Google Interview Question for SDE-3s


Country: United States




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4
of 4 vote

My solution N^3 top down dynamic programming approach but I believe there is an O(n) algorithm that solves it. Anyone have any ideas?

public static long max(int[] nums) {
        
                Long[][] memo = new Long[nums.length][nums.length];
                return max(nums, 0, nums.length - 1, memo);
        
            }
        
            public static long max(int[] nums, int i, int j, Long[][] memo) {
        
                if (memo[i][j] != null) return memo[i][j];
        
                if (i == j) {
                    return nums[i];
                }
        
                long max = Integer.MIN_VALUE;
        
                for (int k = i; k < j; k++) {
        
                    long left = max(nums, i, k, memo);
                    long right = max(nums, k + 1, j, memo);
        
                    for (char c : "*-+".toCharArray()) {
                        long res = apply(c, left, right);
                        max = Math.max(max, res);
                    }
                }
        
                memo[i][j] = max;
        
                return max;
            }

- talal.nabulsi5 August 27, 2017 | Flag Reply
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def find_max_value(numbers)
  one_count = 0
  two_count = 0
  max_val = 1
  numbers.each do |number|
    next if number == 0
    number *= -1 if number < 0 
    if number == 1 
      one_count += 1
    elsif number == 2 
      two_count += 1 
    else
      max_val *= number
    end
  end
  if two_count == 0 && one_count == 1
    min = numbers.sort[1]
    return (max_val/min)*(min+1)  
  end
  get_final_max_value(one_count, two_count, max_val) 
end

def get_final_max_value(one_count, two_count, max_val)
  if one_count == 0 && two_count == 0
    max_val = max_val
  elsif one_count == 0
    max_val *= 2**two_count
  elsif two_count == 0
    max_val *= get_max_val_from_ones_count(one_count)
  else
    if two_count >= one_count
      max_val *= (3**one_count)*(2**(two_count-one_count)) 
    else
      max_val *= (3**two_count)*get_max_val_from_ones_count(one_count-two_count)
    end
  end
  max_val
end

def get_max_val_from_ones_count(one_count)
  quotient, reminder = get_quotient_and_reminder(one_count, 3)
  if reminder == 0
    3**quotient
  elsif reminder == 1
    3**(quotient-1)*4
  else
    3**quotient*2
  end
end

def get_quotient_and_reminder(number, dividend)
  quotient = number / dividend
  reminder = number % dividend
  return quotient, reminder
end

- raunak.gupta29 August 27, 2017 | Flag Reply
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0
of 0 vote

Works for positive and negative numbers.

pair<double, double> Extremes(double v1, double v2)
{
	double min_val = min(v1 + v2, v1 - v2);
	min_val = min(min_val, v1 * v2);
	double max_val = max(v1 + v2, v1 - v2);
	max_val = max(max_val, v1 * v2);
	return pair<double, double>(min_val, max_val);
}

pair<double, double> MinMax(vector<double> const &a, int idx, int size,
	map<pair<int, int>, pair<double, double>> &memo)
{
	if (size == 0 ||
		idx < 0 ||
		idx >= a.size())
	{
		return pair<double, double>(0, 0);
	}
	if (size == 1) {
		return pair<double, double>(a[idx], a[idx]);
	}

	pair<int, int> key = pair<int, int>(idx, size);
	auto it = memo.find(key);
	if (it != memo.end()) {
		return it->second;
	}

	double min_v = numeric_limits<double>::max();
	double max_v = numeric_limits<double>::lowest();

	for (int i = 1; i < size; ++i) {
		pair<double, double> left = MinMax(a, idx, i, memo);
		pair<double, double> right = MinMax(a, idx + i, size - i, memo);

		pair<double, double> p1 = Extremes(left.first, right.first);
		pair<double, double> p2 = Extremes(left.first, right.second);
		pair<double, double> p3 = Extremes(left.second, right.first);
		pair<double, double> p4 = Extremes(left.second, right.second);

		min_v = min(min_v, p1.first);
		min_v = min(min_v, p2.first);
		min_v = min(min_v, p3.first);
		min_v = min(min_v, p4.first);
		max_v = max(max_v, p1.second);
		max_v = max(max_v, p2.second);
		max_v = max(max_v, p3.second);
		max_v = max(max_v, p4.second);
	}

	memo[key] = pair<double, double>(min_v, max_v);
	return memo[key];
}

- Alex August 27, 2017 | Flag Reply
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of 0 vote

