Amazon Interview Question for Quality Assurance Engineers


Country: India
Interview Type: In-Person




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

Can be solved by using the merge sort merging logic.

- azambhrgn January 23, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
1
of 1 vote

int[] ar1 = { 1,3,5,7,9,11,45,67};
                int[] ar2 = { 2,4,6,8,9,11,34,56,789,789,999,1000};
                int[] result;
                int maxarrlen = (ar1.Length > ar2.Length) ? ar1.Length : ar2.Length ;
                int minarrlen = (ar1.Length > ar2.Length) ? ar2.Length : ar1.Length ;
                
                result = new int[maxarrlen+minarrlen];
                int counter = 0;
                int total=maxarrlen + minarrlen;
                for (int i = 0; i < maxarrlen; i++)
                {
                    if (i < minarrlen)
                    {
                        if (ar1[i] >= ar2[i])
                        {
                            result[counter++] = ar2[i];
                            result[counter++] = ar1[i];
                        }
                        else
                        {
                            result[counter++] = ar1[i];
                            result[counter++] = ar2[i];
                        }
                    }
                    else
                    {
                        result[counter++] = ar2[i];
                    }
                }
                foreach (var item in result)
                {
                    Console.Write(string.Format("{0},", item));
                }
                Console.ReadLine();

- Mallikarjun Birajdar February 05, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
1
of 1 vote

int[] A={2,2,4,6,10,14,20};
int[] B={2,3,3,4,8,10};

int lengthC = A.length + B.length;
if(lengthC==0)
{ System.out.println("Length of Both Arrays is 0"); }
else
{
int[] C = new int[lengthC];
for(int i=0,j=0,k=0;k<lengthC;i++,j++,k++)
{
if(j==B.length)
{
C[k]=A[i];
j--;
}
else
if(i==A.length)
{
C[k]=B[j];
i--;
}
else
{
if(A[i]<=B[j])
{
C[k]=A[i];
j--;
}
else
{
C[k]=B[j];
i--;
}
}
System.out.print(C[k]+" ");
}
}

- Puneet Gupta February 11, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

func Merge(A, B []int) []int {
	C := make([]int, len(A)+len(B))
	ia := 0
	ib := 0
	ic := 0
	for ia < len(A) || ib < len(B) {
		if ia == len(A) {
			C[ic] = B[ib]
			ib++
			ic++
			continue
		}
		if ib == len(B) {
			C[ic] = A[ia]
			ia++
			ic++
			continue
		}
		if A[ia] < B[ib] {
			C[ic] = A[ia]
			ia++
		} else {
			C[ic] = B[ib]
			ib++
		}
		ic++
	}
	return C
}

func main() {
	A := []int{3, 6, 6, 7}
	B := []int{3, 3, 8, 10}
	fmt.Println(Merge(A, B))
}

- dmitry.labutin January 23, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Agree with comments above, this task is about the 'merge' subroutine of Merge Sort.
My implementation in Java

static int[] mergeTwoSortedArrays(int[] a, int[] b) {
    // Corner cases
    if (a == null && b == null) {
      return null;
    }
    if (a == null) {
      return Arrays.copyOf(b, b.length);
    }
    if (b == null) {
      return Arrays.copyOf(a, a.length);
    }

    // Common cycle
    int[] c = new int[a.length + b.length];

    int i = 0, j = 0, k = 0;
    while (i < a.length && j < b.length) {
      if (a[i] < b[j]) {
        c[k] = a[i];
        i++;
      } else {
        c[k] = b[j];
        j++;
      }
      k++;
    }

    // ends
    if (i == a.length) {
      while (j < b.length) {
        c[k++] = b[j++];
      }
    } else {
      while (i < a.length) {
        c[k++] = a[i++];
      }
    }

    return c;
  }

- Mike L January 23, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static int[] sort(int[] arr1, int[] arr2) {
		
		if (arr1 == null || arr1 == null)
			return null;
		
		// indexes
		int i1 = 0, i2 = 0, idx = 0;
		
		// new arr length = arr1 + arr2
		int[] arr = new int[arr1.length + arr2.length];

		// loop until reached end of either array
		while(i1 < arr1.length && i2 < arr2.length) {
			// add smallest element, increment idxs
			arr[idx++] = (arr1[i1] < arr2[i2]) ? 
					arr1[i1++] : arr2[i2++];
		}
		
