Amazon Interview Question for Software Engineer / Developers


Country: India
Interview Type: Written Test




Comment hidden because of low score. Click to expand.
4
of 6 vote

sort ( int a[ ] ) {
for(int i =0 , j = a.size()-1 ; i< = j ; ){
if ( a[i] == 0 ){
i++;
}
else{
swap(a[i] , a[j] );
j--;
}
}
}

- william November 04, 2012 | Flag Reply
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0
of 0 votes

nice solution. Just to save few more operations, we can also add one more check before doing swap operation:
else{
if(a[j]==0){
swap(a[i],a[j]);
}
j--;
}

- sp!de? November 05, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

For input 1001001 :
First time a[i] = 1, a[j] = 1 (last element of the array)
swap(1,1) , put 1 in position 0, so a[0] is 1 again.
How does the code work for this input, please explain

- ps December 13, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

Ignore my comment, adding condition if(a[j]==0) makes sense.

- ps December 13, 2012 | Flag
Comment hidden because of low score. Click to expand.
1
of 1 vote

In this approach,lots of unnecessary swaps are being done.. Say I have array as 0,0,1,0,1,1,0,1. Now as per above logic I would swap when i=2 and j=n-1 , but this swap is not required actually(Swapping 1 with 1 doesn't make any sense). So better approach is already posted by amritaansh123 .Please see below that answer.

- Algorithmist March 25, 2013 | Flag
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2
of 2 vote

package algo.solutions;

public class Sort0s1s {

	public void sort(int[] a)
	{
		if(a == null || a.length == 1)
		{
			return;
		}
		int i=0;
		int j = a.length -1;
		while(true)
		{
			while(i < a.length && a[i] == 0)
			{
				i++;
			}
			while(j >= 0 && a[j] == 1)
			{
				j--;
			}
			if(i>j)
			{
				return;
			}
			swap(a,i,j);
		}
	}
	
	private static void swap(int[] a, int i, int j)
	{
		int tmp = a[i];
		a[i] = a[j];
		a[j] = tmp;
	}

}

- sethuraj.32 November 04, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
2
of 2 vote

/crackprogramming.blogspot.com/2012/11/two-color-sort-sort-integer-containing.html

- Anonymous November 04, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
2
of 2 vote

void sort (int a[]) {
	int l, r;
	l = 0; r = n.length - 1;
	while (l < r) {
		while(a[l] != '1')
			l++;
		while(a[r] != '0')
			r--;
		if (l < r)
			swap (a, l, r);
	}
}

- amritaansh123 February 17, 2013 | Flag Reply
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1
of 1 vote

o(n)
Another solution
1) Count the number of Zeros
2) Place zeros in the arrays equal to the count and later place all the ones in the array

- avikodak December 24, 2012 | Flag Reply
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0
of 0 votes

It is much better then using two pointer approach. No Swaps required.

- Algorithmist March 25, 2013 | Flag
Comment hidden because of low score. Click to expand.
1
of 1 vote

public class ArrayWithZeroesNOne {

	public static void main(String[] args){
		int []A={1,0,0,1,1,1,0,1,0,0,0,1,1,1,1,1,0,1,0,1,1,0,0,1};
		int l=A.length;
		int [] b=sortArr(A, l);
		for(int i=0; i<l; i++){
			System.out.print(b[i]+" ");
		}
	}
	
	static int[] sortArr(int[]A, int l){
		for(int i=0, j=l-1; i<j;){
			while(A[i]==0)
				i++;
			while(A[j]!=0)
				j--;
			if(i>=j)
				break;
			int temp=A[i];
			A[i]=A[j];
			A[j]=temp;
		}
		return A;
	}
}

- shashi_kr August 14, 2013 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

int [] sort(int [] arr)
{
     for ( int i= 0; i< arr.Length; i++)
     {
          if (arr[i] != 0)
          {
                for ( int j = i; j< arr.Length; j++)
                {
                     if( arr[j] == 0) break;
                }
                if( j< arr.Length) 
               {
                   int temp = arr[i];
                   arr[i] = arr[j];
                   arr[j] = temp;
               }
               else
                     break;
            }
       }
 }

- VJ November 04, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

(oops forgot the return statement)
int [] sort(int [] arr)

{
     for ( int i= 0; i< arr.Length; i++)
     {
          if (arr[i] != 0)
          {
                for ( int j = i; j< arr.Length; j++)
                {
                     if( arr[j] == 0) break;
                }
                if( j< arr.Length) 
               {
                   int temp = arr[i];
                   arr[i] = arr[j];
                   arr[j] = temp;
               }
               else
                     break;
            }
       }
    return arr;
 }