List<double> allPossibleNum = new List<double>();
double[] array = { 5, 4, 3, 1 };
Array.Sort(array);
Array.Reverse(array);
for (int i=array.Length-1;i > -1; i--)
{
double maxMultiNum = 1;
for (int j = 0; j <= i; j++)
{
maxMultiNum *= array[j];
}

double MaxAddNum = 0;
for (int j = i+1; j < array.Length; j++)
{
MaxAddNum += array[j];
}
allPossibleNum.Add(maxMultiNum * (MaxAddNum>0 ? MaxAddNum : 1));
}

Console.WriteLine("Maximum produced number is " + allPossibleNum.Max());
Console.Read();

- Napster August 27, 2017 | Flag Reply
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0
of 0 vote

List<double> allPossibleNum = new List<double>();
            double[] array = { 5, 4, 3, 1 };
            Array.Sort(array);
            Array.Reverse(array);
            for (int i=array.Length-1;i > -1; i--)
            {
                double maxMultiNum = 1;
                for (int j = 0; j <= i; j++)
                {
                    maxMultiNum *= array[j];
                }

                double MaxAddNum = 0;
                for (int j = i+1; j < array.Length; j++)
                {
                    MaxAddNum += array[j];
                }
                allPossibleNum.Add(maxMultiNum * (MaxAddNum>0 ? MaxAddNum : 1));
            }

            Console.WriteLine("Maximum produced number is " + allPossibleNum.Max());
            Console.Read();

- Napster August 27, 2017 | Flag Reply
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of 0 vote

if you wonder how the dp solutions are built, here the recursion:

term(i,j) = max(term(i,k)+term(k,j), term(i,k)*term(k,j)) if j-i > 1, for all k where i<k<j
            abs(array[i]), if j-i == 1
result = term(0, len(array))

- Chris August 28, 2017 | Flag Reply
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MaxOutput(vector<int> arry, int &maxout)
{
if (arry.size() == 1)
{
maxout = max(maxout, arry[0]);
return;
}
for (int i = 0; i < arry.size()-1; i++)
{
int temp;
for (int k = 0; k < 3; k++)
{
switch (k)
{
case 0: //add
temp = arry[i] + arry[i + 1];
break;
case 1: //sub
temp = arry[i] - arry[i + 1];
break;
case 2: //mul
temp = arry[i] * arry[i + 1];
}

vector<int> subarry;
for (int j = 0; j < arry.size(); j++)
{
if (i == j)
subarry.push_back(temp);
else if (i + 1 != j)
subarry.push_back(arry[j]);
}
MaxOutput(subarry, maxout);
}
}
}

- Namho August 30, 2017 | Flag Reply
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of 0 vote

MaxOutput(vector<int> arry, int &maxout)
{
	if (arry.size() == 1)
	{
		maxout = max(maxout, arry[0]);
		return;
	}
	for (int i = 0; i < arry.size()-1; i++)
	{
		int temp;
		for (int k = 0; k < 3; k++)
		{
			switch (k)
			{
			case 0: //add
				temp = arry[i] + arry[i + 1];
				break;
			case 1: //sub
				temp = arry[i] - arry[i + 1];
				break;
			case 2: //mul
				temp = arry[i] * arry[i + 1];
			}

			vector<int> subarry;
			for (int j = 0; j < arry.size(); j++)
			{
				if (i == j)
					subarry.push_back(temp);
				else if (i + 1 != j)
					subarry.push_back(arry[j]);
			}
			MaxOutput(subarry, maxout);
		}
	}
}

- Namho August 30, 2017 | Flag Reply
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0
of 0 vote