		// One array will still have data. try both
		System.arraycopy(arr1, i1, arr, idx, arr1.length-i1);
		System.arraycopy(arr2, i2, arr, idx, arr2.length-i2);
		
		return arr;
	}

- Anonymous January 24, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

array<int, 3> a { { 1, 3, 7} };
	array<int, 4> b { { 0, 2, 4, 6} };

	array<int, (a.size()+b.size())> c;

	auto cIt = c.rbegin();
	auto aIt = a.rbegin();
	auto bIt = b.rbegin();	

	while (true) {
		if (aIt == a.rend() || bIt == b.rend())
			break;

		if (*aIt > *bIt) {
			*cIt = *aIt;
			cIt++; aIt++; 
		}
		else {
			*cIt = *bIt;
			cIt++; bIt++; 
		}
	}

	while (bIt != b.rend()) {
		*cIt = *bIt;
		cIt++; bIt++;
	}

	while (aIt != a.rend()) {
		*cIt = *aIt;
		cIt++; aIt++;
	}

- this January 24, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class sort {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String [] str1 = br.readLine().split(" ");
        String [] str2 = br.readLine().split(" ");
        int size1 = str1.length;
        int size2 = str2.length;
        int []array1= new int[size1];
        int []array2= new int[size2];
        for(int i=0; i<size1; i++){
            array1[i] = Integer.parseInt(str1[i]);
        }
        for(int j=0; j<size2; j++){
            array2[j] =Integer.parseInt(str2[j]);
        }
        int [] result = mergeArray(array1, array2);
        for(int value: result){
            System.out.print(value+" ");
        }
    }
    public static int [] mergeArray(int [] array1, int [] array2){
        int [] result = null;
        if(array1==null && array2==null){
            return result;
        }
        if(array1==null){
            return array2;
        }
        if(array2 ==null){
            return array1;
        }
        int size1= array1.length;
        int size2 = array2.length;
        int k=0, i=0, j=0;
        result = new int[size1+size2];
        while (i<size1 && j<size2){
            if(array1[i]<array2[j]){
                result[k++] = array1[i++];
            }else{
                result[k++] = array2[j++];
            }
        }
        while (i<size1){
            result[k++] = array1[i++];
        }
        while (j<size2){
            result[k++] = array2[i++];
        }
        return result;
    }
}

- neelabhsingh January 25, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

a=[1,2,3,3,5,8]
b=[2,3,6,9,10]
d={}
for i in (a+b):
try:
d[i] += 1
except:
d[i] =1
for key in d.keys():
ii=1
while ii <= d[key]:
c.append(key)
ii += 1

- Anonymous January 25, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static class Merger
    {
        public static IEnumerable<int> Merge(IEnumerable<int> a, IEnumerable<int> b)
        {
            if (a == null && b == null)
                yield break;

            if (a == null)
            {
                foreach (var item in b)
                    yield return item;
                yield break;
            }
                
            if (b == null)
            {
                foreach (var item in a)
                    yield return item;

                yield break;
            }

            using (var leftEnumerator = a.GetEnumerator())
            using (var rightEnumerator = b.GetEnumerator())
            {
                var leftHasValue = leftEnumerator.MoveNext();
                var rightHasValue = rightEnumerator.MoveNext();

                while (leftHasValue || rightHasValue)
                {
                    if (!leftHasValue)
                    {
                        var val = rightEnumerator.Current;
                        rightHasValue = rightEnumerator.MoveNext();
                        yield return val;
                        continue;
                    }

                    if (!rightHasValue)
                    {
                        var val = leftEnumerator.Current;
                        leftHasValue = leftEnumerator.MoveNext();
                        yield return val;
                        continue;
                    }

                    var leftValue = leftEnumerator.Current;
                    var rightValue = rightEnumerator.Current;

                    if (leftValue <= rightValue)
                    {
                        leftHasValue = leftEnumerator.MoveNext();
                        yield return leftValue;
                    }
                    else
                    {
                        rightHasValue = rightEnumerator.MoveNext();
                        yield return rightValue;
                    }
                }
            }
        }
    }

- aurelian.lica January 25, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static class Merger
{
public static IEnumerable<int> Merge(IEnumerable<int> a, IEnumerable<int> b)
{
if (a == null && b == null)
yield break;

if (a == null)
{
foreach (var item in b)
yield return item;
yield break;
}

if (b == null)
{
foreach (var item in a)
yield return item;

yield break;
}

using (var leftEnumerator = a.GetEnumerator())
using (var rightEnumerator = b.GetEnumerator())
{
var leftHasValue = leftEnumerator.MoveNext();
var rightHasValue = rightEnumerator.MoveNext();

while (leftHasValue || rightHasValue)
{
if (!leftHasValue)
{
var val = rightEnumerator.Current;
rightHasValue = rightEnumerator.MoveNext();
yield return val;
continue;
}

if (!rightHasValue)
{
var val = leftEnumerator.Current;
leftHasValue = leftEnumerator.MoveNext();
yield return val;
continue;
}

var leftValue = leftEnumerator.Current;
var rightValue = rightEnumerator.Current;

if (leftValue <= rightValue)
{
leftHasValue = leftEnumerator.MoveNext();
yield return leftValue;
}
else
{
rightHasValue = rightEnumerator.MoveNext();
yield return rightValue;
}
}
}
}
}

- aurelian.lica January 25, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

C++ solution. Pretty simple.

