CareerCup

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.

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:
Answer:
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

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

#include<bits/stdc++.h>
using namespace std;


class Solution {
public:
int search(vector<int>& nums, int target) {
// support variables
int len = nums.size(), l = 0, r = len, mid;
// finding the pivot
while (l < r) {
mid = (l + r) / 2;
// current element > first elemen: we move the window right
if (nums[mid] >= nums[0]) l = mid + 1;
// otherwise we move it left
else r = mid;
}
// checking for edge case - pivot at the end = sorted vector
if (l == len) l = 0;
// preparing for the next BS: target is between pivot and the last element
if (target >= nums[l] && target <= nums.back()) r = len;
// target is between the first element and the one before pivot
else r = l, l = 0;
// all other cases
// finding the element
while (l < r) {
mid = (l + r) / 2;
// mid matches the target element: we return it
if (nums[mid] == target) return mid;
// mid matches an element smaller than target: we move the window right
else if (nums[mid] < target) l = mid + 1;
// otherwise we move it left
else r = mid;
}
return nums[l] == target ? l : -1;
}
};


int main(){

Solution sol;
int n;
cin>>n;
vector<int> nums;
for(int i=0; i<n; i++){
int x;
cin>>x;
nums.push_back(x);
}
int tar;
cin>>tar;
cout<<sol.search(nums,tar);
}

#include<bits/stdc++.h>
using namespace std;


class Solution {
public:
    int search(vector<int>& nums, int target) {
        // support variables
        int len = nums.size(), l = 0, r = len, mid;
        // finding the pivot
        while (l < r) {
            mid = (l + r) / 2;
            // current element > first elemen: we move the window right
            if (nums[mid] >= nums[0]) l = mid + 1;
            // otherwise we move it left
            else r = mid;
        }
        // checking for edge case - pivot at the end = sorted vector
        if (l == len) l = 0;
        // preparing for the next BS: target is between pivot and the last element
        if (target >= nums[l] && target <= nums.back()) r = len;
        // target is between the first element and the one before pivot
        else r = l, l = 0;
        // all other cases
        // finding the element
        while (l < r) {
            mid = (l + r) / 2;
            // mid matches the target element: we return it
            if (nums[mid] == target) return mid;
            // mid matches an element smaller than target: we move the window right
            else if (nums[mid] < target) l = mid + 1;
            // otherwise we move it left
            else r = mid;
        }
        return nums[l] == target ? l : -1;
    }
};


int main(){

     Solution sol;
     int n;
     cin>>n;
     vector<int> nums;
     for(int i=0; i<n; i++){
       int x;
       cin>>x;
       nums.push_back(x);
     }
     int tar;
     cin>>tar;
     cout<<sol.search(nums,tar);
}