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

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Given non-negative integers K, M, and an array arr[] with N elements find the Mth 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:
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 Mth 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.

  3.  
    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!


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Last Updated : 14 Oct, 2022
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