# Find element at given index after a number of rotations

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 an element at a given index.
Examples :

`Input : arr[] : {1, 2, 3, 4, 5}        ranges[] = { {0, 2}, {0, 3} }        index : 1Output : 3Explanation : 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 givenindex 1.`

Method: Brute-force The brute force approach is to actually rotate the array for all given ranges, and finally, return the element at the given index in the modified array.
Method: Efficient We can do offline processing after saving all ranges.
Suppose, our rotate ranges are : [0..2] and [0..3]
We run through these ranges from the reverse.
After range [0..3], index 0 will have the element which was on index 3.
So, we can change 0 to 3, i.e. if index = left, the index will be changed to right.
After range [0..2], index 3 will remain unaffected.
So, we can make 3 cases :
If index = left, the index will be changed to right.
If the index is not bounded by the range, no effect of rotation.
If the index is in bounds, the index will have the element at index-1.
Below is the implementation :

For better explanation:-

10 20 30 40 50

Index: 1

Rotations: {0,2} {1,4} {0,3}

Answer: Index 1 will have 30 after all the 3 rotations in the order {0,2} {1,4} {0,3}.

We performed {0,2} on A and now we have a new array A1.

We performed {1,4} on A1 and now we have a new array A2.

We performed {0,3} on A2 and now we have a new array A3.

Now we are looking for the value at index 1 in A3.

But A3 is {0,3} done on A2.

So index 1 in A3 is index 0 in A2.

But A2 is {1,4} done on A1.

So index 0 in A2 is also index 0 in A1 as it does not lie in the range {1,4}.

But A1 is {0,2} done on A.

So index 0 in A1 is index 2 in A.

On observing it, we are going deeper into the previous rotations starting from the latest rotation.

{0,3}

|

{1,4}

|

{0,2}

This is the reason we are processing the rotations in reverse order.

Please note that we are not rotating the elements in the reverse order, just processing the index from the reverse.

Because if we actually rotate in reverse order, we might get a completely different answer as in the case of rotations the order matters.

`# Python 3 code to rotate an array`

`# and answer the index query`

`# Function to compute the element`

`# at given index`

`def` `findElement(arr, ranges, rotations, index) :`

`for` `i in` `range(rotations -` `1, -1, -1` `) :`

`# Range[left...right]`

`left =` `ranges[i]`

`right =` `ranges[i]`

`# Rotation will not have`

`# any effect`

`if` `(left <=` `index and` `right >=` `index) :`

`if` `(index ==` `left) :`

`index =` `right`

`else` `:`

`index =` `index -` `1`

`# Returning new element`

`return` `arr[index]`

`# Driver Code`

`arr =` `[ 1, 2, 3, 4, 5` `]`

`# No. of rotations`

`rotations =` `2`

`# Ranges according to`

`# 0-based indexing`

`ranges =` `[ [ 0, 2` `], [ 0, 3` `] ]`

`index =` `1`

`print(findElement(arr, ranges, rotations, index))`

Time Complexity: O(R), R is the number of rotations

Auxiliary Space: O(1)

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