
int n=sc.nextInt();
int[] a=new int[n];
int c=0;
for(int i=0;i<n;i++)
a[i]=sc.nextInt();
for(int i=0;i<n;i++)
{
if(a[i]!=0)
{
System.out.print(a[i]+" ");
c++;
}}
while(c!=n)
{
System.out.print("0 ");
c++;}
}}

- Sindhujadurairaj August 15, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

moveZeroes(std::vector<int>& v)
{
auto it = std::remove(v.begin(), v.end(), 0);
int nzeroes = std::distance(it, v.end());
std::fill_n(it, nzeroes, 0);
}

- Anonymous August 15, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

for (counter = 0; counter < a.length; counter++) {
	if (a[counter] === 0) { a.splice(counter,1); a.push(0)}
}

- David Scheiner August 18, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

for (counter = 0; counter < a.length; counter++) {
	if (a[counter] === 0) { a.splice(counter,1); a.push(0)}
}

- David Scheiner August 18, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I think this can be achieved by one loop
public void MoveZeroes(int[] arr) {
int pos = 0;
int zeroCOunt=0;
int a[] = new int[arr.length];
for (int i=0; i<arr.length; i++) {
if (arr[i] != 0){
a[pos]=arr[i];
pos++;
} else{
a[arr.length-1-zeroCOunt] = 0;
zeroCOunt++;
}
}
System.out.println(Arrays.toString(a));
}

- Amit August 19, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

This can be achieved within one loop
public void MoveZeroes(int[] arr) {
int pos = 0;
int zeroCOunt=0;
int a[] = new int[arr.length];
for (int i=0; i<arr.length; i++) {
if (arr[i] != 0){
a[pos]=arr[i];
pos++;
} else{
a[arr.length-1-zeroCOunt] = 0;
zeroCOunt++;
}

}

System.out.println(Arrays.toString(a));
}

- amitamit281 August 19, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Work backwards (Python):

arr = [1,3,0,8,12,0,4,0,7]

i = len(arr) - 1

while i >= 0:
    if arr[i] == 0:
        del arr[i]
        arr.append(0)
    i -= 1

- FadyMak August 19, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Work backwards instead (using Python):

arr = [1,3,0,8,12,0,4,0,7]

i = len(arr) - 1

while i >= 0:
    if arr[i] == 0:
        del arr[i]
        arr.append(0)
    i -= 1

- Fady Makram August 19, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static void moveZeroLast() {
int a[] = { 1, 7, 0, 4, 5, 0, 7, 7, 0, 8 };
int count = 0;
for (int i = 0; i < a.length; i++) {
if (a[i] == 0) {
count++;
} else {
a[i - count] = a[i];
}
if (i + count >= a.length - 1) {
a[i] = 0;
}

}
}

- rathor.rajeev August 20, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Using Python:

def place_zeros(n,ls):
    x=0
    for i in range(n):
        if ls[i]==0:
            x=ls[i]
            ls.remove(x)
            ls.append(x)
            return ls

- Jon Snow August 23, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def place_zeros(n,ls):
    x=0
    for i in range(n):
        if ls[i]==0:
            x=ls[i]
            ls.remove(x)
            ls.append(x)
            return ls

- J August 23, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class AllZerosAtEnd {

static int[] a = { 1, 3, 0, 0,0,0,8, 12, 0, 4, 0, 7 };

public static void main(String[] args) {

System.out.print("Before :");
for(int k=0;k<a.length;k++){
System.out.print(a[k]+", ");
}
System.out.println();
moveZeroToEnd(a);
}

static void moveZeroToEnd(int[] a) {

int zeroStartIndex = 0, nonZeroIndex = 0;
while (a[zeroStartIndex] != 0)
zeroStartIndex++;

// find next non zero start Index
nonZeroIndex = zeroStartIndex + 1;
int zeroCountLength = 1;
while (a[nonZeroIndex] == 0) {
nonZeroIndex++;
zeroCountLength++;
}

for (int i = 0; i < zeroCountLength && nonZeroIndex < a.length; i++) {
if(a[nonZeroIndex]!=0){
a[zeroStartIndex++] = a[nonZeroIndex];
a[nonZeroIndex++] = 0;
i--;
} else{
zeroCountLength++;
nonZeroIndex++;
}
}

System.out.print("After :");
for(int k=0;k<a.length;k++){
System.out.print(a[k]+", ");
}
}
}

