Python3 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}
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)

1. Efficient Approach:
2. To optimize the problem, observe the following points:
3. 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.
4. The Mth element of the array after K left rotations is
   

   { (K + M - 1) % N }th

   element in the original array.
5.  
   Below is the implementation of the above approach:
6. 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 
   

7. 
   Output

   5

8. Time complexity: O(1)
   Auxiliary Space: O(1)
9. Please refer complete article on
10. Find the Mth element of the Array after K left rotations
11. for more details!

kartik

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