## Amdocs Interview Question for Testing / Quality Assurances

• 0

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0
of 0 vote

1) We could use in-place sort algorithm (quicksort for example). space: O(1) time: O(n*lgn).
2) Use in-place merge algorithm.

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0
of 0 vote

bitonic sort

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0
of 0 vote

// java code

public class MergeShort {

/**
* @param args
*/
/**
* @param args
*/
public static void main(String[] args) {
int[] a={4,7,9,20,1,3,10,15};
int i=0, length=8, j, k, temp;
// for(i=0;i<length;i++){
// System.out.println(i+":"+a[i]);
// }
for(j=length/2; j<length; j++){
for(; i<j; i++){
if(a[i] >= a[j]){
temp=a[j];
for(k=j; k>i; k--){
a[k]=a[k-1];
}
a[i]=temp;
i++;
break;
}
}
}
System.out.println("#######################");
for(i=0;i<length;i++){
System.out.println(i+":"+a[i]);
}

}

}

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0
of 0 vote

I've wrote just logic using some hard coded values like array length.

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0
of 0 vote

Are not we using extra space 'temp'?? [do not use any extra space] It is clearly specified in given question.

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0
of 2 vote

We can do it inplace using heapssort method

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0

I am downvoting this answer, because there are lots of ways to in-place sort a list, but this brief answer doesn't explain why heapsort would be superior to, say, quicksort, given the partial ordering.

Of the most common divide-and-conquer sorts, mergesort is the one that breaks down the problem by recursively creating two sorted lists, so it would seem like the most appropriate sort to adopt for this problem. Unfortunately, it's really hard to do a mergesort efficiently on array-based lists when you don't have contiguous storage at your disposal for the merged result. If the original list were a linked list instead of an array, then you could do it pretty trivial with no extra storage, but the problem explicitly says this an array.

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0
of 0 vote

``````#include<stdio.h>
void mergetwoSortedRangeArray(int *, int );
void swap(int *, int *);

int main(void){

int a={1,3,6,8,-5,-2,3,8};
int len = sizeof(a)/sizeof(int);
mergetwoSortedRangeArray(a,len);
for(int i=0; i<len; i++){
printf("%d 	",a[i]);
}

return 0;
}

void mergetwoSortedRangeArray(int a[], int len){
int i=0, j=len/2;
while(i<j && j<=len){
if(a[j]<a[i]){
for(int k=i; k<j; k++){
swap(&a[k], &a[j]);
}
i++;
j++;
}
else{
i++ ;
}
}
printf("\n");
}

void swap(int *x, int *y){
*x ^= *y;
*y ^= *x;
*x ^= *y;
}``````

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0
of 0 vote

\$arr = array(1,3,6,8,-5,-2,3,8);

\$arr_len = count(\$arr);

\$mid = \$arr_len/2;
\$i=0;

while (\$mid < \$arr_len) {
if (\$arr[\$i] < \$arr[\$mid]) {
\$i++;
} else {
\$temp = \$arr[\$i];
\$arr[\$i] = \$arr[\$mid];
\$arr[\$mid] = \$temp;
\$mid++;
}
}

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0
of 0 vote

#include<stdio.h>
#include<conio.h>

int _tmain(int argc, _TCHAR* argv[])
{
int a,len,*p,temp=0,i,j;
a=(1,3,6,8,-5,-2,3,8);
p=a;

len=sizeof(a)/sizeof(a);

printf("length of the array is : %d",len);
printf("\n");
for( i=0;i<len/2;i++)
{
for( j=len/2;j<len;j++)
{
if(*(p+i) > *(p+j) || *(p+i)==*(p+j))
{
int k=0;
k=k+i;
temp=*(p+j);
printf("%d ",*(p+j));
while(k<=j)
{
*(p+k+1)=*(p+k);

k++;

}
*(p+i)=temp;

}

}
}
for(i=0;i<len;i++)
{
printf("%d ",*(p+i));
}

getch();

return 0;
}
--------------------------------------------------------------------------------------------------
check and plz let me knw tat what is the problem in this code?

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0

and plz dont consider line-> printf("%d ",*(p+j));

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0
of 0 vote

