## Samsung Interview Question for Senior Software Development Engineers

Country: United States
Interview Type: Phone Interview

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If the array is regular array without any special property then it does not add any value for multiple times rotation. A regular linear search will be the answer.

If the array is sorted and rotated an unknown number of times and find an integer, still binary search algorithm with minor change will solve this problem.

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``````Step 1:
Find the number of times array was rotated.
Step 2:
Split the array into two parts
step 3: apply binary search on same.
ex: original array
1,2,3,4,5,6,7,8,9
rotated array:
6,7,8,9,1,2,3,4,5,
the number to find 7:
Step 1: array was rotated by 4 times.
step 2:
part 1 array: 6,7,8,9
part 2 array: 1,2,3,4,5
step 3:
Do a binary search in part 1``````

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``````#include<bits/stdc++.h>
using namespace std;
int findPivotIndex(int * arr,int n){
int start=0,end=n-1;
while(start<end){
int mid=(start+end+1)/2;
if(arr[mid]>arr[0]){
start=mid+1;
}
else if(arr[mid]<arr[0]){
end=mid-1;
}
}
}
int findElement(int * arr,int start,int end,int ele){
int index=-1;
while(start<end){
int mid=(start+end)/2;
if(ele>arr[mid]){
start=mid+1;
}
else if(ele<arr[mid]){
end=mid-1;
}else if(ele==arr[mid]){
return mid;
}
}

return index;
}
int main(){
int n;
cin>>n;
int arr[n];
for(int i=0;i<n;i++){
cin>>arr[i];
}
int element;//element to be searched
cin>>element;

int pivot=findPivotIndex(arr,n),index=-1;

if(pivot>0 && element<=arr[pivot-1] && element>=arr[0]){
index=findElement(arr,0,pivot-1,element);
}else if(element>=arr[pivot] && element<=arr[n-1]){
index=findElement(arr,pivot,n-1,element);
}else{
index=-1;
}

cout<<"Index of element is "<<index<<endl;

return 0;
}``````

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