Amazon Interview Question for Quality Assurance Engineers


Country: India
Interview Type: In-Person




Comment hidden because of low score. Click to expand.
3
of 5 vote

merge sort from right to left, placing the result in "a" as you go. return a.

- Anonymous February 24, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
3
of 5 vote

Set two pointers at the very last number of the two arrays. Compare with the two numbers at the pointers, put the bigger one at the end of first array and move that pointer one left. Continue this process until both pointer move at the head of the arrays.

- spy February 24, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

You mean, use an insertion sort correct?
I concur.

- ArxoClay March 02, 2014 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

no, this is an usual mergesort, but from the other direction

- Markus March 06, 2014 | Flag
Comment hidden because of low score. Click to expand.
1
of 1 vote

public static void main(String[] args)
{
	int[] arr1=new int[]{7,10,11,20,0,0,0}//0 at the end should be assumed as empty space
	int arr2 = new int[]{1,3,7};
	
	int i2=arr2.length() -1;
	int i1 = arr1.length()  -1 - i2;
	int j = arr1.length()-1;

	while(i1>0 && i2 > 0 && j > 0)
	{
		if(arr1[i1] > arr2[i2])  
			 arr[j--] = arr[i1--];
		else
			 arr[j--] = arr[i2--];
	}

	if(i2>=0)
	{
		while(i2>=0)
		{
			arr[j--] = arr[i2--]
		}
	}

	if(i1>=0)
	{
		while(i1>=0)
		{
			arr[j--] = arr[i1--]
		}
	}


}

- Inquisitive March 06, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

/*
a[] is the bigger array; 
b[] is the smaller array; 
n is the size of bigger array;
m is the size of smaller array
*/
merge(int a[], int b[], int n, int m)
{
	int c[n];
	int i = n-m-1;
	int j = m-1;
	int k = n-1;
	while(i>=0 && j>=0 && k>=0)
	{
		if(a[i] > b[j])
		{
			c[k] = a[i];
			i--;
		}
		else
		{
			c[k] = b[j];
			j--;
		}
		k--;
	}
	if(i<0)
	{
		while(j>=0 && k>=0)
		{
			c[k] = b[j];
			k--;
			j--;
		}
	}
	else if(j<0)
	{
		while(i>=0 && k>=0)
		{
			c[k] = a[i];
			k--;
			i--;
		}
	}
}

- chaitanyapss February 25, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public int[] mergeSortedArray(int[] array1, int sizeA, int[] array2)
{
int i=sizeA;
int j = array2.length-1;
int k = array1.length -1;

while(k>=0 && i>=0 && j>=0)
{
if(array1[i] > array2[j])
{
array1[k]=array1[i];
k--;
i--;
}
else
{
if(j>=0)
{
array1[k]=array2[j];
}
k--;
j--;
}

}
return array1;
}

- bankertapan March 04, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

The empty space in a is equal to the number of element in b, so we can merge from last element without extra space.

void mergeSortAB(int *a, int *b)
{
	int ia = 3;
	int ib = 2;
	int last = 6;

	while (ib >= 0)
	{
		if (ia >= 0 && a[ia] > b[ib])
		{
			a[last--] = a[ia--];
		}
		else
		{
			a[last--] = b[ib--];
		}
	}
}

- LiMingjie0719 March 05, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

static int a[7] = {4, 10, 11, 20, 0, 0, 0};
static int a_len = 4;
static int b[3] = {1, 3, 7};

void merge_sort()
{
int store_pos = 6;
int a_cur = a_len-1;
int b_cur = 2;
while(a_cur >= 0 && b_cur >= 0) {
if (a[a_cur] > b[b_cur]) {
a[store_pos--] = a[a_cur--];
} else {
a[store_pos--] = b[b_cur--];
}
}

if (a_cur < 0) {
memmove(a, b, (b_cur+1)*sizeof(int));
}
}

- eric March 22, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

pos is the positon of largest element in longarray
j is the positon of largest element in small arry i.e. size

