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Interview Question for Software Engineer at Facebook:
Oct 24, 2011
1 of
1 people found this helpful
Given an array of integers, now we want to erase all 0's (can be other value), and we want the result array condensed, meaning no empty cell in the array.
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Answers & Comments (9)
Oct 29, 2011
Assuming that the array is named iArray and the unwanted number is named delthis.
I think this should work:
public condense(int[] iArray, int delthis){
ArrayList iList = new ArrayList();
for (int i=0; i<iArray.length; i++){
if (iArray[i] != delthis)
iList.add(iArray[i]);
}
return iList.toArray();
}
I think this should work:
public condense(int[] iArray, int delthis){
ArrayList iList = new ArrayList();
for (int i=0; i<iArray.length; i++){
if (iArray[i] != delthis)
iList.add(iArray[i]);
}
return iList.toArray();
}
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Oct 29, 2011
Test on Java 6:
public Integer[] condense(int[] iArray, Integer delThis){
ArrayList<Integer> iList = new ArrayList<Integer>();
for (int i=0; i<iArray.length; i++){
if (iArray[i] != delThis)
iList.add(iArray[i]);
}
return iList.toArray(new Integer[iList.size()]);
}
public Integer[] condense(int[] iArray, Integer delThis){
ArrayList<Integer> iList = new ArrayList<Integer>();
for (int i=0; i<iArray.length; i++){
if (iArray[i] != delThis)
iList.add(iArray[i]);
}
return iList.toArray(new Integer[iList.size()]);
}
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Nov 2, 2011
void erase( int arr[] , int size , int &new_size )
{
int p1 = 0, p2 = 0;
while ( p1 < size && p2 < size )
{
if ( arr[p1] == 0 ) {
while ( p2 < size && arr[p2] == 0 ) {
++ p2;
}
if ( p2 == size ) {
break;
}
arr[p1] = arr[p2];
++ p1;
++ p2;
}
else {
++ p1;
++ p2;
}
}
new_size = p1+1;
}
{
int p1 = 0, p2 = 0;
while ( p1 < size && p2 < size )
{
if ( arr[p1] == 0 ) {
while ( p2 < size && arr[p2] == 0 ) {
++ p2;
}
if ( p2 == size ) {
break;
}
arr[p1] = arr[p2];
++ p1;
++ p2;
}
else {
++ p1;
++ p2;
}
}
new_size = p1+1;
}
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Nov 2, 2011
test on gcc
void erase( int arr[] , int size , int &new_size )
{
int p1 = 0, p2 = 0;
while ( p1 < size && p2 < size )
{
if ( arr[p1] == 0 ) {
while ( p2 < size && arr[p2] == 0 ) {
++ p2;
}
if ( p2 == size ) {
break;
}
arr[p1] = arr[p2];
arr[p2] = 0;
++ p1;
++ p2;
}
else {
++ p1;
++ p2;
}
}
new_size = p1;
}
void erase( int arr[] , int size , int &new_size )
{
int p1 = 0, p2 = 0;
while ( p1 < size && p2 < size )
{
if ( arr[p1] == 0 ) {
while ( p2 < size && arr[p2] == 0 ) {
++ p2;
}
if ( p2 == size ) {
break;
}
arr[p1] = arr[p2];
arr[p2] = 0;
++ p1;
++ p2;
}
else {
++ p1;
++ p2;
}
}
new_size = p1;
}
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0 of 2 people found this helpful
Nov 2, 2011
// This is erase/remove idiom one liner. Too much code with C.
std::vector v;
v.erase(std::remove(v.begin(), v.end(), 0), v.end());
std::vector v;
v.erase(std::remove(v.begin(), v.end(), 0), v.end());
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2 of 2 people found this helpful
Dec 9, 2011
#include<iostream>
using namespace std;
int main()
{
int a[] = { 0,2,3,0,4,0,5,0,0,0};
int i,nz = 0;
for(i = 0; i<=9 ; i++)
{
if (a[i] == 0)
nz++;
else
a[i-nz] = a[i];
}
for(i = 0; i <=9-nz ; i++)
cout << a[i]<<" ";
return 0;
}
using namespace std;
int main()
{
int a[] = { 0,2,3,0,4,0,5,0,0,0};
int i,nz = 0;
for(i = 0; i<=9 ; i++)
{
if (a[i] == 0)
nz++;
else
a[i-nz] = a[i];
}
for(i = 0; i <=9-nz ; i++)
cout << a[i]<<" ";
return 0;
}
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Mar 30, 2012
Python:
-------------
s = [1, 2, 4, 0, 5, 0, 1, 0, 0, 1, 7, 0, 8, 0, 2, 4, 5, 1, 2, 8, 0, 5, 6, 8]
for k in xrange(len(s), 0, -1):
if not s[k-1]: s.pop(k-1)
--------------
after:
[1, 2, 4, 5, 1, 1, 7, 8, 2, 4, 5, 1, 2, 8, 5, 6, 8]
-------------
s = [1, 2, 4, 0, 5, 0, 1, 0, 0, 1, 7, 0, 8, 0, 2, 4, 5, 1, 2, 8, 0, 5, 6, 8]
for k in xrange(len(s), 0, -1):
if not s[k-1]: s.pop(k-1)
--------------
after:
[1, 2, 4, 5, 1, 1, 7, 8, 2, 4, 5, 1, 2, 8, 5, 6, 8]
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Apr 3, 2012
void erase(int arr[], int sz, int errasedVal) {
int i = 0, j = -1;
while(i < sz && j < sz) {
if(arr[i] == errasedVal && j == -1) j = i;
else if(arr[i] != errasedVal && j != -1) {
cout << arr[i]<< " " << j << endl;
swap(arr[i], arr[j]);
j++;
while(arr[j] != errasedVal && j < sz) j++;
}
i++;
}
}
int i = 0, j = -1;
while(i < sz && j < sz) {
if(arr[i] == errasedVal && j == -1) j = i;
else if(arr[i] != errasedVal && j != -1) {
cout << arr[i]<< " " << j << endl;
swap(arr[i], arr[j]);
j++;
while(arr[j] != errasedVal && j < sz) j++;
}
i++;
}
}
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0 of 1 people found this helpful
by MOHIT DHINGRA:
#include<conio.h>
void main()
{
clrscr();
int i,j,k.a[20];
cout<<"enter the size of array"';
cin>>n;
cout<<"enter the elements of array";
for(i=0;i<n;i++){
cin>>a[i];
}
cout<<"entered array is";
for(i=0;i<n;i++)
{
cout<<a[i];
}
for(i=0;i<n;i++)
{ if(a[i]==0) /*operation for deleting zero's*/
for(j=i;j<n;j++)
{a[j]=a[j+1];
}}
cout<<"array left is";
for(i=0;i<n;i++)
{ cout<<a[i];
}
getch();
}