Java Program to Find the Mth element of the Array after K left rotations

Given non-negative integers K, M, and an array arr[] with N elements find the M^th element of the array after K left rotations.

Examples:

Input: arr[] = {3, 4, 5, 23}, K = 2, M = 1
Output: 5
Explanation: 
The array after first left rotation a₁[ ] = {4, 5, 23, 3}
The array after second left rotation a₂[ ] = {5, 23, 3, 4}
st element after 2 left rotations is 5.

Input: arr[] = {1, 2, 3, 4, 5}, K = 3, M = 2
Output: 5 
Explanation: 
The array after 3 left rotation has 5 at its second position.

Naive Approach: The idea is to Perform Left rotation operation K times and then find the M^th element of the final array.

Time Complexity: O(N * K)
Auxiliary Space: O(N)

Efficient Approach: To optimize the problem, observe the following points:

1. If the array is rotated N times it returns the initial array again.
   

   For example, a[ ] = {1, 2, 3, 4, 5}, K=5 then the array after 5 left 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.

2. The Mth element of the array after K left rotations is
   

   { (K + M – 1) % N }th

   element in the original array.


Below is the implementation of the above approach:

Java
// Java program for the above approach
import java.util.*;
class GFG{

// Function to return Mth element of
// array after k left rotations
public static int getFirstElement(int[] a, int N,
                                  int K, int M)
{

    // The array comes to original state
    // after N rotations
    K %= N;

    // Mth element after k left rotations
    // is (K+M-1)%N th element of the
    // original array
    int index = (K + M - 1) % N;

    int result = a[index];

    // Return the result
    return result;
}

// Driver code
public static void main(String[] args)
{

    // Array initialization
    int a[] = { 3, 4, 5, 23 };

    // Size of the array
    int N = a.length;

    // Given K rotation and Mth element
    // to be found after K rotation
    int K = 2, M = 1;

    // Function call
    System.out.println(getFirstElement(a, N, K, M));
}
}

// This code is contributed by divyeshrabadiya07
Output:

5

Time complexity: O(1)
Auxiliary Space: O(1)

Please refer complete article on Find the Mth element of the Array after K left rotations for more details!