``````C# code
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace Sort2HalfSortedArrays
{
class Program
{
static void Main(string[] args)
{
int[] array = { 1, 3, 6, 8, -5, -2, 3, 8 };
int[] sortedArray = Sort(array); ;
Console.WriteLine("Sorted Array");
for (int i = 0; i < sortedArray.Length; i++)
{
Console.Write("{0} ", sortedArray[i]);
}

}
public static int[] Sort(int[] array)
{
int i = 0;
while (array[i] < array[i + 1])
{
if (i + 1 == array.Length - 1)
{
return array;
}
else
{
i++;
}
}

int j = array.Length - 1;
while ((i >= 0) && i < j)
{
if (array[i] >= array[j])
{
int k = i;
while (k < j)
{
int temp = array[k];
array[k] = array[k + 1];
array[k + 1] = temp;
k++;
}
i--;
j -= 2;
}
else
{
j--;
}
}
return array;
}

}
}

output:
Sorted Array
-5,-2,1,3,3,6,8,8

Comment if any issues in this.``````

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0
of 0 vote

Basically what we are doing is comparing the first element of both sorted portion

Scenario 1) in case second portion elemnt is smaller , we are bringing it to front and shifting the elements to right making room for it

Scenario 2) in case first portion elemnt is smaller, just goto second element in first portion, remain at first element in second portion

1,3,6,8,-5,-2,3,8
-5,1,3,6,8,-2,3,8
-5,-2,1,3,6,8,3,8
-5,-2,1,3,6,8,3,8
-5,-2,1,3,3,6,8,8

detail explanation

array length = 8
i = 0 ;
j = length/2; => 4

indices 0 1 2 3 4 5 6 7
values 1 3 6 8 -5 -2 3 8

compare if (a[i] > a[j]) i.e a and a ( 1 > -5 in our example)
place a[j] in a[i] and shift every indices by one
i.e. each a[i+1]= a[i] till a[j];
i++, j++

indices 0 1 2 3 4 5 6 7
values -5 1 3 6 8 -2 3 8

now compare a with a ( 1 > -2) , replace a with a ans shift

indices 0 1 2 3 4 5 6 7
values -5 -2 1 3 6 8 3 8

i++, j++;

compare a with a this time a[i] < a[j], no swap and shifting is required, but dont increase j only i

indices 0 1 2 3 4 5 6 7
values -5 -2 1 3 6 8 3 8
i++; j is not increased

now a with a still a is less than a so increase i only
indices 0 1 2 3 4 5 6 7
values -5 -2 1 3 6 8 3 8

now a with a, swap and shift, we get

indices 0 1 2 3 4 5 6 7
values -5 -2 1 3 3 6 8 8
i++, j++

now a with a, no change required, increase only i;

similalry done

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0
of 0 vote

Here is my algo:
Consider an array: [1, 3, 6, 8, -5, -2, 3, 8]
We have to maintain two pointers 'endOfTheFirstHalf' and 'endOfTheSecondHalf'. Both pointers will indicate the maximum element in given half.
1) At the beginning the first pointer is indicating the element at index 3 = 8 whereas second pointer index 7 which is also equal 8.
2) Move second pointer till you find element which is smaller then max element in first half indicated by first pointer
in this case it is index 6 = 3;
3) Shift left all elements located between firstPtr and secPtr by one to make place for max element e.g. after first swap original array looks like this: [1, 3, 6, -5, -2, 3, 8, 8] (elements -5, -2, 3 were shift left by one position)
4) Assign max element firstPtr to its position - move element at index 3 = 8 to index 6 where previously was an element = 3
5) Repeat the process till all elements in first half are smaller the min element in sec half. Here is the output from my program:
[1, 3, 6, 8, -5, -2, 3, 8]
[1, 3, 6, -5, -2, 3, 8, 8]
[1, 3, -5, -2, 3, 6, 8, 8]
[1, -5, -2, 3, 3, 6, 8, 8]
[-5, -2, 1, 3, 3, 6, 8, 8]

