Skip to content
Related Articles

Related Articles

Improve Article
Find the Mth element of the Array after K left rotations
  • Difficulty Level : Easy
  • Last Updated : 29 Apr, 2021

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.

     
    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 %=
      
        # Mth element after k left rotations 
        # is (K+M-1)%N th element of the 
        # original array 
        index = (K + M - 1) %
      
        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)

    Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

    In case you wish to attend live classes with industry experts, please refer DSA Live Classes




    My Personal Notes arrow_drop_up
Recommended Articles
Page :