There are Many Efficient Solutions Than this
Amazon - First Round -1
*******************
Count the number of duplicate Elements present in the array
Sample Input:
10
1 2 2 4 4 5 5 6 6 10
Sample Output:
4
Explanation :
2 present twice count is incremented
4 present twice count is incremented
5 present twice count is incremented
6 present twice count is incremented
#include <stdio.h>
#include <limits.h>
int main() {
int n,max=0,hash[1000]={0},count=0;
scanf("%d",&n);
int arr[n],i;
for(i=0;i<n;i++)
{
scanf("%d",&arr[i]);
hash[arr[i]]++; // the values are stored in the hash table
if(arr[i]>max)
{
max = arr[i]; // for finding the max value of array
}
}
for(i=0;i<=max;i++)
{
if(hash[i]>1) //if a number present more than once
count++;
}
printf("%d",count);
return 0;
}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Amazon- First Round -2
*****************
Given two arrays merge and sort the arrays
Sample Input:
5 --->size of the array
2 4 6 8 10 ---->Elements of array 1
1 3 5 7 9 ---->Elements of array 2
Sample Output:
// The vector is similar to the array
// push_back() is used to insert the value into the vector
// size() is used to find the length of the array
#include<bits/stdc++.h>
using namespace std;
#define lli long long int
vector<lli> mergeArray(vector<lli> a,vector<lli> b)
{
vector<lli> res;
lli size1=a.size(),size2=b.size();
lli i=0,j=0;
while((i<size1)&&(j<size2))
{
if(a[i]<b[j])
{
res.push_back(a[i]);
i++; // if smallest element the values are pushed into this
}
else
{
res.push_back(b[j]);
j++; // if largest element the values are pushed into this
}
}
while(i<size1)
{
res.push_back(a[i]); // in some Testcases only these two conditions needed
i++;
}
while(j<size2)
{
res.push_back(b[j]);
j++;
}
return res;
}
int main()
{
vector<lli> a,b;
lli n;
cin>>n;
for(lli i=0;i<n;i++)
{
lli k;
cin>>k;
a.push_back(k); //Reading the first array
}
for(lli i=0;i<n;i++)
{
lli k;
cin>>k;
b.push_back(k); //Reading the second array
}
vector<lli> res=mergeArray(a,b);
n+=n; //The vectors are passed to the function
for(lli i=0;i<n;i++)
{
cout<<res[i]<<" ";
}
}
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Amazon Second Round - 1
*****************
Even numbers at Even Positions and Odd numbers at odd numbers at Positions if Exceeds the array is Untouched
http://www.crazyforcode.com/rearrange-array-even-numbers-odd-numbers/
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Amazon Second Round - 2
*******************
Given a string Count the number of unequalities and make as Palindrome
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Amazon Third Round - 1
*****************
Search in the array:
http://www.geeksforgeeks.org/efficient-search-in-an-array-where-difference-between-adjacent-is-1/
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Amazon Third Round - 2
*****************
Sort the array:
http://www.geeksforgeeks.org/sort-an-array-of-0s-1s-and-2s/
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Amazon - Fourth Round
*****************
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Amazon - Fifth Round
*****************
Amazon - First Round -1
*******************
Count the number of duplicate Elements present in the array
Sample Input:
10
1 2 2 4 4 5 5 6 6 10
Sample Output:
4
Explanation :
2 present twice count is incremented
4 present twice count is incremented
5 present twice count is incremented
6 present twice count is incremented
#include <stdio.h>
#include <limits.h>
int main() {
int n,max=0,hash[1000]={0},count=0;
scanf("%d",&n);
int arr[n],i;
for(i=0;i<n;i++)
{
scanf("%d",&arr[i]);
hash[arr[i]]++; // the values are stored in the hash table
if(arr[i]>max)
{
max = arr[i]; // for finding the max value of array
}
}
for(i=0;i<=max;i++)
{
if(hash[i]>1) //if a number present more than once
count++;
}
printf("%d",count);
return 0;
}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Amazon- First Round -2
*****************
Given two arrays merge and sort the arrays
Sample Input:
5 --->size of the array
2 4 6 8 10 ---->Elements of array 1
1 3 5 7 9 ---->Elements of array 2
Sample Output:
1 2 3 4 5 6 7 8 9 10
// The vector is similar to the array
// push_back() is used to insert the value into the vector
// size() is used to find the length of the array
#include<bits/stdc++.h>
using namespace std;
#define lli long long int
vector<lli> mergeArray(vector<lli> a,vector<lli> b)
{
vector<lli> res;
lli size1=a.size(),size2=b.size();
lli i=0,j=0;
while((i<size1)&&(j<size2))
{
if(a[i]<b[j])
{
res.push_back(a[i]);
i++; // if smallest element the values are pushed into this
}
else
{
res.push_back(b[j]);
j++; // if largest element the values are pushed into this
}
}
while(i<size1)
{
res.push_back(a[i]); // in some Testcases only these two conditions needed
i++;
}
while(j<size2)
{
res.push_back(b[j]);
j++;
}
return res;
}
int main()
{
vector<lli> a,b;
lli n;
cin>>n;
for(lli i=0;i<n;i++)
{
lli k;
cin>>k;
a.push_back(k); //Reading the first array
}
for(lli i=0;i<n;i++)
{
lli k;
cin>>k;
b.push_back(k); //Reading the second array
}
vector<lli> res=mergeArray(a,b);
n+=n; //The vectors are passed to the function
for(lli i=0;i<n;i++)
{
cout<<res[i]<<" ";
}
}
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Amazon Second Round - 1
*****************
Even numbers at Even Positions and Odd numbers at odd numbers at Positions if Exceeds the array is Untouched
http://www.crazyforcode.com/rearrange-array-even-numbers-odd-numbers/
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Amazon Second Round - 2
*******************
Given a string Count the number of unequalities and make as Palindrome
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Amazon Third Round - 1
*****************
Search in the array:
http://www.geeksforgeeks.org/efficient-search-in-an-array-where-difference-between-adjacent-is-1/
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Amazon Third Round - 2
*****************
Sort the array:
http://www.geeksforgeeks.org/sort-an-array-of-0s-1s-and-2s/
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Amazon - Fourth Round
*****************
- Testing --->An application sends an automatic Birthday wishes Write test case for that
- Troubleshooting ---> An ecommerce Website Located in 14 countries Example(an website is located in many countries In one country alone in one Account book is not downloading troubleshoot that)
- Linux Comands
- Dbms (SQL Querries)
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Amazon - Fifth Round
*****************
- Project
- Testing
- Troubleshooting
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