C++ 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}
st 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; 
} 


Output: 

5

 

Time complexity: O(1)
Auxiliary Space: O(1)

Please refer complete article on Find the Mth element of the Array after K left rotations for more details!


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Article Tags :

• C++
• rotation
• school-programming

Practice Tags :

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