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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