- sanjeet August 26, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class AllZerosAtEnd {

	static int[] a = { 1, 3, 0, 0,0,0,8, 12, 0, 4, 0, 7 };

	public static void main(String[] args) {

		System.out.print("Before :");
		for(int k=0;k<a.length;k++){
			System.out.print(a[k]+",  ");
		}
		System.out.println();
		moveZeroToEnd(a);
	}

	static void moveZeroToEnd(int[] a) {

		int zeroStartIndex = 0, nonZeroIndex = 0;
		while (a[zeroStartIndex] != 0)
			zeroStartIndex++;

		nonZeroIndex = zeroStartIndex + 1;
		int zeroCountLength = 1;
		while (a[nonZeroIndex] == 0) {
			nonZeroIndex++;
			zeroCountLength++;
		}

		for (int i = 0; i < zeroCountLength && nonZeroIndex < a.length; i++) {
			if(a[nonZeroIndex]!=0){
				a[zeroStartIndex++] = a[nonZeroIndex];
				a[nonZeroIndex++] = 0;
				i--;
			} else{
				zeroCountLength++;
				nonZeroIndex++;
			}
		}
		
		System.out.print("After  :");
		for(int k=0;k<a.length;k++){
			System.out.print(a[k]+",  ");
		}
	}

}

- Anonymous August 26, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class ArrayTest {

public static void main(String[] args) throws Exception{
int[] input = {1,3,0,8,12,0,4,0,7};

for(int i = 0; i < input.length; i++) {
if (input[i] == 0){
for (int j = i; j < input.length -1; j++){
input[j] = input[j+1];
}
input[input.length - 1] = 0;
}
System.out.println(input[i]);
}
}
}

- satishvarmavvu August 29, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Python solution:

def shift_zeroes(l):
shift = 0
for i in range(0, len(l)):
if l[i] == 0:
shift += 1
else:
temp = l[i-shift]
l[i-shift] = l[i]
l[i] = temp

return l

- Matthew September 15, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Python solution:

def shift_zeroes(l):
    shift = 0
    for i in range(0, len(l)):
        if l[i] == 0:
            shift += 1
        else:
            temp = l[i-shift]
            l[i-shift] = l[i]
            l[i] = temp

    return l

- Matthew September 15, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Python solution:

def shift_zeroes(l):
    shift = 0
    for i in range(0, len(l)):
        if l[i] == 0:
            shift += 1
        else:
            temp = l[i-shift]
            l[i-shift] = l[i]
            l[i] = temp

    return l

- Matthew September 15, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def appendZeros(listIntegers):
    numZeros = listIntegers.count(0)
    listIntegers = filter(lambda a: a != 0, listIntegers)
    for i in range(0, numZeros):
        listIntegers.append(0)
    return listIntegers

int_list = [4,6,2,0,5,0,6,8,2]

print (appendZeros(int_list))

- sona_python_solution September 20, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def appendZeros(listIntegers):
numZeros = listIntegers.count(0)
listIntegers = filter(lambda a: a != 0, listIntegers)
for i in range(0, numZeros):
listIntegers.append(0)
return listIntegers

int_list = [4,6,2,0,5,0,6,8,2]

print (appendZeros(int_list))

- sona_python_solution September 20, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def appendZeros(listIntegers):
        numZeros = listIntegers.count(0)
        listIntegers = filter(lambda a: a != 0, listIntegers)
        for i in range(0, numZeros):
            listIntegers.append(0)
        return listIntegers

int_list = [4,6,2,0,5,0,6,8,2]

print (appendZeros(int_list))

- Anonymous September 20, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def appendZeros(listIntegers):
    numZeros = listIntegers.count(0)
    listIntegers = filter(lambda a: a != 0, listIntegers)
    for i in range(0, numZeros):
        listIntegers.append(0)
    return listIntegers

int_list = [4,6,2,0,5,0,6,8,2]

print (appendZeros(int_list))

- sg_coffee September 20, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include <iostream>
#include <vector>
using namespace std;
void move_zero(int * ,int);
int main()
{
int test_m[]={1,3,0,8,12,0,4,0,7};
const int arr_len=sizeof(test_m)/sizeof(int);
move_zero(test_m,arr_len);

for(int i=0;i< arr_len;i++)
cout << test_m[i] << endl;
}
void move_zero(int test_m[] ,const int arr_len)
{
int count=0;
for(int i=0;i< arr_len;i++)
{ if(test_m[i] != 0)
{
test_m[count++]=test_m[i];
}