C++ Code
========
#include <iostream>
#include <algorithm>
using namespace std;

void PrintArray(int *a, int n);

int main(void) {
int n=0, r=0, a[100]={0};

cout << "Enter number of array elements:";
cin >> n;
for(int i=0; i<n; i++ ){
    cout << "a[" << i << "]=";
    cin >> a[i];
}

sort(a, a+n);
PrintArray(a, n);

for(int i=0; i<n; i++){
    if(a[i]==1){
    for(int j=0; j<n; j++){
        if(a[j]!=0 && i != j){
            a[j]+=a[i];
            a[i]=0;
            break;
            }
        }
    PrintArray(a, n);
    }
}

for(int i=0; i<n; i++){
    if(a[i]!=0){
        r = (r==0)? a[i]:r*a[i];
    }
}

cout << "Max=" << r << endl;
return 0;
}

void PrintArray(int *a, int n){
for(int i=0; i<n; i++)
    cout << a[i] << " ";
cout << endl;

Python Code
==========
a = [int(x) for x in input().split()]
print(a)
a.sort()
print(a)

for i in range(0, len(a)):
    if a[i] == 1:
        for j in range(0, len(a)):
            if i != j and a[j] != 0:
                a[j] += a[i]
                a[i] = 0
                break
        print(a)

r = 0
for i in a:
    if a[i] != 0:
        r = a[i] if r == 0 else r*a[i]

print("max=%d", r)

Bug:
if a[i] != 0:
IndexError: list index out of range

- Anonymous September 02, 2017 | Flag Reply
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0
of 0 vote

My solution in O(n) only for positive numbers.

import java.util.ArrayList;

public class ArrayOfNumbersMax {
	
	
	public static void main (String[] args) {
		int[] a = {1,3,4,5};
		System.out.println(max(a));
		
	}
	
	public static int max (int[] a) {
		
		ArrayList<Integer> arr = new ArrayList<Integer> ();
		
		addNumbersReplaceOnes (a, arr);
		
		int result = 1;
		for (int num : arr) {
			result *= num;
		}
		return result;
	}

	private static void addNumbersReplaceOnes(int[] a, ArrayList<Integer> arr) {
		int numOfOnes = 0;
		int min = Integer.MAX_VALUE;
		for (int n : a) {			

			if (n == 1) {
				numOfOnes ++;
			} else {
				if (n < min) {
					min = n;
				} else {
					arr.add(n);
				}
			}
		}
		ArrayList<Integer> temp = new ArrayList<Integer> ();
		while (numOfOnes >= 2) {
			temp.add(2);
			numOfOnes--;
		}
		if (numOfOnes == 1) {
			if (temp.size() > 0) {
				temp.set(0, 3);
			} else {
				min++;
			}
		}
		arr.add(min);
		arr.addAll(temp);
	}

}

- baryaacov September 06, 2017 | Flag Reply
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0
of 0 vote

Does this work for all?

public static int max(int[] a)
	{
		Arrays.sort(a);
		int max=a[0];
		
		for(int i=1; i< a.length; i++)
		{
			max = Math.max(max*a[i], Math.max(max+a[i], max-a[i]));
		}
		return max;
	}

- clk October 04, 2017 | Flag Reply
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0
of 0 vote

// Input:
// p : array of integers
// n : number of elements in array
//
// Question:
// use operators {+,-,*} along with () to get the maximum value from the list of integers, assuming they are positive
//
// Solution
// for negative numbers, they are prefixed with (-?) in order to make them positive & increase result. after that, its the same is first solution
// all values of 1 needs to be added to the smallest numbers. as we pick the smallest & add 1, it might no longer be smallest, hence we add the next '1' to the next smallest.
// so in fact, we sort all numbers & then add the '1' values to the smallest numbers. this can be done in loops on array bcz as we add the '1', the values increase & match values that were previously larger.
// logic: if we have K times 5 & K times 1 then the max valule is obviously (1+5)^K - nothing can beat that.
// p.s.: if 0 is a valid input value, it'll need to be added so it wont zero everything.

