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 Mth element of the array after K left rotations.
Examples:
Input: arr[] = {3, 4, 5, 23}, K = 2, M = 1
Output: 5
Explanation:Â
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: 5Â
Explanation:Â
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)
- 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.
- The Mth element of the array after K left rotations is
{ (K + M – 1) % N }th
element in the original array.
Efficient Approach: To optimize the problem, observe the following points:
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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 |
5
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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!



