Find element at given index after a number of rotations ( Efficient method )

Question - 

An array consisting of N integers is given. There are several Right Circular Rotations of range[L..R] that we perform. After performing these rotations, we need to find element at a given index.

Example - 

Input : arr[] : {1, 2, 3, 4, 5}
        ranges[] = { {0, 2}, {0, 3} }
        index : 1
Output : 3
Explanation : After first given rotation {0, 2}
                arr[] = {3, 1, 2, 4, 5}
              After second rotation {0, 3} 
                arr[] = {4, 3, 1, 2, 5}
After all rotations we have element 3 at given
index 1. 

Approach - 

In first method we rotate array with given condition so time is very high. In this method we change the value of index according to array rotation.

Implementation - 

#include<iostream>

using namespace std;

int findvalue(int arr[],int n,int rotation,int range[][2],int index)

{

    

    for(int i=rotation-1;i>=0;i--){

        int left=range[i][0];

        int right=range[i][1];

    if (left <= index && right >= index) { 

        if(index==left)

            index=right;

        else

            index--;

                }

    }

    return arr[index];

}

int main()

{

    int n;

    cout<<"Enter the number of elements : ";cin>>n;

    int arr[n];

    cout<<"Enter elements : "<<endl;

    for(int i=0;i<n;i++)

        cin>>arr[i];

    int rotation;

    cout<<"Ënter the number of rotation : ";cin>>rotation;

    int range[rotation][2];

    for(int i=0;i<rotation;i++){

        cout<<"Range "<<i+1<<" : ";

        cin>>range[i][0];

        cin>>range[i][1];

    }

    int index;

    cout<<"Enter index : ";cin>>index;

    

    int value=findvalue(arr,n,rotation,range,index);

    cout<<endl<<"Value is : "<<value<<endl;

}

code on compiler

Time complexity - O(n)

Space - O(1)