在数组的 K 次右旋转之后找到第 M 个元素

给定非负整数K 、 M和一个由N个元素组成的数组arr[ ] ，任务是在K次右转后找到数组的第 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右旋转操作K次，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)
有效的方法：
为了优化问题，需要进行以下观察：

• 如果数组旋转N次，它会再次返回初始数组。

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

• 因此，第K^次旋转后的数组中的元素与原始数组中索引K%N处的元素相同。
• 如果K >= M ，则经过 K 次右转后数组的第 M 个元素为
  

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

• 如果K < M ，则经过 K 次右转后数组的第 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 for the above approach
#include 
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;
  int index;
  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()
{
   
  // Array initialization
  int a[] = { 1, 2, 3, 4, 5 };
  int N = sizeof(a) / sizeof(a[0]);
  int K = 3, M = 2;
 
  // Function call
  cout << getFirstElement(a, N, K, M);
  return 0;
}
 
// This code is contributed by GSSN Himabindu

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

# Python 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
  index = 0
  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
 
# Array initialization
a = [ 1, 2, 3, 4, 5 ]
N = len(a)
K,M = 3,2
 
# Function call
print(getFirstElement(a, N, K, M))
 
# This code is contributed by shinjanpatra

Javascript

 
// 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
  let index
  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)
 
  let result = a[index]
 
  // Return the result
  return result
}
 
// Driver Code
 
// Array initialization
let a = [ 1, 2, 3, 4, 5 ]
let N = a.length
let K = 3, M = 2
 
// Function call
document.write(getFirstElement(a, N, K, M))
 
// This code is contributed by shinjanpatra
 

输出：

4

时间复杂度： O(1)
辅助空间： O(1)有关详细信息，请参阅关于数组的 K 次右旋转后第 M 个元素的完整文章！