int main()
{

int longarr[7] = { 4,10,11,20 };
int smallarr[3]= { 1,3,7 };


int i =0,pos=3, n=7, j=2;


while(j>=0  && pos >=0)
{
	if ( longarr[pos] > smallarr[j])
	{	
		longarr[--n] = longarr[pos];
		pos--;
	}
	else
	{
	longarr[--n] = smallarr[j];
	j--;
	}
}

if(pos < 0)
{
	while( j>=0)
	longarr[j--]=smallarr[j];
}

for(int k=0;k<7;k++)
cout<<longarr[k]<<"\t";

return 0;
}

- its.me.siddh April 02, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

What is mean by empty, it will initialized value or garbage value. Please clarify.

- jare April 09, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Merge and then sort :
int a[] = {4, 10, 11, 20, 0, 0, 0};
int b[] = {1,3,7};
int c = a.length+b.length;
int d[] = new int[c];
int num = 0;
for(int i=0;i<a.length;i++){
d[num] = a[i];
num++;
}

for(int j=0;j<b.length;j++){
d[num] = b[j];
num++;
}


for(int i=0;i<d.length;i++){
int temp=0;
for(int j=0;j<d.length-1;j++){
if(d[j]>d[j+1]){
temp = d[j];
d[j] = d[j+1];
d[j+1] = temp;

}
}

}

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

- Sathish June 14, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

a = [4, 10, 11, 20] 
b = [1,3,7]

count = 0
for num in b:
    while True:
        if num < a[count]:
            a.insert(count, num)
            count +=1
            break;
        count +=1
print a


OUTPUT:
[1, 3, 4, 7, 10, 11, 20]

- Sunil August 23, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

int[] big = new int[] {4,10,11,20,0,0,0};
		int[] small = new int[] {1,3,7};
		int smallLength = small.length -1 ;
		int bigvalue = big.length -1 -smallLength-1;
		int bigLength = big.length -1;
		
		while( bigLength>=0 && smallLength>=0 && bigvalue>=0 )
		if(small[smallLength]>big[bigvalue]){
			big[bigLength] = small[smallLength];
			bigLength--;
			smallLength--;
		}
		
		else if (big[bigvalue]>small[smallLength]){
			
			big[bigLength] = big[bigvalue];
			bigLength--;
			bigvalue--;
		}
		
		if (smallLength>=0){
			while(smallLength>=0){
				if(small[smallLength]<big[bigLength]){
					big[bigLength] = small[smallLength];
				}
				smallLength--;
				bigLength--;
				
				
			}

}

- Neels September 14, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

int[] big = new int[] {4,10,11,20,0,0,0};
		int[] small = new int[] {1,3,7};
		int smallLength = small.length -1 ;
		int bigvalue = big.length -1 -smallLength-1;
		int bigLength = big.length -1;
		
		while( bigLength>=0 && smallLength>=0 && bigvalue>=0 )
		if(small[smallLength]>big[bigvalue]){
			big[bigLength] = small[smallLength];
			bigLength--;
			smallLength--;
		}
		
		else if (big[bigvalue]>small[smallLength]){
			
			big[bigLength] = big[bigvalue];
			bigLength--;
			bigvalue--;
		}
		
		if (smallLength>=0){
			while(smallLength>=0){
				if(small[smallLength]<big[bigLength]){
					big[bigLength] = small[smallLength];
				}
				smallLength--;
				bigLength--;
				
				
			}

}

- Neels September 14, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

int[] big = new int[] {4,10,11,20,0,0,0};
		int[] small = new int[] {1,3,7};
		int smallLength = small.length -1 ;
		int bigvalue = big.length -1 -smallLength-1;
		int bigLength = big.length -1;
		
		while( bigLength>=0 && smallLength>=0 && bigvalue>=0 )
		if(small[smallLength]>big[bigvalue]){
			big[bigLength] = small[smallLength];
			bigLength--;
			smallLength--;
		}
		
		else if (big[bigvalue]>small[smallLength]){
			
			big[bigLength] = big[bigvalue];
			bigLength--;
			bigvalue--;
		}
		
		if (smallLength>=0){
			while(smallLength>=0){
				if(small[smallLength]<big[bigLength]){
					big[bigLength] = small[smallLength];
				}
				smallLength--;
				bigLength--;