#include <vector>
#include <type_traits>
#include <iostream>

template<typename IteratorType, typename ResultType>
void pushRest(IteratorType && it, IteratorType && end, ResultType && result)
{
	while(it != end)
		result.push_back(*it++);
}

template<typename IteratorType, typename ValueType, typename ResultType>
IteratorType pushUntilGreaterEqual(IteratorType && it, IteratorType && end, ValueType && comparand, ResultType && result)
{
	while(it != end)
	{
		auto value = *it;

		if(value >= comparand)
			break;

		result.push_back(value);
		it++;
	}

	return it;
}

template<typename IteratorType, typename ValueType>
IteratorType skipDuplicates(IteratorType && it, IteratorType && end, ValueType && value)
{
	while(it != end && value == *it)
		it++;

	return it;
}

template<typename T>
typename std::remove_reference<T>::type merge(T &&left, T &&right)
{
	typename std::remove_reference<T>::type result;

	auto lit = left.cbegin();
	auto rit = right.cbegin();

	auto lend = left.cend();
	auto rend = right.cend();

	for(;;)
	{
		if(lit == lend)
		{
			// done with left, push rest of right
			pushRest(rit, rend, result);
			break;
		}
		else if(rit == rend)
		{
			// done with right, push rest of left
			pushRest(lit, lend, result);
			break;
		}
		else
		{
			auto leftValue = *lit;
			auto rightValue = *rit;

			if(leftValue == rightValue)
			{
				// equal values, push one and advance until not equal
				result.push_back(leftValue);
				lit = skipDuplicates(lit, lend, leftValue);
				rit = skipDuplicates(rit, rend, rightValue);
			}
			else if(leftValue < rightValue)
			{
				// left is lower, push values from left until this isn't so
				lit = pushUntilGreaterEqual(lit, lend, rightValue, result);
			}
			else if(rightValue < leftValue)
			{
				// right is lower, push values from right until this isn't so
				rit = pushUntilGreaterEqual(rit, rend, leftValue, result);
			}
		}
	}

	return result;
}

int main(void)
{
	std::vector<int> a {0, 0, 1, 5, 7, 19, 19, 19, 19, 21};
	std::vector<int> b {-50, -45, -12, 19, 19, 19, 19, 19, 19, 19, 20, 22, 5000};

	auto result = merge(a, b);

	for(auto value : result)
		std::cout << value << " ";

	std::cout << std::endl;

	return 0;
}

- snichols@therealm.io January 25, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

//Autor : prizkaboom 
static public void GetNewSortArray(int A[],int B[])
 {
 
 
   ArrayList C = new ArrayList();
   int i=0,j=0;
   
   while(i<A.length && j<B.length)
   {
     
         if(A[i]< B[j])
         {
             C.add(A[i]);
             i++;
         
         }
         
         if(A[i]>B[j]){
         
           C.add(B[j]);
           j++;
         }
         
          if(A[i]==B[j])
          {
         
           C.add(B[j]);
           C.add(A[i]);
           j++;
           i++;
         }
         
     
   } 
   
    while(j<B.length)
   {
        
         C.add(B[j]);
           j++;
         
   } 
    
     while(i<A.length)
   {
         
         C.add(A[i]);
           i++;
         
   }  
       System.out.println(C);
     

 }

- prizkaboom January 26, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

d={}
d['a']= 1; d['b'] = 2; d['c'] = 3; d['d'] = 4
n = 2
d.keys()[d.values().index(n)]

- mary January 26, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

d={}
d['a']= 1; d['b'] = 2; d['c'] = 3; d['d'] = 4
n = 2
d.keys()[d.values().index(n)]

- mary January 26, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include <iostream>
#include <vector>

using namespace std;

void fv(vector<int>&v)
{
    unsigned int i = 0;
        while(i<=v.size()) {
            if(v[i]==-1) {
                ++i;
                continue;
            }
            for(unsigned int j=i+1;j<=v.size();j++ ) {
                if(v[i]==v[j])
                v[j]=-1;
            }
            ++i;
        }
}

void pv(vector<int>&v){
        for(vector<int>::iterator it=v.begin();it!=v.end();it++)
        cout<<*it<<" ";
}

vector<int> merge(vector<int>&v1, vector<int>&v2) {
    for(unsigned int i=0;i<=v1.size();i++)
    for(unsigned int j=0;j<=v2.size();j++)
    {
        if(v1[i]==v2[j])
        v2[j]=-1;
    }
    
    vector<int>v3;
    unsigned int i;
    if(v1.size()>v2.size())
    i=v1.size();
    else
    i=v2.size();
    
    for(unsigned int j = 0; j<i; j++){
        if(v1.size()>j && v1[j]!=-1)
        v3.push_back(v1[j]);
        if(v2.size()>j && v2[j]!=-1)
        v3.push_back(v2[j]);
    }
    
    return v3;
    
}

int main()
{
vector<int> v1 = {1,1,1,2,5};
vector<int> v2 = {2,2,2,5,9,11,19};

fv(v1);
fv(v2);

vector<int>v3 = merge(v1, v2);
pv(v3);

- SREE.gooogle February 03, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Can construct a binary search tree with one array and insert one element at a time from the other array. I guess the insertion complexity would be nlogm where n is the length of the BST array and m of the other.

