## Google Interview Question for Software Engineers

• 6

Country: United States
Interview Type: In-Person

Comment hidden because of low score. Click to expand.
3
of 3 vote

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Interviewee's solution

``````#include <iostream>
#include <vector>
using namespace std;

void printLargerSmaller(const vector<int> &v) {
int n = v.size();
vector<int>leftMax(n, INT_MIN);
for(int i=1;i<n; ++i)     leftMax  = max(leftMax[i-1], v[i-1]);
int rightMin =INT_MAX;
for(int i=n-1;i>=0; i--)  {
if(i<n-1)rightMin = min(rightMin, v[i+1]);
if(v >leftMax && v < rightMin) cout << v << " ";
}
}
int main() {
vector<int>v = {3,4,7,1,8,12};
printLargerSmaller(v);
return 0;
}``````

Comment hidden because of low score. Click to expand.
2
of 2 vote

Here's my solution using stack

``````void order(int * num, int size){
int max = 0;

stack<int> s;
for(unsigned int i = 0; i < size; i++){
if(num[i] > max){
max = num[i];
s.push(max);
}
while(!s.empty() && num[i] < s.top()){
s.pop();
}
}

while(!s.empty()){
printf("%d ", s.top()) ;
s.pop();
}
printf("\n");

}

int main(){
int  a[] = {6, 5, 4, 3, 9, 100, 87, 64, 34, 101};
int  b[] = {3, 4, 7, 1, 8, 12};
order(a, 10) ;
order(b, 6) ;

return 0;
}``````

Comment hidden because of low score. Click to expand.
1
of 1 vote

``````def greaterThanLeftLessThanRight(array):
if array is None:
raise TypeError('Array cannot be None')

if len(array) == 0:
return None

leftArray = [0 for _ in range(len(array))]
rightArray = [0 for _ in range(len(array))]

leftMax = array
rightMin = array[-1]

# build array of greatest element seen so far at each index from the left
for i in range(1, len(array)):
leftArray[i] = leftMax
if array[i] > leftMax:
leftMax = array[i]

# build array of smallest element seen so far at each index from the right
for i in range(len(array) - 2, -1, -1):
rightArray[i] = rightMin
if array[i] < rightMin:
rightMin = array[i]

# iterate over array - checking conditions to print
for i in range(len(array)):
if i == 0:
if array[i] < rightArray[i]:
print array[i]
elif i == len(array) - 1:
if array[i] > leftArray[i]:
print array[i]
elif array[i] > leftArray[i] and array[i] < rightArray[i]:
print array[i]``````

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

``````public class Numbers {

public static void main(String[] args) {

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

printFitNumbers(a);
}

public static void printFitNumbers(int[] a) {

int count = 0;

for (int i = 0; i < a.length; i++) {

int k;

for (k = 0; k < i; k++)
if (a[k] >= a[i]) {
count++;
break;
}
if (count == 0 && (i + 1 < a.length)) {
for (int j = i + 1; j < a.length; j++)
if (a[i] >= a[j]) {
count++;
break;
}
}
if (count == 0)
System.out.print(a[i] + " ");

count = 0;
}
}
}``````

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

``````public int[] solution(int [] input){

int maxTillNow = Integer.MIN_VALUE;
int minAfterNow = Integer.MAX_VALUE;
int index = 0;

List<Integer> result = new ArrayList<Integer>();
for(int i = 0;i<input.length;i++){

if(i>=index){
minAfterNow = Integer.MAX_VALUE;
for(int j=i+1;j<input.length;j++){
if(input[j]<minAfterNow){
minAfterNow=input[j];
index=j;
}

}
}

if(input[i]>maxTillNow && input[i]<minAfterNow){

}

maxTillNow = Math.max(maxTillNow, input[i]);

}

int[] returnVal = new int[result.size()];
int idx = 0;
for(int r : result){
returnVal[idx]=r;
idx++;
}

return returnVal;

}``````

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

this is a kind of quick sort problem.

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

Simple Recursive solution is following

``````FindElement(int[] arr, int MaxTillNow, int currentIndex)
{
if(currentIndex == 0)
{
return FindElement(arr, arr[currentIndex], currentIndex + 1);
}
if(currentIndex == arr.Length-1)
{
return arr[currentIndex];
}
int minTillNow = FindElement(arr, Math.Max(MaxTillNow, arr[currentIndex]), currentIndex + 1)
if(arr[currentIndex] > MaxTillNow && arr[currentIndex] < minTillNow)
{
Console.WriteLine(arr[currentIndex]);
}
return Math.Min(minTillNow, arr[currentIndex]);
}``````

Complexity : O(N), we are only doing one pass to the array.
Space complexity is O(1).

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

Use Quicksort .

Comment hidden because of low score. Click to expand.
-1
of 1 vote

``````{
public void printArray(int[] array) {

for(int i = 1; i < array.length-1; i++){
if(array[i] > array[i-1] && array[i] < array[i+1])
{
System.out.println(array[i]);
}
}
}``````

}

Comment hidden because of low score. Click to expand.
-1
of 1 vote

The trick here is that there can only be one - a maximum

``````package com.prajnainc;

import java.util.HashSet;
import java.util.Set;

public class Peaks {
/* Second question Given an array of integers, print all the numbers that meet the following requirement - when the
* number is greater than every number on its left and smaller than every number on the right.
*/

/* There can only be one - so just find the maximum */
static int a[] = {1,3,-1,5,7,9,1,3};

public static void main(String args[]) {
int max_i = 0;

for(int i = 1; i<a.length; i++) {
if(a[i] > a[max_i]) {
max_i = i;
}

}
System.out.println("a["+max_i+"]="+a[max_i]);

}
}``````

Comment hidden because of low score. Click to expand.
-1
of 1 vote

The trick here is that there can only be one. so just find the maximum

``````package com.prajnainc;

import java.util.HashSet;
import java.util.Set;

public class Peaks {
/* Second question Given an array of integers, print all the numbers that meet the following requirement - when the
* number is greater than every number on its left and smaller than every number on the right.
*/

/* There can only be one - so just find the maximum */
static int a[] = {1,3,-1,5,7,9,1,3};

public static void main(String args[]) {
int max_i = 0;

for(int i = 1; i<a.length; i++) {
if(a[i] > a[max_i]) {
max_i = i;
}

}
System.out.println("a["+max_i+"]="+a[max_i]);

}
}``````

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

``[1,2,6,4,9,15]``

Shouldn't that return

``[2,9``

?

Comment hidden because of low score. Click to expand.
-2
of 2 vote

``````{
public void printArray(int[] array) {

for(int i = 1; i < array.length-1; i++){
if(array[i] > array[i-1] && array[i] < array[i+1])
{
System.out.println(array[i]);
}
}
}``````

}

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