Code:

``````public static void merge2(int arr[]) {

int endOfSecHalf = arr.length - 1;
int endOfFirstHalf = arr.length / 2 - 1;
System.out.println(Arrays.toString(arr));
while (endOfFirstHalf >= 0) {
// find first smaller element in sec half then the max el in first half
while (arr[endOfSecHalf] >= arr[endOfFirstHalf] && endOfSecHalf > endOfFirstHalf) {
endOfSecHalf--;
}
// all elements in sec half are bigger the elements in the first
// half
if (endOfSecHalf == endOfFirstHalf) {
// break;
return;
}
int temp = arr[endOfFirstHalf];
for (int i = endOfFirstHalf + 1; i <= endOfSecHalf; i++) {
arr[i - 1] = arr[i];
}
arr[endOfSecHalf] = temp;
endOfSecHalf--;
endOfFirstHalf--;
System.out.println(Arrays.toString(arr));
}
}``````

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

It is a simple case of Insertion Sort. Where we don't need any extra space.

void SortMyArray(int A[], int size)
{
for (int i=1; i < n; i++)
{
//we have to maintain A[i], usually we have to store it in a key variable, while we will not use in this case (no extra space)

j = i-1;

while ( j >= 0 && A[j] > A[i] )
{
A[j+1] = A[j];
j = j-1;
}
A[j+1] = A[i];
}
}

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

This can be solved with Insertion sort algorithm.

void SortMyArray(int A[], int n)
{
int i, j;
for( i=1; i < n; i++)
{
//We usually maintain A[i] in a variable which is not required in this case(no extra space)

j = i-1;

while ( j >= 0 && A[j] > A[i] )
{
A[j+1] = A[j];
j = j-1;
}
A[j+1] = A[i];
}
}

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

I have solved this problem in c#. Compare the first and mid element on the array, like that we compare the all the array elements. When we find smaller number in the second set of array, need to swap the array elements.

void mergetwoSortedRangeArray(int[] arr)
{
int i=0;
int j=(int)arr.Length/2;
while(i<j && j< arr.Length -1)
{

if (arr[i] > arr[j])
{
rotateArray(arr, i, j);
j++;
}
i++;
}
}

void rotateArray(int[] arr, int startInd, int endInd)
{
int newInd = startInd, temp = 0;
int? prevValue = null;

if ((endInd - startInd) > 1)
{
for (int i = startInd; i <= endInd; i++)
{
if (newInd == endInd)
newInd = startInd;
else
newInd++;

temp = arr[newInd];
if (prevValue == null)
arr[newInd] = arr[i];
else
arr[newInd] = (int)prevValue;
prevValue = temp;
}
}
}

Let me know if anyone find better solution

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0

For input array: {1, 3, 6, 8, -5, -2, 3, 8 }
{1, 1, 1, 1, 1, 1, 3, 8}

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0

I have tested this algorithm with different inputs. It works without any issue. Can you check it out what is wrong on this algorithm?

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1
of 1 vote

Your rotateArray doesn't "rotate" a range with just two elements. In that case endInd = startEnd + 1. So, endInd - startInd is 1, and rotateArray does nothing. Change the test to
if ((endInd - startInd) >= 1)
The other mistake you have is that mergeSortedRangeArray skips the very last element.
Change the while condition to
while (i < j && j < arr.length)
Try you current implementation on {1, 4, 2, 3}.
Unfortunately your solution is O(n^2). It is just an insert sort. It is possible to do the merge in O(n) time.

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

``````#include<iostream>
using namespace std;
int main()
{
int i,j,temp,n=8;
int a={1,3,6,8,-5,-2,3,8};
i=0,j=4;
while(i<j && j<n)
{
if(a[j] <= a[i])
{
temp=a[j];
for(int k=j-1;k>=i;k--)
{
a[k+1]=a[k];
}
a[i]=temp;
i++;
j++;
}
else if(a[i]<a[j])i++;
}

for(i=0;i<n;i++)cout<<a[i]<<" ";
cout<<endl;
}``````

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