- Anonymous February 04, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

class FindSortedArray():
    array1=list()
    array2=list()
    sortedarray=list()
    def __init__(self,array1,array2):
        self.array1 = array1
        self.array2 = array2
    def AddElemIntoSortedArray(self,elem):
        for i in range(len(self.sortedarray)):
            if self.sortedarray[i] > elem:
                self.sortedarray.insert(i,elem)
                return
        self.sortedarray.append(elem)

    def SortArrays(self):
        for i in range(max(len(self.array1),len(self.array2))):
            if i <= len(self.array1)-1 :
                arry1_element=self.array1[i]
            else:
                arry2_element=self.array2[i]
                self.AddElemIntoSortedArray(arry2_element)
                continue

            if i <= len(self.array2)-1 :
                arry2_element=self.array2[i]
            else:
                arry1_element=self.array1[i]
                self.AddElemIntoSortedArray(arry1_element)
                continue
            if arry1_element < arry2_element :
                self.AddElemIntoSortedArray(arry1_element)
                self.AddElemIntoSortedArray(arry2_element)
            elif arry1_element >= arry2_element:
                self.AddElemIntoSortedArray(arry2_element)
                self.AddElemIntoSortedArray(arry1_element)
        return self.sortedarray
if __name__ == "__main__":
    s=FindSortedArray([5,10,3,9],[1,40,4,30,9,9,9])
    print s.SortArrays()

- ankitpatel2100 February 05, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

int[] ar1 = { 1,3,5,7,9,11,45,67};
                int[] ar2 = { 2,4,6,8,9,11,34,56,789,789,999,1000};
                int[] result;
                int maxarrlen = (ar1.Length > ar2.Length) ? ar1.Length : ar2.Length ;
                int minarrlen = (ar1.Length > ar2.Length) ? ar2.Length : ar1.Length ;
                
                result = new int[maxarrlen+minarrlen];
                int counter = 0;
                int total=maxarrlen + minarrlen;
                for (int i = 0; i < maxarrlen; i++)
                {
                    if (i < minarrlen)
                    {
                        if (ar1[i] >= ar2[i])
                        {
                            result[counter++] = ar2[i];
                            result[counter++] = ar1[i];
                        }
                        else
                        {
                            result[counter++] = ar1[i];
                            result[counter++] = ar2[i];
                        }
                    }
                    else
                    {
                        result[counter++] = ar2[i];
                    }
                }
                foreach (var item in result)
                {
                    Console.Write(string.Format("{0},", item));
                }
                Console.ReadLine();

- Mallikarjun Birajdar February 05, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

int[] A={2,2,4,6,10,14,20};
int[] B={2,3,3,4,8,10};
int lengthC = A.length + B.length;
if(lengthC==0) { System.out.println("Length of Both Arrays is 0"); }
else { int[] C = new int[lengthC];
for(int i=0,j=0,k=0;k<lengthC;i++,j++,k++)
{
if(j==B.length)
{
C[k]=A[i];
j--;
}
else
if(i==A.length)
{
C[k]=B[j];
i--;
}
else
{
if(A[i]<=B[j])
{
C[k]=A[i];
j--;
}
else
{
C[k]=B[j];
i--;
}
}
System.out.print(C[k]+" ");
}

- Puneet Gupta February 11, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

int[] A={2,2,4,6,10,14,20};
int[] B={2,3,3,4,8,10};

int lengthC = A.length + B.length;
if(lengthC==0)
{ System.out.println("Length of Both Arrays is 0"); }
else
{
int[] C = new int[lengthC];
for(int i=0,j=0,k=0;k<lengthC;i++,j++,k++)
{
if(j==B.length)
{
C[k]=A[i];
j--;
}
else
if(i==A.length)
{
C[k]=B[j];
i--;
}
else
{
if(A[i]<=B[j])
{
C[k]=A[i];
j--;
}
else
{
C[k]=B[j];
i--;
}
}
System.out.print(C[k]+" ");
}
}

- puneetgupta2212 February 11, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

a1 = [4,6,7,9,10]
a2 = [3,4,8,10,45,67]
d = a1
maxVal=d[len(d) -1]
minVal=d[0]

for i in range(len(a2)):
  if a2[i] < minVal:
    d.insert(0, a2[i])
  elif(a2[i] > maxVal):
    d.append(a2[i])
  else:
   for j in range(i,len(d)):
    if a2[i] > d[j] and a2[i] < d[j + 1]:
      d.insert(j,a2[i])
      break
    elif(a2[i] == d[j]):
      d.insert(j,a2[i])
      break

#quicksort(nl, 0, len(nl) -1)
print(d)

