Hackerrank - Circular Array Rotation Solution
2 min read

Hackerrank - Circular Array Rotation Solution

Hackerrank - Circular Array Rotation Solution

John Watson knows of an operation called a right circular rotation on an array of integers. One rotation operation moves the last array element to the first position and shifts all remaining elements right one. To test Sherlock's abilities, Watson provides Sherlock with an array of integers. Sherlock is to perform the rotation operation a number of times then determine the value of the element at a given position.

For each array, perform a number of right circular rotations and return the value of the element at a given index.

For example, array , number of rotations,  and indices to check, .
First we perform the two rotations:

Now return the values from the zero-based indices  and  as indicated in the  array.

Function Description

Complete the circularArrayRotation function in the editor below. It should return an array of integers representing the values at the specified indices.

circularArrayRotation has the following parameter(s):

  • a: an array of integers to rotate
  • k: an integer, the rotation count
  • queries: an array of integers, the indices to report

Input Format

The first line contains  space-separated integers, , , and , the number of elements in the integer array, the rotation count and the number of queries.
The second line contains  space-separated integers, where each integer  describes array element  (where ).
Each of the  subsequent lines contains a single integer denoting , the index of the element to return from .

Constraints

Output Format

For each query, print the value of the element at index  of the rotated array on a new line.

Sample Input 0

3 2 3
1 2 3
0
1
2

Sample Output 0

2
3
1

Explanation 0

After the first rotation, the array becomes .
After the second (and final) rotation, the array becomes .

Let's refer to the array's final state as array . For each query, we just have to print the value of  on a new line:

  1. m = 1, .
  2. m = 2, .
  3. m = 3, .

Solution in Python

n,k,q = map(int,input().split())
k = k%n
a = list(map(int,input().split()))
for _ in range(q):
    m = int(input())
    print(a[m-k])

Explanation

Instead of making changes to our original array we subtract the value of k from m. This gives us the correct index of our required value.

Enjoying these posts? Subscribe for more


Adblocker detected! Please consider reading this notice.

We've detected that you are using AdBlock Plus or some other adblocking software which is preventing the page from fully loading.

That's okay. But without advertising-income, we can't keep making this site awesome.

We don't have any banner, Flash, animation, obnoxious sound, or popup ad. We do not implement these annoying types of ads!

We need money to operate the site, and almost all of it comes from our online advertising.

Please add thepoorcoder.com to your ad blocking whitelist or disable your adblocking software.

×