数组的K个右旋转后的第M个元素

给定非负整数K ， M和由N个元素组成的数组arr [] ，任务是在K个右旋转后找到数组的^第M^个元素。
例子：

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. 

天真的方法：
解决问题的最简单方法是执行K次右旋转操作，然后找到最终数组的^第M^个元素。
时间复杂度： O(N * K)
辅助空间： O(N)
高效方法：
为了优化该问题，需要进行以下观察：

• 如果将数组旋转N次，它将再次返回初始数组。

 
 

For example, a[ ] = {1, 2, 3, 4, 5}, K=5 
Modifed array after 5 right rotation a₅[ ] = {1, 2, 3, 4, 5}. 

 

• 因此，第K^次旋转后数组中的元素与原始数组中索引为K％N的元素相同。
• 如果K > = M ，则向右旋转K后，数组的第M个元素为

 
 

{ (N-K) + (M-1) } ^th element in the original array. 

 

• 如果K ，则向右旋转K后，数组的第M个元素为：

 
 

(M – K – 1) th  element in the original array. 

 

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include
using namespace std;
  
// Function to return Mth element of
// array after k right rotations
int getFirstElement(int a[], int N,
                    int K, int M)
{
    // The array comes to original state
    // after N rotations
    K %= N;
    int index;
  
    // If K is greater or equal to M
    if (K >= M)
  
        // Mth element after k right
        // rotations is (N-K)+(M-1) th
        // element of the array
        index = (N - K) + (M - 1);
  
    // Otherwise
    else
  
        // (M - K - 1) th element
        // of the array
        index = (M - K - 1);
  
    int result = a[index];
  
    // Return the result
    return result;
}
  
// Driver Code 
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; 
}

Java

// Java program to implement
// the above approach
class GFG{
   
// Function to return Mth element of
// array after k right rotations
static int getFirstElement(int a[], int N,
                           int K, int M)
{
    // The array comes to original state
    // after N rotations
    K %= N;
    int index;
   
    // If K is greater or equal to M
    if (K >= M)
   
        // Mth element after k right
        // rotations is (N-K)+(M-1) th
        // element of the array
        index = (N - K) + (M - 1);
   
    // Otherwise
    else
   
        // (M - K - 1) th element
        // of the array
        index = (M - K - 1);
   
    int result = a[index];
   
    // Return the result
    return result;
}
   
// Driver Code 
public static void main(String[] args) 
{ 
    int a[] = { 1, 2, 3, 4, 5 }; 
     
    int N = 5; 
     
    int K = 3, M = 2; 
     
    System.out.println(getFirstElement(a, N, K, M)); 
}
} 
  
// This code is contributed by Ritik Bansal

Python3

# Python3 program to implement
# the above approach
  
# Function to return Mth element of
# array after k right rotations
def getFirstElement(a, N, K, M):
  
    # The array comes to original state
    # after N rotations
    K %= N
  
    # If K is greater or equal to M
    if (K >= M):
  
        # Mth element after k right
        # rotations is (N-K)+(M-1) th
        # element of the array
        index = (N - K) + (M - 1)
  
    # Otherwise
    else:
  
        # (M - K - 1) th element
        # of the array
        index = (M - K - 1)
  
    result = a[index]
  
    # Return the result
    return result
  
# Driver Code
if __name__ == "__main__":
      
    a = [ 1, 2, 3, 4, 5 ]
    N = len(a)
  
    K , M = 3, 2
  
    print( getFirstElement(a, N, K, M))
  
# This code is contributed by chitranayal

C#

// C# program to implement
// the above approach
using System;
class GFG{
  
// Function to return Mth element of
// array after k right rotations
static int getFirstElement(int []a, int N,
                        int K, int M)
{
    // The array comes to original state
    // after N rotations
    K %= N;
    int index;
  
    // If K is greater or equal to M
    if (K >= M)
  
        // Mth element after k right
        // rotations is (N-K)+(M-1) th
        // element of the array
        index = (N - K) + (M - 1);
  
    // Otherwise
    else
  
        // (M - K - 1) th element
        // of the array
        index = (M - K - 1);
  
    int result = a[index];
  
    // Return the result
    return result;
}
  
// Driver Code 
public static void Main() 
{ 
    int []a = { 1, 2, 3, 4, 5 }; 
      
    int N = 5; 
      
    int K = 3, M = 2; 
      
    Console.Write(getFirstElement(a, N, K, M)); 
}
} 
  
// This code is contributed by Code_Mech

输出：

4

时间复杂度： O(1)
辅助空间： O(1)