- Vivek Dubey March 04, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

a1 = [4,6,7,9,10]
a2 = [3,4,8,10,45,67]
d = a1
maxVal=d[len(d) -1]
minVal=d[0]

for i in range(len(a2)):
  if a2[i] < minVal:
    d.insert(0, a2[i])
  elif(a2[i] > maxVal):
    d.append(a2[i])
  else:
   for j in range(i,len(d)):
    if a2[i] > d[j] and a2[i] < d[j + 1]:
      d.insert(j,a2[i])
      break
    elif(a2[i] == d[j]):
      d.insert(j,a2[i])
      break

#quicksort(nl, 0, len(nl) -1)
print(d)

- Vivek Dubey March 04, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

import java.util.*;
class ArrayMerge2<T extends Comparable<T>>
{

public void buildStack(Stack<T> s1,T [] ar1)
{
for(int i=ar1.length-1;i>=0;i--)
{
if(!s1.empty())
{
if(s1.peek()==ar1[i])
{}
else
s1.push(ar1[i]);
}
else
s1.push(ar1[i]);
}

}//buildStack

public void mergeStack(Stack<T> s1,Stack<T> s2,Vector<T> v1) 
{
T x,y;
while(!s1.empty()&&!s2.empty())
{
x= s1.peek();
y= s2.peek();
if(x.compareTo(y)==0)
{
v1.addElement(x);
x= s1.pop();
y= s2.pop();
}
else if(x.compareTo(y)<0)
{
v1.addElement(x);
x= s1.pop();
}
else
{
v1.addElement(y);
y=s2.pop();
}
}

}//mergeStack



public static void main(String args[])
{
Stack<Integer> s1=new Stack<Integer>();
ArrayMerge2<Integer> am1=new ArrayMerge2<Integer>();
Integer[] ar1={1,2,3,4,4,5,6,7,8,9,11,13,13,14,15,26};
am1.buildStack(s1,ar1);
System.out.println("Stack contents"+s1);

Stack<Integer> s2=new Stack<Integer>();
Integer[] ar2={1,5,6,7,8,9,10,11,12,13,13,14,15,22,26};
am1.buildStack(s2,ar2);
System.out.println("Stack contents"+s2);

Vector<Integer> v1=new Vector<Integer>(3,2);
//Vector v1=new Vector(3,2);
am1.mergeStack(s1,s2,v1);
System.out.println("Vector contents"+v1);


}



}//end of class

- Anonymous March 28, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

private static void merger2Arrays(int[] a, int[] b) {

		int[] c = new int[a.length + b.length];
		char value = 0;
		int counterC = 0;
		int counterA = 0;
		int counterB = 0;

		// loop for Array C to put the char
		for (int loopC = 0; loopC < c.length; loopC++) {

			// Check if A array is already moved
			if (counterA > a.length - 1) {
				// then move all B into C
				c[loopC] = b[counterB];
				counterB++;
			}
			// Check if B array is already moved
			else if (counterB > b.length - 1) {
				// then move all B into C
				c[loopC] = a[counterA];
				counterA++;
			}

			// If A is smaller or equal to
			else if (a[counterA] <= b[counterB]) {
				// then add value of a in c
				c[loopC] = a[counterA];
				// increment the counter of A and C
				counterA++;
			}
			// If B is smaller
			else if (a[counterA] > b[counterB]) {
				// then add value of a in c
				c[loopC] = b[counterB];
				// increment the counter of B
				counterB++;
			} else {
				System.out.println("uncovered path");
			}

		}
		for (int d : c) {
			System.out.println(d);
		}
		// End of method
	}

	public static void main(String[] args) {
		int[] a = { 1, 2};
		int[] b = { 0, 3 };

		merge(a, b);
	}

- Nitesh April 15, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Python::

Descending order::
>>> def man(l1, l2):
c = []
for i in l2:
c.append(i)
for j in l1:
c.append(j)
c.sort(reverse=True)
return c

>>> man([2,3,4,2,3], [4,5,6,5,6])
[6, 6, 5, 5, 4, 4, 3, 3, 2, 2]

Ascending Order::
>>> def man(l1, l2):
c = []
for i in l2:
c.append(i)
for j in l1:
c.append(j)
c.sort()
return c

>>> man([2,3,4,2,3], [4,5,6,5,6])
[2, 2, 3, 3, 4, 4, 5, 5, 6, 6]
>>>

- Madhu Mohan May 17, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class Learn {
	public static void main(String args[]) {
		int arr1[] = { 1, 2, 3, 4, 5, 6, 7, 8 };
		int arr2[] = { 6, 7, 8, 9 };
		int arr3[] = new int[arr1.length + arr2.length];
		int i = 0, j = 0, k = 0;
		while (i < arr1.length && j < arr2.length) {
			if (arr1[i] < arr2[j]) {
				arr3[k] = arr1[i];
				++i;
				++k;
			} else if (arr2[j] < arr1[i]) {
				arr3[k] = arr2[j];
				++j;
				++k;
			} else {
				arr3[k] = arr1[i];
				++i;
				++j;
				++k;

