向左旋转K后，找到数组的第M个元素

给定非负整数K ， M和具有N个元素的数组arr [] ，在向左旋转K^个元素后，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 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.

天真的方法：想法是执行K次左旋转操作，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：要优化问题，请注意以下几点：

1.  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 then the array after 5 left 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。

2. 左旋转K后，数组的第M个元素为

   { (K + M – 1) % N }th

   原始数组中的元素。

 

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; 
  
    // 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

输出：

5

时间复杂度： O(1)
辅助空间： O(1)