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}
1st 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:

C++








// C++ program for the above approach  #include <bits/stdc++.h>  using namespace std;     // Function to return Mth element of  // array after k left rotations  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  int main()  {      // Array initialization      int a[] = { 3, 4, 5, 23 };         // Size of the array      int N = sizeof(a) / sizeof(a[0]);         // Given K rotation and Mth element      // to be found after K rotation      int K = 2, M = 1;         // Function call      cout << getFirstElement(a, N, K, M);      return 0;  }

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

Python3








# Python3 program for the above approach     # Function to return Mth element of  # array after k left rotations  def getFirstElement(a, N, K, 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      index = (K + M - 1) % N         result = a[index]         # Return the result      return result     # Driver Code  if __name__ == '__main__':             # Array initialization      a = [ 3, 4, 5, 23 ]         # Size of the array      N = len(a)         # Given K rotation and Mth element      # to be found after K rotation      K = 2     M = 1        # Function call      print(getFirstElement(a, N, K, M))     # This code is contributed by mohit kumar 29

C#








// C# program for the above approach  using System;    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      Console.Write(getFirstElement(a, N, K, M)); } }    // This code is contributed by rutvik_56

Javascript








<script>    // Javascript program for the above approach         // Function to return Mth element of     // array after k left rotations     function getFirstElement(a , N , K , 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         var index = (K + M - 1) % N;            var result = a[index];            // Return the result         return result;     }        // Driver code                   // Array initialization         var a = [ 3, 4, 5, 23 ];            // Size of the array         var N = a.length;            // Given K rotation and Mth element         // to be found after K rotation         var K = 2, M = 1;            // Function call         document.write(getFirstElement(a, N, K, M));    // This code contributed by gauravrajput1     </script>

Output: 

5

 

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

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