			}
		}
		if (j < arr2.length) {
			while (j < arr2.length) {
				arr3[k] = arr2[j];
				++j;
				++k;
			}
		} else {
			while (i < arr1.length) {
				arr3[k] = arr1[i];
				++i;
				++k;
			}
		}
		for (int x : arr3) {
			System.out.println(x);
		}
	}
}

- Deepak September 13, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class Learn {
	public static void main(String args[]) {
		int arr1[] = { 1, 2, 3, 4, 5, 6, 7, 8 };
		int arr2[] = { 6, 7, 8, 9 };
		int arr3[] = new int[arr1.length + arr2.length];
		int i = 0, j = 0, k = 0;
		while (i < arr1.length && j < arr2.length) {
			if (arr1[i] < arr2[j]) {
				arr3[k] = arr1[i];
				++i;
				++k;
			} else if (arr2[j] < arr1[i]) {
				arr3[k] = arr2[j];
				++j;
				++k;
			} else {
				arr3[k] = arr1[i];
				++i;
				++j;
				++k;

			}
		}
		if (j < arr2.length) {
			while (j < arr2.length) {
				arr3[k] = arr2[j];
				++j;
				++k;
			}
		} else {
			while (i < arr1.length) {
				arr3[k] = arr1[i];
				++i;
				++k;
			}
		}
		for (int x : arr3) {
			System.out.println(x);
		}
	}
}

- Deepak September 13, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class SortedArrayFromGiven2SortedArrays {


	public void getArr(int[] array1, int[] array2){
		int arr3[] = new int[array1.length + array2.length];
		int i = 0, j = 0, k = 0;
		while (i < array1.length && j < array2.length) {
			if (array1[i] < array2[j]) {
				arr3[k] = array1[i];
				++i;
				++k;
			} else if (array2[j] < array1[i]) {
				arr3[k] = array2[j];
				++j;
				++k;
			} else if(array2[j]==array1[i]){
				arr3[k] = array1[i];
				++k;
				arr3[k] = array2[j];
				++i;
				++j;
				++k;

			}
		}
		if (j < array2.length) {
			while (j < array2.length) {
				arr3[k] = array2[j];
				++j;
				++k;
			}
		} else {
			while (i < array1.length) {
				arr3[k] = array1[i];
				++i;
				++k;
			}
		}
		for (int x : arr3) {
			System.out.println(x);
		}
		System.out.println("");
	}
	public static void main(String args[]){
		SortedArrayFromGiven2SortedArrays s = new SortedArrayFromGiven2SortedArrays();
		int[] ar1 = {1,4,5,7,9,12};
		int[]ar2 = {2,3,6,7,8,10,11,13,14,15};
				s.getArr(ar1, ar2);

		int[] ar3 = {};
				s.getArr(ar3, ar2);

		int[] ar4 = {0,0,0,0,0,0,0,0,0,0,0,0};
				s.getArr(ar1, ar4);

		int[] ar5 = {-1,4,5,7,9,12};
		int[] ar6 = {-3,-2,2,3,6,7,8,10,11,13,14,15};
		s.getArr(ar5, ar6);
	}
}

- anonymousNg November 12, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

O(m+n) solution s.t. m is first array's length and n is the second's array length

public static Integer[] merge(Integer[] arr1, Integer[] arr2) {
		int n = arr1.length+arr2.length;
		Integer[] arr = new Integer[n];
		int arr1Index = 0;
		int arr2Index = 0;
		for(int i=0;i<n;i++) {
			if(arr1Index==arr1.length) {
				arr[i] = arr2[arr2Index++];
			} else if(arr2Index==arr2.length) {
				arr[i] = arr1[arr1Index++];
			} else if(arr1[arr1Index]<arr2[arr2Index]) {
				arr[i] = arr1[arr1Index++];
			} else {
				arr[i] = arr2[arr2Index++];				
			}
		}
		return arr;
	}

- engeloded November 27, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public void newarray()
{
int []a={1,2,3,4,5,6,7,10,100,1000};
int []b={4,5,6,7,10,45};
int x=0;
if (a.length>b.length)
{
int [] c= new int[a.length];
for(int i=0;i<a.length;i++)
{
for(int j=0;j<b.length;j++)
{
if(a[i]==b[j])
{
c[j]=a[i];
x++;
}
}
}
for(int j=0;j<x;j++)
System.out.print(c[j]+"\t");
}
else
{
int [] c= new int[b.length];
for(int i=0;i<b.length;i++)
{
for(int j=0;j<a.length;j++)
{
if(b[i]==a[j])
{
c[j]=b[i];
x++;
}
}
}
for(int j=0;j<x;j++)
System.out.print(c[j]+"\t");
}
}