}

- Neels September 14, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

int[] big = new int[] {4,10,11,20,0,0,0};
		int[] small = new int[] {1,3,7};
		int smallLength = small.length -1 ;
		int bigvalue = big.length -1 -smallLength-1;
		int bigLength = big.length -1;
		
		while( bigLength>=0 && smallLength>=0 && bigvalue>=0 )
		if(small[smallLength]>big[bigvalue]){
			big[bigLength] = small[smallLength];
			bigLength--;
			smallLength--;
		}
		
		else if (big[bigvalue]>small[smallLength]){
			
			big[bigLength] = big[bigvalue];
			bigLength--;
			bigvalue--;
		}
		
		if (smallLength>=0){
			while(smallLength>=0){
				if(small[smallLength]<big[bigLength]){
					big[bigLength] = small[smallLength];
				}
				smallLength--;
				bigLength--;

}

- Neels September 14, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

int[] big = new int[] {4,10,11,20,0,0,0};
		int[] small = new int[] {1,3,7};
		int smallLength = small.length -1 ;
		int bigvalue = big.length -1 -smallLength-1;
		int bigLength = big.length -1;
		
		while( bigLength>=0 && smallLength>=0 && bigvalue>=0 )
		if(small[smallLength]>big[bigvalue]){
			big[bigLength] = small[smallLength];
			bigLength--;
			smallLength--;
		}
		
		else if (big[bigvalue]>small[smallLength]){
			
			big[bigLength] = big[bigvalue];
			bigLength--;
			bigvalue--;
		}
		
		if (smallLength>=0){
			while(smallLength>=0){
				if(small[smallLength]<big[bigLength]){
					big[bigLength] = small[smallLength];
				}
				smallLength--;
				bigLength--;			
				
			}

- Anonymous September 14, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

int[] big = new int[] {4,10,11,20,0,0,0};
		int[] small = new int[] {1,3,7};
		int smallLength = small.length -1 ;
		int bigvalue = big.length -1 -smallLength-1;
		int bigLength = big.length -1;
		
		while( bigLength>=0 && smallLength>=0 && bigvalue>=0 )
		if(small[smallLength]>big[bigvalue]){
			big[bigLength] = small[smallLength];
			bigLength--;
			smallLength--;
		}
		
		else if (big[bigvalue]>small[smallLength]){
			
			big[bigLength] = big[bigvalue];
			bigLength--;
			bigvalue--;
		}
		
		if (smallLength>=0){
			while(smallLength>=0){
				if(small[smallLength]<big[bigLength]){
					big[bigLength] = small[smallLength];
				}
				smallLength--;
				bigLength--;			
				
			}

- Anonymous September 14, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def merge_arr(a1,a2):
    a = []
    a = a1+a2
    return sorted(a)

            
                
    

print merge_arr([4,10,11,12],[1,3,2,4])

- parashar.002 January 15, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class MergeNew {
	
	static int[] merge(int[] a, int[] b) {
		int i= a.length-4 ;
		int k = a.length-1;
		int j = b.length-1;
		while( i != -1 && j != -1) {
			if(a[i] > b[j]) {
				a[k] = a[i]; 
				i--;
			} else {
				a[k] = b[j];
				j--;
			}
			k--;
		}
		
		while(i != -1) {
			a[k] = a[i];
			i--;
			k--;
		}
		
		while(j != -1) {
			a[k] = b[j];
			j--;
			k--;
		}
		return a;
	}
	
	public static void main(String[] args) {
		
		int a[] = {4, 10, 11, 20, 0, 0, 0}; 
		int b[] = {1,3,7};
		
		int[] c = (merge(a,b));
		
		for(int k : c) {
			System.out.print(k + " ");
		}
		
	}

}

- AK March 16, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

Are we allowed to use O(1) extra space?

- trein February 24, 2014 | 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