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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)

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