Mth element after K Right Rotations of an Array
Last Updated :
16 Oct, 2023
Given non-negative integers K, M, and an array arr[ ] consisting of N elements, the task is to find the Mth element of the array after K right rotations.
Examples:
Input: arr[] = {3, 4, 5, 23}, K = 2, M = 1
Output: 5
Explanation:
The array after first right rotation a1[ ] = {23, 3, 4, 5}
The array after second right rotation a2[ ] = {5, 23, 3, 4}
1st element after 2 right rotations is 5.
Input: arr[] = {1, 2, 3, 4, 5}, K = 3, M = 2
Output: 4
Explanation:
The array after 3 right rotations has 4 at its second position.
Naive Approach:
The simplest approach to solve the problem is to Perform Right Rotation operation K times and then find the Mth element of the final array.
Algorithm:
- Define a function called leftrotate that takes a vector and an integer d as input. The function should reverse the elements of the vector from the beginning up to index d, then from index d to the end, and finally the entire vector.
- Define a function called rightrotate that takes a vector and an integer d as input. The function should call leftrotate with the vector and the difference between the size of the vector and d as arguments.
- Define a function called getFirstElement that takes an integer array a, its size N, and two integers K and M as input. The function should do the following:
- Initialize a vector v with the elements of array a.
- Right rotate the vector v K times by calling rightrotate in a loop with v and the integer value 1 as arguments, K times.
- Return the Mth element of the rotated vector v.
- In the main function, initialize an integer array a and its size N, and two integers K and M with appropriate values.
- Call the function getFirstElement with an array a, N, K, and M as arguments and print the returned value.
Below is the implementation of the approach:
C++
#include <bits/stdc++.h>
using namespace std;
void leftrotate(vector< int >& v, int d)
{
reverse(v.begin(), v.begin() + d);
reverse(v.begin() + d, v.end());
reverse(v.begin(), v.end());
}
void rightrotate(vector< int >& v, int d)
{
leftrotate(v, v.size() - d);
}
int getFirstElement( int a[], int N, int K, int M)
{
vector< int > v;
for ( int i = 0; i < N; i++)
v.push_back(a[i]);
while (K--) {
rightrotate(v, 1);
}
return v[M - 1];
}
int main()
{
int a[] = { 1, 2, 3, 4, 5 };
int N = sizeof (a) / sizeof (a[0]);
int K = 3, M = 2;
cout << getFirstElement(a, N, K, M);
return 0;
}
|
Java
import java.util.Arrays;
public class GFG {
static void leftRotate( int [] arr, int d) {
int n = arr.length;
reverse(arr, 0 , d - 1 );
reverse(arr, d, n - 1 );
reverse(arr, 0 , n - 1 );
}
static void rightRotate( int [] arr, int d) {
int n = arr.length;
leftRotate(arr, n - d);
}
static void reverse( int [] arr, int start, int end) {
while (start < end) {
int temp = arr[start];
arr[start] = arr[end];
arr[end] = temp;
start++;
end--;
}
}
static int getFirstElement( int [] arr, int K, int M) {
int [] rotatedArray = Arrays.copyOf(arr, arr.length);
for ( int i = 0 ; i < K; i++) {
rightRotate(rotatedArray, 1 );
}
return rotatedArray[M - 1 ];
}
public static void main(String[] args) {
int [] arr = { 1 , 2 , 3 , 4 , 5 };
int K = 3 ;
int M = 2 ;
System.out.println(getFirstElement(arr, K, M));
}
}
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Python3
def left_rotate(v, d):
v[:d] = v[:d][:: - 1 ]
v[d:] = v[d:][:: - 1 ]
v[:] = v[:: - 1 ]
def right_rotate(v, d):
left_rotate(v, len (v) - d)
def get_first_element(a, K, M):
v = list (a)
while K > 0 :
right_rotate(v, 1 )
K - = 1
return v[M - 1 ]
a = [ 1 , 2 , 3 , 4 , 5 ]
K = 3
M = 2
print (get_first_element(a, K, M))
|
C#
using System;