#include <stdlib.h>
#include <stdio.h>
#include <assert.h>
#include <math.h>


int comp_int(const void *_p1, const void *_p2)
{
const int p1 = *((int*)_p1);
const int p2 = *((int*)_p2);
if (p1 < p2)
return -1;
else if (p1 == p2)
return 0;
else
return 1;
}

int max_from_set(int *p, int n)
{
int count_1;
int start_1, start;
int res;
int i;

if (n == 0)
return 0;
#ifdef SUPPORT_NEGATIVE
// convert negative values to positive. this can also be done withn the sorting comparison routine bcz it must compare all values for sorting.
for (i=0; i<n ; i++) {
if (p[i] < 0)
p[i] = -p[i];

}
#endif
// sort all values.
qsort(p, n, sizeof(*p), comp_int);
// skip zeros & count
for (i=0; (i<n) && (p[i] == 0); i++);
if (i == n) {
// only zeroes in input
return 0;
}
start_1 = i;
// counting instances of value '1' & find index of first value > 1.
for (count_1=0; i<n && (p[i] == 1); i++, count_1++);
// compute the max value of 1's (after we dropped 0). so we add the '1' to themselves, to get '2' that will multiply
// go for 2*2*2*2....*2 with optional single '3' incase we have odd number of '1'
if ((count_1 >= 1) && (count_1 <= 3)) {
res = count_1;
} else if (count_1 % 2) {
// even number (>= 5) of '1' - go for 2*2*2...*2 = 2 ^ (count_1/2)
res = 3*pow(2, (count_1-3)/2);
} else {
// odd number (>= 4) of '1' - go for 3*2*2...*2 = 3 * 2 ^ (count_1/2)
res = pow(2, count_1/2);
}
if (i == n) {
// only '1' values
return res;
} else {
// we add into number(s) greater than 1
start = i;
}
// we now have a sorted array with count_1 times the value '1' & p[start] > 1 unless no such value exist (i.e.: i == n)
// for the sorted array : [0,....0,1,...1,(x>1), ....]
// | |
// start_1----/ |
// start ------------/
if (count_1 == 1)
p[start] ++; // add '1' to the smallest value
// multiply all values in range [start .. n)
for (i=start; i<n ; i++) {
res *= p[i];
}
return res;
}

- Eitan BA October 27, 2017 | Flag Reply
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0
of 0 vote

O(n) soln. -

public static void main(String[] args) {
    int[] arr = {1,1,1,5};
    int n = max(arr);
    System.out.println(n);
  }
  
  public static int max(int[] arr){
  	
    int n = arr.length;
    int[] dp = new int[n];
    
    dp[0] = arr[0];
    dp[1] = Math.max(arr[0] + arr[1], arr[0] * arr[1]);
    dp[1] = Math.max(dp[1], arr[1] != 0 ? arr[0] / arr[1] : 0);
    dp[1] = Math.max(dp[1], arr[0] - arr[1]);
    
    int prev = -1;
    
    for(int i = 2; i < n; i++){
      	prev = dp[i-2];
    	dp[i] = Math.max(dp[i-1] + arr[i], dp[i-1] * arr[i]);
      	dp[i] = Math.max(dp[i], arr[i] != 0? dp[i-1] / arr[i]: 0);
	    dp[i] = Math.max(dp[i], dp[i-1] - arr[i]);
      	
      	dp[i] = Math.max(dp[i], prev*(arr[i-1] * arr[i]));
      	dp[i] = Math.max(dp[i], prev*(arr[i-1] + arr[i]));
      	dp[i] = Math.max(dp[i], prev*(arr[i] != 0? arr[i-1] / arr[i] : 0));
        dp[i] = Math.max(dp[i], prev*(arr[i-1] - arr[i]));
      
      	dp[i] = Math.max(dp[i], prev+(arr[i-1] * arr[i]));
      	dp[i] = Math.max(dp[i], prev+(arr[i-1] + arr[i]));
      	dp[i] = Math.max(dp[i], prev+(arr[i] != 0? arr[i-1] / arr[i]: 0));
        dp[i] = Math.max(dp[i], prev+(arr[i-1] - arr[i]));
      
