Find Mth element after K Right Rotations of an Array
Last Updated :
23 Jul, 2025
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:
a. Initialize a vector v with the elements of array a.
b. Right rotate the vector v K times by calling rightrotate in a loop with v and the integer value 1 as arguments, K times.
c. Return the Mth element of the rotated vector v.
4. In the main function, initialize an integer array a and its size N, and two integers K and M with appropriate values.
5. Call the function getFirstElement with array a, N, K, and M as arguments and print the returned value.
Below is the implementation of the approach:
C++
// C++ program to find the Mth element
// of the array after K right rotations.
#include <bits/stdc++.h>
using namespace std;
// In-place rotates s towards left by d
void leftrotate(vector<int>& v, int d)
{
reverse(v.begin(), v.begin() + d);
reverse(v.begin() + d, v.end());
reverse(v.begin(), v.end());
}
// In-place rotates s towards right by d
void rightrotate(vector<int>& v, int d)
{
leftrotate(v, v.size() - d);
}
// Function to return Mth element of
// array after k right rotations
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]);
// Right rotate K times
while (K--) {
rightrotate(v, 1);
}
// return Mth element
return v[M - 1];
}
// Driver code
int main()
{
// Array initialization
int a[] = { 1, 2, 3, 4, 5 };
int N = sizeof(a) / sizeof(a[0]);
int K = 3, M = 2;
// Function call
cout << getFirstElement(a, N, K, M);
return 0;
}
// Java program to find the Mth element
// of the array after K right rotations.
import java.util.Arrays;
public class GFG {
// In-place rotates array towards left by d
static void leftRotate(int[] arr, int d) {
reverse(arr, 0, d - 1);
reverse(arr, d, arr.length - 1);
reverse(arr, 0, arr.length - 1);
}
// In-place rotates array towards right by d
static void rightRotate(int[] arr, int d) {
leftRotate(arr, arr.length - d);
}
// Function to reverse elements in array from start to end indices
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--;
}
}
// Function to return Mth element of array after K right rotations
static int getFirstElement(int[] arr, int K, int M) {
int[] rotatedArray = Arrays.copyOf(arr, arr.length);
// Right rotate K times
while (K > 0) {
rightRotate(rotatedArray, 1);
K--;
}
// Return Mth element
return rotatedArray[M - 1];
}
// Driver code
public static void main(String[] args) {
// Array initialization
int[] arr = { 1, 2, 3, 4, 5 };
int K = 3, M = 2;
// Function call
System.out.println(getFirstElement(arr, K, M));
}
}
# C++ program to find the Mth element
# of the array after K right rotations.
# In-place rotates s towards left by d
def left_rotate(arr, d):
return arr[d:] + arr[:d]
# In-place rotates s towards right by d
def right_rotate(arr, d):
n = len(arr)
d = d % n # Handle the case where d > n
return left_rotate(arr, n - d)
# Function to return Mth element of
# array after k right rotations
def get_first_element(arr, K, M):
N = len(arr)
# Right rotate K times
for _ in range(K):
arr = right_rotate(arr, 1)
# Return Mth element
return arr[M - 1]
a = [1, 2, 3, 4, 5]
K = 3
M = 2
print(get_first_element(a, K, M))
// C# program to find the Mth element
// of the array after K right rotations.
