Javascript Program to Find Mth element after K Right Rotations of an Array
Last Updated :
21 Apr, 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 named leftrotate that takes a reference to a vector of integers v and an integer d as input parameters.
- Reverse the first d elements of the vector v.
- Reverse the remaining elements of the vector v starting from index d to the end.
- Reverse the entire vector v.
- Define a function named rightrotate that takes a reference to a vector of integers v and an integer d as input parameters.
- Call the leftrotate function with the vector v and the length of the vector minus d as input parameters.
- Define a function named getFirstElement that takes an array of integers a, its size N, an integer K and an integer M as input parameters.
- Declare a vector of integers v.
- Add all the elements of the array a to the vector v.
- Loop K times.
- Call the rightrotate function with the vector v and 1 as input parameters.
- Return the Mth element of the vector v minus 1 as the output of the function.
- In the main function, define an array of integers a, its size N, an integer K, and an integer M.
- Initialize the values of the array a, N, K, and M.
- Call the getFirstElement function with the array a, N, K, and M as input parameters and print its output.
Below is the implementation of the approach:
Javascript
function leftrotate(arr, d) {
arr.push.apply(arr, arr.splice(0, d));
}
function rightrotate(arr, d) {
leftrotate(arr, arr.length - d);
}
function getFirstElement(arr, N, K, M) {
let v = [];
for (let i = 0; i < N; i++)
v.push(arr[i]);
while (K--) {
rightrotate(v, 1);
}
return v[M - 1];
}
let a = [ 1, 2, 3, 4, 5 ];
let N = a.length;
let K = 3, M = 2;
console.log(getFirstElement(a, N, K, M));
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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 <iostream>
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 = 5;
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));
}
}
|
Python
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
a = [1, 2, 3, 4, 5]
N = 5
K = 3
M = 2
print(getFirstElement(a, N, K, M))
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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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C#
using System;
public class GFG
{
public 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;
int M = 2;
Console.WriteLine(GetFirstElement(a, N, K, M));
}
}
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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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