在 K 次左旋转后找到数组的第 M 个元素

给定非负整数K 、 M和具有N 个元素的数组arr[]在K 次左旋转后找到数组的^第M^个元素。

例子：

Input: arr[] = {3, 4, 5, 23}, K = 2, M = 1
Output: 5
Explanation: 
The array after first 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.

朴素的方法：想法是执行左旋转操作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：要优化问题，请注意以下几点：

1. 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 then the array after 5 left rotation a₅[ ] = {1, 2, 3, 4, 5}.

   因此，经过第 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。

2. K次左旋转后数组的第M个元素是

   { (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

   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;
     
           // Mth element after k left rotations
           // is (K+M-1)%N th element of the
           // original array
           var index = (K + M - 1) % N;
     
           var result = a[index];
     
           // Return the result
           return result;
       }
     
       // Driver code
         
     
           // Array initialization
           var a = [ 3, 4, 5, 23 ];
     
           // Size of the array
           var N = a.length;
     
           // Given K rotation and Mth element
           // to be found after K rotation
           var K = 2, M = 1;
     
           // Function call
           document.write(getFirstElement(a, N, K, M));
     
   // This code contributed by gauravrajput1 
     

   输出：

   5

   时间复杂度： O(1)
   辅助空间： O(1)

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