- Manjuraj December 13, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public void newarray()
		{
			int []a={1,2,3,4,5,6,7,10,100,1000};
			int []b={4,5,6,7,10,45};
			int x=0;
			if (a.length>b.length)
			{
				int [] c= new int[a.length];
				for(int i=0;i<a.length;i++)
				{
					for(int j=0;j<b.length;j++)
					{
					    if(a[i]==b[j])
					    {
					          c[j]=a[i];
					          x++;
					    }
					}
				}
				for(int j=0;j<x;j++)
				System.out.print(c[j]+"\t");
			}
			else
			{
				int [] c= new int[b.length];
				for(int i=0;i<b.length;i++)
				{
					for(int j=0;j<a.length;j++)
					{
					    if(b[i]==a[j])
					    {
					          c[j]=b[i];
					          x++;
					    }
					}
				}
				for(int j=0;j<x;j++)
				System.out.print(c[j]+"\t");
			}

}

- Manjuraj December 13, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public void newarray()
{
int []a={1,2,3,4,5,6,7,10,100,1000};
int []b={4,5,6,7,10,45};
int x=0;
if (a.length>b.length)
{
int [] c= new int[a.length];
for(int i=0;i<a.length;i++)
{
for(int j=0;j<b.length;j++)
{
if(a[i]==b[j])
{
c[j]=a[i];
x++;
}
}
}
for(int j=0;j<x;j++)
System.out.print(c[j]+"\t");
}
else
{
int [] c= new int[b.length];
for(int i=0;i<b.length;i++)
{
for(int j=0;j<a.length;j++)
{
if(b[i]==a[j])
{
c[j]=b[i];
x++;
}
}
}
for(int j=0;j<x;j++)
System.out.print(c[j]+"\t");
}
}

- Manjuraj December 13, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public void newarray()
{
int []a={1,2,3,4,5,6,7,10,100,1000};
int []b={4,5,6,7,10,45};
int x=0;
if (a.length>b.length)
{
int [] c= new int[a.length];
for(int i=0;i<a.length;i++)
{
for(int j=0;j<b.length;j++)
{
if(a[i]==b[j])
{
c[j]=a[i];
x++;
}
}
}
for(int j=0;j<x;j++)
System.out.print(c[j]+"\t");
}

else
{
int [] c= new int[b.length];
for(int i=0;i<b.length;i++)
{
for(int j=0;j<a.length;j++)
{
if(b[i]==a[j])
{
c[j]=b[i];
x++;
}
}
}
for(int j=0;j<x;j++)
System.out.print(c[j]+"\t");
}
}

- Manjuraj December 13, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class array{

public static void main(String[] args)
{
int []a={1,2,3,4,5,6,7,10,100,1000};
int []b={4,5,6,7,10,45};
int x=0;
if (a.length>b.length)
{
int [] c= new int[a.length];
for(int i=0;i<a.length;i++)
{
for(int j=0;j<b.length;j++)
{
if(a[i]==b[j])
{
c[j]=a[i];
x++;
}
}
}
for(int j=0;j<x;j++)
System.out.print(c[j]+"\t");
}

else
{
int [] c= new int[b.length];
for(int i=0;i<b.length;i++)
{
for(int j=0;j<a.length;j++)
{
if(b[i]==a[j])
{
c[j]=b[i];
x++;
}
}
}
for(int j=0;j<x;j++)
System.out.print(c[j]+"\t");
}
}
}

- Manjuraj December 13, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

class Array{ public void NewArray()
{
int []a={1,2,3,4,5,6,7,10,100,1000};
int []b={4,5,6,7,10,45};
int x=0;
if (a.length>b.length)
{
int [] c= new int[a.length];
for(int i=0;i<a.length;i++)
{
for(int j=0;j<b.length;j++)
{
if(a[i]==b[j])
{
c[j]=a[i];
x++;
}
}
}
for(int j=0;j<x;j++)
System.out.print(c[j]+"\t");
}

else
{
int [] c= new int[b.length];
for(int i=0;i<b.length;i++)
{
for(int j=0;j<a.length;j++)
{
if(b[i]==a[j])
{
c[j]=b[i];
x++;
}
}
}
for(int j=0;j<x;j++)
System.out.print(c[j]+"\t");
}
}
public static void main(String[] args)
{
Array a = new Array();
l.NewArray();
}
}

- Anonymous December 13, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Class array {
public void newarray()
{
int []a={1,2,3,4,5,6,7,10,100,1000};
int []b={4,5,6,7,10,45};
int x=0;
if (a.length>b.length)
{
int [] c= new int[a.length];
for(int i=0;i<a.length;i++)
{
for(int j=0;j<b.length;j++)
{
if(a[i]==b[j])
{
c[j]=a[i];
x++;
}
}
}
for(int j=0;j<x;j++)
System.out.print(c[j]+"\t");
}

else
{
int [] c= new int[b.length];
for(int i=0;i<b.length;i++)
{
for(int j=0;j<a.length;j++)
{
if(b[i]==a[j])
{
c[j]=b[i];
x++;
}
}
}
for(int j=0;j<x;j++)
System.out.print(c[j]+"\t");
}
}
public static void main(String[] args)
{
array l = new array ();
l.newarray();
}

