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!