Find the Mth element of the Array after K left rotations

Last Updated : 14 Oct, 2022

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:

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.
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> 