using System.Linq;
class GFG
{
static void leftrotate( ref int [] v, int d)
{
Array.Reverse(v, 0, d);
Array.Reverse(v, d, v.Length - d);
Array.Reverse(v);
}
static void reftrotate( ref int [] v, int d)
{
leftrotate( ref v, v.Length - d);
}
static int getFirstElement( int [] a, int K, int M)
{
int [] v = a.ToArray();
while (K > 0)
{
reftrotate( ref v, 1);
K--;
}
return v[M - 1];
}
static void Main( string [] args)
{
int [] a = { 1, 2, 3, 4, 5 };
int N = a.Length;
int K = 3, M = 2;
Console.WriteLine(getFirstElement(a, K, M));
}
}
|
Javascript
function leftrotate(v, d) {
const reversedFirstPart = v.slice(0, d).reverse();
const reversedSecondPart = v.slice(d).reverse();
const reversedArray = reversedFirstPart.concat(reversedSecondPart).reverse();
for (let i = 0; i < v.length; i++) {
v[i] = reversedArray[i];
}
}
function rightrotate(v, d) {
leftrotate(v, v.length - d);
}
function getFirstElement(a, N, K, M) {
let v = [];
for (let i = 0; i < N; i++) {
v.push(a[i]);
}
while (K--) {
rightrotate(v, 1);
}
return v[M - 1];
}
let a = [1, 2, 3, 4, 5];
let N = a.length;
let K = 3;
let M = 2;
console.log(getFirstElement(a, N, K, M));
|
Time Complexity: O(N * K)
Auxiliary Space: O(N)
Efficient Approach:
To optimize the problem, the following observations need to be made:
- If the array is rotated N times it returns the initial array again.
For example, a[ ] = {1, 2, 3, 4, 5}, K=5
Modified array after 5 right rotation a5[ ] = {1, 2, 3, 4, 5}.
- Therefore, the elements in the array after Kth rotation is the same as the element at index K%N in the original array.
- If K >= M, the Mth element of the array after K right rotations is
{ (N-K) + (M-1) } th element in the original array.
- If K < M, the Mth element of the array after K right rotations is:
(M – K – 1) th element in the original array.
Below is the implementation of the above approach:
C++
#include<bits/stdc++.h>
using namespace std;
int getFirstElement( int a[], int N,
int K, int M)
{
K %= N;
int index;
if (K >= M)
index = (N - K) + (M - 1);
else
index = (M - K - 1);
int result = a[index];
return result;
}
int main()
{
int a[] = { 1, 2, 3, 4, 5 };
int N = sizeof (a) / sizeof (a[0]);
int K = 3, M = 2;
cout << getFirstElement(a, N, K, M);
return 0;
}
|
Java
class GFG{
static int getFirstElement( int a[], int N,
int K, int M)
{
K %= N;
int index;
if (K >= M)
index = (N - K) + (M - 1 );
else
index = (M - K - 1 );
int result = a[index];
return result;
}
public static void main(String[] args)
{
int a[] = { 1 , 2 , 3 , 4 , 5 };
int N = 5 ;
int K = 3 , M = 2 ;
System.out.println(getFirstElement(a, N, K, M));
}
}
|
Python3
def getFirstElement(a, N, K, M):
K % = N
if (K > = M):
index = (N - K) + (M - 1 )
else :
index = (M - K - 1 )
result = a[index]
return result
if __name__ = = "__main__" :
a = [ 1 , 2 , 3 , 4 , 5 ]
N = len (a)
K , M = 3 , 2
print ( getFirstElement(a, N, K, M))
|
C#
using System;
class GFG{
static int getFirstElement( int []a, int N,
int K, int M)
{
K %= N;
int index;
if (K >= M)
index = (N - K) + (M - 1);
else
index = (M - K - 1);
int result = a[index];
return result;
}
public static void Main()
{
int []a = { 1, 2, 3, 4, 5 };
int N = 5;
int K = 3, M = 2;
Console.Write(getFirstElement(a, N, K, M));
}
}
|
Javascript
<script>
function getFirstElement(a, N,
K, M)
{
K %= N;
let index;
if (K >= M)
index = (N - K) + (M - 1);
else
index = (M - K - 1);
let result = a[index];
return result;
}
let a = [ 1, 2, 3, 4, 5 ];
let N = 5;
let K = 3, M = 2;
document.write(getFirstElement(a, N, K, M));
</script>
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Time Complexity: O(1)
Auxiliary Space: O(1)
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