}

- Anonymous December 13, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Class array {
public void newarray()
{
int []a={1,2,3,4,5,6,7,10,100,1000};
int []b={4,5,6,7,10,45};
int x=0;
if (a.length>b.length)
{
int [] c= new int[a.length];
for(int i=0;i<a.length;i++)
{
for(int j=0;j<b.length;j++)
{
if(a[i]==b[j])
{
c[j]=a[i];
x++;
}
}
}
for(int j=0;j<x;j++)
System.out.print(c[j]+"\t");
}

else
{
int [] c= new int[b.length];
for(int i=0;i<b.length;i++)
{
for(int j=0;j<a.length;j++)
{
if(b[i]==a[j])
{
c[j]=b[i];
x++;
}
}
}
for(int j=0;j<x;j++)
System.out.print(c[j]+"\t");
}
}
public static void main(String[] args)
{
array l = new array ();
l.newarray();
}

}

- Anonymous December 13, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Class array {
public void newarray()
{
int []a={1,2,3,4,5,6,7,10,100,1000};
int []b={4,5,6,7,10,45};
int x=0;
if (a.length>b.length)
{
int [] c= new int[a.length];
for(int i=0;i<a.length;i++)
{
for(int j=0;j<b.length;j++)
{
if(a[i]==b[j])
{
c[j]=a[i];
x++;
}
}
}
for(int j=0;j<x;j++)
System.out.print(c[j]+"\t");
}

else
{
int [] c= new int[b.length];
for(int i=0;i<b.length;i++)
{
for(int j=0;j<a.length;j++)
{
if(b[i]==a[j])
{
c[j]=b[i];
x++;
}
}
}
for(int j=0;j<x;j++)
System.out.print(c[j]+"\t");
}
}
public static void main(String[] args)
{
array l = new array ();
l.newarray();
}

}

- manjuraj December 13, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Class array {
public void newarray()
{
int []a={1,2,3,4,5,6,7,10,100,1000};
int []b={4,5,6,7,10,45};
int x=0;
if (a.length>b.length)
{
int [] c= new int[a.length];
for(int i=0;i<a.length;i++)
{
for(int j=0;j<b.length;j++)
{
if(a[i]==b[j])
{
c[j]=a[i];
x++;
}
}
}
for(int j=0;j<x;j++)
System.out.print(c[j]+"\t");
}

else
{
int [] c= new int[b.length];
for(int i=0;i<b.length;i++)
{
for(int j=0;j<a.length;j++)
{
if(b[i]==a[j])
{
c[j]=b[i];
x++;
}
}
}
for(int j=0;j<x;j++)
System.out.print(c[j]+"\t");
}
}
public static void main(String[] args)
{
array l = new array ();
l.newarray();
}

}

- manjuraj December 13, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

w

- manjuraj December 13, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

l1=[-8,-6,-5,-5,1,2,3,4,4,5,6,8]
l=[1,2,7,100]
l2=[]
i=0
j=0
while(len(l)>0 and len(l1)>0):
if l[i]>l1[j]:
l2.append(l1[j])
l1.pop(j)
else:
l2.append(l[i])
l.pop(i)
if len(l)>0:
l2.extend(l)
elif len(l1)>0:
l2.extend(l1)
for k in range(0,len(l2)):
print l2[k],

- Anonymous June 20, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

l1=[-8,-6,-5,-5,1,2,3,4,4,5,6,8,100]
l=[1,2,7,10]
l2=[]
i=0
j=0
while(len(l)>0 and len(l1)>0):
if l[i]>l1[j]:
l2.append(l1[j])
l1.pop(j)
else:
l2.append(l[i])
l.pop(i)
if len(l)>0:
l2.extend(l)
elif len(l1)>0:
l2.extend(l1)
for k in range(0,len(l2)):
print l2[k],

- Shweta June 20, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

A1=[1,2,3,4]

B1=[1,3,5,8]

C1 = []

for i in A1:
    if i in B1:
        B1.remove(i)
        C1.append(i)
    else:
        C1.append(i)
C1=C1[:]+B1[:]
print(C1[:])

- blankspace August 03, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

a=[3,2,1,4,5,10,9]
b=[13,4,12]
c=[]
len_of_c=len(a)+len(b)
for i in range(0,len_of_c):
    try:
        if len(a) == 0:
            a_m=0
        else:
            a_m = max(a)
        if len(b) == 0:
            b_m=0
        else:
            b_m = max(b)

        if  a_m > b_m:
            c.append(a_m)
            a.remove(a_m)
        else:
            c.append(b_m)
            b.remove(b_m)
    except:
        print("List Empty")
c.reverse()
print(c)

- Shashi October 04, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
-2
of 2 vote

a = [10,10,5,5,6]
b = [4,5,7,5,4]
c = []
for i in a:
for j in b:
if i>j and j not in c:
c.append(j)
else:
c.append(i)
print c

- Nitish January 24, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.


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