- VJ November 04, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

# include<stdio.h>

int main()

{

int arr[]={0,1,0,1,0,1,1,0,0,0,1};

int len= sizeof(arr)/sizeof(int);

int i,j,c;
c=0;
for(i=0;i<len;i++)
{

if(arr[i]==0)
c++;


}

for(i=0;i<c;i++)
arr[i]=0;
for(;i<len;i++)
arr[i]=1;


for(i=0;i<len;i++)

printf("\n the array is %d", arr[i]);
}
~

- Gaurav November 04, 2012 | Flag Reply
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0
of 0 vote

private static int[] Sort(int[] set) {
// TODO Auto-generated method stub
for(int i = 0; i < set.length - 1;){
if(set[i] == 0){
i++;
} else if (set[i] == 1){
for(int j = i + 1; j < set.length; j++){
if (set[j] == 0){
int temp = set[j];
set[j] = set[i];
set[i] = temp;
}
}
i++;
}
}
return set;
}

- Anonymous November 07, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

private static int[] Sort(int[] set) {
		// TODO Auto-generated method stub
		for(int i = 0; i < set.length - 1;){
			if(set[i] == 0){
				i++;
			} else if (set[i] == 1){
				for(int j = i + 1; j < set.length; j++){
					if (set[j] == 0){
						int temp = set[j];
						set[j] = set[i];
						set[i] = temp;
					}
				}
				i++;
			}
		}
		return set;
	}

- Anonymous November 07, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static void sortArrayWithZeroAndOne(int [] a)
{
if (a.Length == 0)
return;
else
{
//as we know array is of only two types of value zero and one, so creating another array which hold two int
int[] tempArray = new int[2];
for (int i = 0; i < a.Length; i++)
{
tempArray[a[i]]++;
}
for (int i = 0; i < tempArray[0]; i++)
{
a[i] = 0;
}
for (int i = tempArray[0]; i < a.Length; i++)
{
a[i] = 1;
}

foreach (int i in a)
{
Console.Write(i+ ",");
}

}

}

- Anonymous November 11, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

See the simplest and working solution

public static void main(String[] args)
	{
		int[] arr = { 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1 };
		sort(arr);

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

	private static void sort(final int[] arr)
	{
		int i = 0, j = arr.length - 1;
		while (i < j)
		{
			while (arr[i] == 0 && i < arr.length)
				i++;
			while (arr[j] == 1 && j >= 0)
				j--;
			if (i < j)
			{
				int temp = arr[i];
				arr[i] = arr[j];
				arr[j] = temp;
			}
		}
	}

- Prashant Singh November 25, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

use strict;
my @arr=qw(0 1 0 1 0 1 1 0 1 1 0 1 0 1 0 0 1 1 0 1);
my ($i,$cnt);
foreach(@arr) {
$cnt++ if ($_ eq 1);
}
print "@arr\n";
print "1's are : $cnt\n0's are : ", (scalar(@arr)-$cnt), "\n";
for (my $i=0; $i<scalar(@arr); $i++) {
$arr[$i] = 0 if ($i < (scalar(@arr) - $cnt));
$arr[$i] = 1 if ($i >= (scalar(@arr) - $cnt));
}
print "@arr\n";

- chinna December 06, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

not with strict programming.
Sort or algorithm below for the logic:

i = a[0], j = a[1]
while( (a[j] != 0) && j <= end of array)
{
if( (a[i] == 1)
{
swap(a[i],a[j]);
i++;
}
j++;
}

checked with such strings: 1100101,...
As per logic: j should point to location where array has element with 0.
if a[i] has value 1 than swap the values and do i++;

correct me if i am wrong

- Anonymous January 07, 2013 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

just one correction in above code:
do i++ also along with j++ in else case of inner if condition.
this to increament 'i' in case a[i] != 1 also.

- Anonymous January 07, 2013 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

@ avikodak
I have written an algorithm with complexity O(n)

-- 1 -- Traverse Array A
-- 2-- For every '1' increase sum by 1 and set A[count-sum] =1 (at the tail of the array)
-- 3-- For every '0' set A[i-sum] =1 (at the rear end of the array)

arr = array('0','0','1','0','1','1','0','1','0');
 count = sizeof(arr);
sum = 0;
for(i = 0 ; i < count ; i++){ 
        if(!arr[i]){
            newArr[i-sum] = 0;
        }else{            
            sum++;
            newArr[count-sum] = 1;
        }
}

output newArr has  ('0','0','0','0','0','1','1','1','1');

- pjae April 18, 2013 | Flag Reply


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