using System;
using System.Collections.Generic;
class GFG
{
// In-place rotates s towards left by d
static void LeftRotate(List<int> v, int d)
{
v.Reverse(0, d);
v.Reverse(d, v.Count - d);
v.Reverse();
}
// In-place rotates s towards right by d
static void RightRotate(List<int> v, int d)
{
LeftRotate(v, v.Count - d);
}
// Function to return Mth element of
// array after k right rotations
static int GetFirstElement(int[] a, int N, int K, int M)
{
List<int> v = new List<int>();
for (int i = 0; i < N; i++)
v.Add(a[i]);
// Right rotate K times
while (K > 0)
{
RightRotate(v, 1);
K--;
}
// return Mth element
return v[M - 1];
}
// Driver code
static void Main(string[] args)
{
// Array initialization
int[] a = { 1, 2, 3, 4, 5 };
int N = a.Length;
int K = 3, M = 2;
// Function call
Console.WriteLine(GetFirstElement(a, N, K, M));
}
}
// Function to left rotate array towards left by d
function leftRotate(arr, d) {
return arr.slice(d).concat(arr.slice(0, d));
}
// Function to right rotate array towards right by d
function rightRotate(arr, d) {
const n = arr.length;
d = d % n; // Handle the case where d > n
return leftRotate(arr, n - d);
}
// Function to return Mth element of array after K right rotations
function getFirstElement(arr, K, M) {
const N = arr.length;
// Right rotate K times
for (let i = 0; i < K; i++) {
arr = rightRotate(arr, 1);
}
// Return Mth element
return arr[M - 1];
}
const a = [1, 2, 3, 4, 5];
const K = 3;
const M = 2;
console.log(getFirstElement(a, 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++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to return Mth element of
// array after k left rotations
int getFirstElement(int a[], int N,
int K, int M)
{
// The array comes to original state
// after N rotations
K %= N;
int index;
if (K >= M)
// Mth element after k right
// rotations is (N-K)+(M-1) th
// element of the array
index = (N - K) + (M - 1);
// Otherwise
else
// (M - K - 1) th element
// of the array
index = (M - K - 1);
int result = a[index];
// Return the result
return result;
}
// Driver Code
int main()
{
// Array initialization
int a[] = { 1, 2, 3, 4, 5 };
int N = sizeof(a) / sizeof(a[0]);
int K = 3, M = 2;
// Function call
cout << getFirstElement(a, N, K, M);
return 0;
}
// This code is contributed by GSSN Himabindu
// Java program to implement
// the above approach
import java.io.*;
class GFG{
// Function to return Mth element of
// array after k right rotations
static int getFirstElement(int a[], int N,
int K, int M)
{
// The array comes to original state
// after N rotations
K %= N;
int index;
// If K is greater or equal to M
if (K >= M)
// Mth element after k right
// rotations is (N-K)+(M-1) th
// element of the array
index = (N - K) + (M - 1);
// Otherwise
else
// (M - K - 1) th element
// of the array
index = (M - K - 1);
int result = a[index];
// Return the result
return result;
}
// Driver Code
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));
}
}
// This code is contributed by Ritik Bansal
# Python program for the above approach
# Function to return Mth element of
# array after k left rotations
def getFirstElement(a, N, K, M):
# The array comes to original state
# after N rotations
K %= N
index = 0
if (K >= M):
# Mth element after k right
# rotations is (N-K)+(M-1) th
# element of the array
index = (N - K) + (M - 1)
# Otherwise
else:
# (M - K - 1) th element
# of the array
index = (M - K - 1)
result = a[index]
# Return the result
return result
# Driver Code
# Array initialization
a = [ 1, 2, 3, 4, 5 ]
N = len(a)
K,M = 3,2
# Function call
print(getFirstElement(a, N, K, M))
# This code is contributed by shinjanpatra
using System;
using System.Linq;
class GFG {
// Function to return Mth element of
// array after k left rotations
static int getFirstElement(int []a, int N,
int K, int M)
{
// The array comes to original state
// after N rotations
K %= N;
int index;
if (K >= M)
// Mth element after k right
// rotations is (N-K)+(M-1) th
// element of the array
index = (N - K) + (M - 1);
// Otherwise
else
// (M - K - 1) th element
// of the array
index = (M - K - 1);
int result = a[index];
// Return the result
return result;
}
/* Driver program to test above
functions */
public static void Main()
{
int []arr = {1, 2, 3, 4, 5};
int N = arr.Length;
int K = 3, M = 2;
Console.Write(getFirstElement(arr, N, K, M));
}
}
// This code is contributed by Aarti_Rathi
<script>
// JavaScript program for the above approach
// Function to return Mth element of
// array after k left rotations
function getFirstElement(a, N, K, M)
{
// The array comes to original state
// after N rotations
K %= N
let index
if (K >= M)
// Mth element after k right
// rotations is (N-K)+(M-1) th
// element of the array
index = (N - K) + (M - 1)
// Otherwise
else
// (M - K - 1) th element
// of the array
index = (M - K - 1)
let result = a[index]
// Return the result
return result
}
// Driver Code
// Array initialization
let a = [ 1, 2, 3, 4, 5 ]
let N = a.length
let K = 3, M = 2
// Function call
document.write(getFirstElement(a, N, K, M))
// This code is contributed by shinjanpatra
</script>
Time complexity: O(1)
Auxiliary Space: O(1)
Please refer complete article on Mth element after K Right Rotations of an Array for more details!
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