      	dp[i] = Math.max(dp[i], prev-(arr[i-1] * arr[i]));
      	dp[i] = Math.max(dp[i], prev-(arr[i-1] + arr[i]));
      	dp[i] = Math.max(dp[i], prev-(arr[i] != 0? arr[i-1] / arr[i]: 0));
        dp[i] = Math.max(dp[i], prev-(arr[i-1] - arr[i]));
      
      	if(arr[i-1] * arr[i] != 0)
      		dp[i] = Math.max(dp[i], prev/(arr[i-1] * arr[i]));
      	if(arr[i-1] + arr[i] != 0)
      		dp[i] = Math.max(dp[i], prev/(arr[i-1] + arr[i]));
      	if((arr[i] != 0? arr[i-1] / arr[i]: 0) != 0)
      		dp[i] = Math.max(dp[i], prev/(arr[i] != 0? arr[i-1] / arr[i]: 0));
        if(arr[i-1] - arr[i] != 0)
      		dp[i] = Math.max(dp[i], prev/(arr[i-1] - arr[i]));
    }
    return dp[n-1];
  }

- sudip.innovates November 03, 2017 | Flag Reply
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0
of 0 vote

in Python, though I think there is a bug when the first number is a negative.

def get_arithmetics(v1, v2):
      return [
            v1 * v2,
            v1 + v2,
            v1 - v2,
            v1 % v2,
      ]

def get_max(input):      
      totals = [1, input[0]]

      for index, num in enumerate(input[1:]):
            index = index+1            
            rhs = max(get_arithmetics(input[index-1], num))
            max_potentials = [max(get_arithmetics(totals[index], num))]
            max_potentials.append(totals[index-1] * rhs)
            
            if (len(totals)>2):
                  max_potentials += get_arithmetics(totals[index-1], rhs)
            totals.append(max(max_potentials))
      return totals[-1]


assert get_max([3, 4, 5, 1]) == 72, "[3, 4, 5, 1] == 72"
assert get_max([1, 1, 1, 5]) == 15, "[1, 1, 1, 5] == 15"
assert get_max([8, 5, -1]) == 48, "8, 5, -1] == 48"

- yessah February 28, 2019 | Flag Reply
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of 0 vote

def get_arithmetics(v1, v2):
      return [
            v1 * v2,
            v1 + v2,
            v1 - v2,
            v1 % v2,
      ]

def get_max(input):      
      totals = [1, input[0]]

      for index, num in enumerate(input[1:]):
            index = index+1            
            rhs = max(get_arithmetics(input[index-1], num))
            max_potentials = [max(get_arithmetics(totals[index], num))]
            max_potentials.append(totals[index-1] * rhs)

            if (len(totals)>2):
                  max_potentials += get_arithmetics(totals[index-1], rhs)
            totals.append(max(max_potentials))
      return totals[-1]


assert get_max([3, 4, 5, 1]) == 72, "[3, 4, 5, 1] == 72"
assert get_max([1, 1, 1, 5]) == 15, "[1, 1, 1, 5] == 15"
assert get_max([8, 5, -1]) == 48, "8, 5, -1] == 48"

- yessah March 01, 2019 | Flag Reply
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-1
of 1 vote

public static void main(String args[]) throws Exception {
		int arr[] = { 1, 1, 1, 5 };
		System.out.println(maxValue(arr));
	}

	public static int maxValue(int arr[]) {
		int matr[][] = new int[arr.length][arr.length];
		for (int index = 0; index < arr.length; index++)
			matr[index][index] = arr[index];
		for (int len = 2; len <= arr.length; len++) {
			for (int i = 0; i < arr.length - len + 1; i++) {
				int j = i + len - 1;
				for (int k = i + 1; k <= j; k++) {
					matr[i][j] = max(matr[i][j], matr[i][k - 1] + matr[k][j], matr[i][k - 1] - matr[k][j], matr[i][k - 1] * matr[k][j]);
				}
			}
		}
		return matr[0][arr.length - 1];
	}

	public static int max(int... arr) {
		return Arrays.stream(arr).max().getAsInt();
	}

