C++ Program to Find Mth element after K Right Rotations of an Array
Last Updated :
14 Sep, 2023
Given non-negative integers K, M, and an array arr[ ] consisting of N elements, the task is to find the Mth element of the array after K right rotations.
Examples:
Input: arr[] = {3, 4, 5, 23}, K = 2, M = 1
Output: 5
Explanation:
The array after first right rotation a1[ ] = {23, 3, 4, 5}
The array after second right rotation a2[ ] = {5, 23, 3, 4}
1st element after 2 right rotations is 5.
Input: arr[] = {1, 2, 3, 4, 5}, K = 3, M = 2
Output: 4
Explanation:
The array after 3 right rotations has 4 at its second position.
Naive Approach:
The simplest approach to solve the problem is to Perform Right Rotation operation K times and then find the Mth element of the final array.
Algorithm:
- Define a function called leftrotate that takes a vector and an integer d as input. The function should reverse the elements of the vector from the beginning up to index d, then from index d to the end, and finally the entire vector.
- Define a function called rightrotate that takes a vector and an integer d as input. The function should call leftrotate with the vector and the difference between the size of the vector and d as arguments.
- Define a function called getFirstElement that takes an integer array a, its size N, and two integers K and M as input. The function should do the following:
a. Initialize a vector v with the elements of array a.
b. Right rotate the vector v K times by calling rightrotate in a loop with v and the integer value 1 as arguments, K times.
c. Return the Mth element of the rotated vector v.
4. In the main function, initialize an integer array a and its size N, and two integers K and M with appropriate values.
5. Call the function getFirstElement with array a, N, K, and M as arguments and print the returned value.
Below is the implementation of the approach:
C++
#include <bits/stdc++.h>
using namespace std;
void leftrotate(vector< int >& v, int d)
{
reverse(v.begin(), v.begin() + d);
reverse(v.begin() + d, v.end());
reverse(v.begin(), v.end());
}
void rightrotate(vector< int >& v, int d)
{
leftrotate(v, v.size() - d);
}
int getFirstElement( int a[], int N, int K, int M)
{
vector< int > v;
for ( int i = 0; i < N; i++)
v.push_back(a[i]);
while (K--) {
rightrotate(v, 1);
}
return v[M - 1];
}
int main()
{
int a[] = { 1, 2, 3, 4, 5 };
int N = sizeof (a) / sizeof (a[0]);
int K = 3, M = 2;
cout << getFirstElement(a, N, K, M);
return 0;
}
|
Python3
def leftrotate(arr, d):
arr[:d] = reversed (arr[:d])
arr[d:] = reversed (arr[d:])
arr[:] = reversed (arr)
def rightrotate(arr, d):
leftrotate(arr, len (arr) - d)
def getFirstElement(a, N, K, M):
v = list (a)
while K > 0 :
rightrotate(v, 1 )
K - = 1
return v[M - 1 ]
if __name__ = = "__main__" :
a = [ 1 , 2 , 3 , 4 , 5 ]
N = len (a)
K = 3
M = 2
print (getFirstElement(a, N, K, M))
|
Time Complexity: O(N * K)
Auxiliary Space: O(N)
Efficient Approach:
To optimize the problem, the following observations need to be made:
- If the array is rotated N times it returns the initial array again.
For example, a[ ] = {1, 2, 3, 4, 5}, K=5
Modified array after 5 right 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.
- If K >= M, the Mth element of the array after K right rotations is
{ (N-K) + (M-1) } th element in the original array.
- If K < M, the Mth element of the array after K right rotations is:
(M – K – 1) th element in the original array.
Below is the implementation of the above approach:
C++
#include<bits/stdc++.h>
using namespace std;
int getFirstElement( int a[], int N,
int K, int M)
{
K %= N;
int index;
if (K >= M)
index = (N - K) + (M - 1);
else
index = (M - K - 1);
int result = a[index];
return result;
}
int main()
{
int a[] = { 1, 2, 3, 4, 5 };
int N = sizeof (a) / sizeof (a[0]);
int K = 3, M = 2;
cout << getFirstElement(a, N, K, M);
return 0;
}
|
Time complexity: O(1)
Auxiliary Space: O(1)
Please refer complete article on Mth element after K Right Rotations of an Array for more details!
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