}
for( ;count < arr_len ;count++)
test_m[count]=0;


}

- agrawaankit60 September 22, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

a = [1,3,0,8,0,0,12,50,9,90,5,0,0,0,4,0,7]
i=0
j=len(a)-1
while i <= j:
    if a[i]==0:
        a.append(a.pop(i))
        j-=1
    else:
        i+=1

print a

- Kshitij Srivastava October 20, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

a = [1,3,0,8,0,0,12,50,9,90,5,0,0,0,4,0,7]
i=0
j=len(a)-1
while i <= j:
    if a[i]==0:
        a.append(a.pop(i))
        j-=1
    else:
        i+=1

print a

- Kshitij Srivastava October 20, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

void moveZeros(vector<int>& v)
{
   auto ite = remove(v.begin(), v.end(), 0);
   fill(ite, v.end(), 0);

    return 0;
}

- Anonymous November 03, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

//
// Given an unsorted integer array, place all zeros to the
// end of the array without changing the sequence of non-zero
// elements. ( ie. [1,3,0,8,12,0,4,0,7] => [1,3,8,12,4,7,0,0,0] )
//
// Run with VM arguments -ea to enable assert testing
//
// (c) 2016 Perry Anderson, All Rights Reserved, worldwide.
//
//

import java.util.Vector;
import java.util.Arrays;

public class LPMain002 {

	static Integer[] sortOut(Integer[] data) {
		
		Vector<Integer> results = new Vector<Integer>();
		int zeroCount = 0;
		
		for (int i = 0; i < data.length; i++) {
			if ( data[i]==0)
				zeroCount++;
			else
				results.add(data[i]);
		}
		
		while (zeroCount-->0) 
			results.add(new Integer(0));
		
		return results.toArray(new Integer[results.size()]);
	}

	public static void main(String[] args) {

		Integer[] data1 = new Integer[] { 1, 3, 0, 8, 12, 0, 4, 0, 7 };
		Integer[] data2 = new Integer[] { 1, 3, 8, 12, 4, 7, 0, 0, 0 };
	
		assert Arrays.equals(sortOut(data1), data2);
		assert !Arrays.equals(sortOut(data1), data1);
		assert Arrays.equals(sortOut(data2), data2);
		assert !Arrays.equals(sortOut(data2), data1);
		
	}

}

- perry.anderson November 21, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

int n = 9, A[] = {1, 3, 0, 8, 12, 0, 4, 0, 7};
  int i = 0, j = n - 1;
  while(i < j) {
    while(i < j && A[i]) ++i;
    while(j > i && !A[j]) --j;
    swap(A[i], A[j]);
  }
  for(int a: A) 
    cout << a << " ";

- Mahmoud Sayed January 23, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static int[] MoveZeros(int[] arr)
        {
            int cnt = 0;
            for (int i = 0; i < arr.Length; i++)
            {
                if(arr[i] != 0)
                {
                    arr[i - cnt] = arr[i];
                }
                else
                {
                    cnt++;
                }
            }


            for (int i = 1; i <= cnt; i++)
            {
                arr[arr.Length - i] = 0;
            }
            return arr;
        }

- vh.vahan February 15, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

Implementation done in Go that runs in linear time. Leverages the fact that slices are mutable in the Go language. A similar implementation can also be done in Ruby.

However, this approach will not work in Java or C where arrays are immutable.

func moveZeros(){
	arr := []int{1,3,0,8,1,2,0,4,0,7}
	fmt.Println(arr)
	for i := 0; i < len(arr); i++{
		if(arr[i] == 0){
			arr := append(arr[:i], arr[i+1:]...)
			arr = append(arr, 0)
		}
	}
	fmt.Println(arr)
}

- JH August 12, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

func moveZeros(){
	arr := []int{1,3,0,8,1,2,0,4,0,7}
	fmt.Println(arr)
	for i := 0; i < len(arr); i++{
		if(arr[i] == 0){
			arr := append(arr[:i], arr[i+1:]...)
			arr = append(arr, 0)
		}
	}
	fmt.Println(arr)
}

- JH August 12, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

Iterate through the array, keeping track of how many zeroes you've seen. For non-zero values, set a[i-count]=a[i]. As you approach the end, you'll want to see if i + numZeroes >= length. If it is, set a[i] to 0.

Time: O(n)
Space: O(1)

- Andy August 12, 2016 | 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