- koustav.adorable August 27, 2017 | Flag Reply
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-1
of 3 vote

72 for first Ex. is not Max.
See this (1+3)*4*5=80

- Nikolay V. Ilin August 27, 2017 | Flag Reply
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-1
of 1 vote

easy solution

package com.indus.training.persist.impl;

import java.util.Arrays;

public class MaxArray {

	public static void main(String[] args) {
		max mObj = new max();
		int[] a = { 1, 3,5,1,1};
		System.out.println(mObj.maxval(a));

	}

}

class max {
	public int maxval(int[] ex) {
		int j = 0;

		for (int i = 0; i < ex.length; i++) {
			if (ex[i] == 1) {
				j++;
			}

		}
		if (j == 1) {
			Arrays.sort(ex);
			int[] mex = Arrays.copyOfRange(ex, 1, ex.length);
			mex[0] = mex[0] + 1;
			int m = 1;
			for (int k = 0; k < mex.length; k++) {
				m = m * mex[k];
			}
			return m;

		} else {
			Arrays.sort(ex);
			int[] nEx = Arrays.copyOfRange(ex, j, ex.length);
			int m = 1;
			for (int k = 0; k < nEx.length; k++) {
				m = m * nEx[k];
			}
			m = m * j;
			return m;
		}

	}
}

- Sudheer.Renu August 29, 2017 | Flag Reply
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-1
of 1 vote

int max(int t[])
{
    int max = 0;
    int tmp = 0;
    sort(t);
    for(int i=0;i<len;i++) {
        for(int j=0;j<=i;j++)
           tmp+=t[j]
        for(int j=i+1;j<len;j++)
            tmp=tmp*t[j];
        if (tmp > max)
            max = tmp;
    }
    return max;
}

- KP August 30, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

Following code works for positive and negative numbers in O(n) time.

public static long getMaxValue(int[] nums) {
	int length = nums.length;
	long leftVal = nums[0];
	long rightVal = nums[length - 1];
		
	for (int leftIndex = 1, rightIndex = length - 2; leftIndex < length; leftIndex++, rightIndex--) {
		long left = leftVal * nums[leftIndex];
		if (!(Math.max(left, leftVal + nums[leftIndex]) == left)) {
			left = leftVal + nums[leftIndex];
		}
		if (!(Math.max(left, leftVal - nums[leftIndex]) == left)) {
			left = leftVal - nums[leftIndex];
		}
		leftVal = left;

		long right = rightVal * nums[rightIndex];
		if (!(Math.max(right, rightVal + nums[rightIndex]) == right)) {
			right = rightVal + nums[rightIndex];
		}
		if (!(Math.max(right, rightVal - nums[rightIndex]) == right)) {
			right = rightVal - nums[rightIndex];
		}
		rightVal = right;
	}

	return Math.max(leftVal, rightVal);
}

- asaqib September 13, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

public static long maxOutput(int[] in, int start, int end)
{
if(end < start || end >= in.length)
{
return Long.MIN_VALUE;
}

if(end == start)
{
return in[end];
}

long result = Long.MIN_VALUE;

for(int i = start; i < end; i++)
{
long left = maxOutput(in, start, i);
long right = maxOutput(in, i+1, end);
long mul = left * right;
long add = left + right;
long sub = left - right;

result = Math.max(result, Math.max(mul, Math.max(add, sub)));
}
return result;
}

- Anonymous September 22, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

public static long maxOutput(int[] in, int start, int end)
{
if(end < start || end >= in.length)
{
return Long.MIN_VALUE;
}

if(end == start)
{
return in[end];
}

long result = Long.MIN_VALUE;

for(int i = start; i < end; i++)
{
long left = maxOutput(in, start, i);
long right = maxOutput(in, i+1, end);
long mul = left * right;
long add = left + right;
long sub = left - right;

result = Math.max(result, Math.max(mul, Math.max(add, sub)));
}
return result;
}

- guru September 22, 2017 | Flag Reply


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