通过根据给定条件向左或向右旋转来减少给定的 [1, N] 数组

给定一个前N个自然数的排序数组arr[]和一个整数X ，任务是在执行以下操作(N – 1)次后打印最后一个剩余元素：

• 如果X的值为1 ，则将数组右旋1 个单位并删除最后一个元素。
• 如果X的值为2 ，则将数组左旋1 个单位并删除第一个元素。

例子：

Input: N = 5, arr[] = {1, 2, 3, 4, 5}, X = 1
Output: 3
Explanation:
The operations are performed as follows:

1. Rotating the array right by 1 unit modifies the array to {5, 1, 2, 3, 4}. Deleting the array element modifies the array to {5, 1, 2, 3}.
2. Rotating the array right by 1 unit modifies the array to {3, 5, 1, 2}. Deleting the array element modifies the array to {3, 5, 1}.
3. Rotating the array right by 1 unit modifies the array to {1, 3, 5}. Deleting the array element modifies the array to {1, 3}.
4. Rotating the array right by 1 unit modifies the array to {3, 1}. Deleting the array element modifies the array to {3}.

Therefore, the last remaining element is 3.

Input: N = 5, arr[] = {1, 2, 3, 4, 5}, X = 2
Output: 3

朴素方法：解决给定问题的最简单方法是将范围[1, N]的所有数字压入一个双端队列，并根据给定的X值执行给定的操作(N – 1)次，然后打印最后剩下的元素。

时间复杂度： O(N)
辅助空间： O(1)

有效方法：可以根据以下观察优化给定问题：

1. 如果X的值为1 ，那么最后一个左元素将是2的最小幂之间的差大于N 和N。
2. 如果X的值为2 ，那么最后一个左元素将为2*(N – D) + 1，其中D表示小于或等于N的2的最大幂。

请按照以下步骤解决问题：

• 将大于N的 2 的幂存储在变量中，例如nextPower。
• 如果X的值为1 ，则打印值(nextPower – N)作为结果。
• 否则，打印值2*(N – nextPower / 2) + 1作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the last element
// left after performing N - 1 queries
// of type X
int rotate(int arr[], int N, int X)
{
    // Stores the next power of 2
    long long int nextPower = 1;
 
    // Iterate until nextPower is at most N
    while (nextPower <= N)
        nextPower *= 2;
 
    // If X is equal to 1
    if (X == 1)
        return nextPower - N;
 
    // Stores the power of 2 less than or
    // equal to N
    long long int prevPower = nextPower / 2;
 
    // Return the final value
    return 2 * (N - prevPower) + 1;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int X = 1;
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << rotate(arr, N, X);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG
{
   
  // Function to find the last element
// left after performing N - 1 queries
// of type X
static int rotate(int arr[], int N, int X)
{
   
    // Stores the next power of 2
    long nextPower = 1;
 
    // Iterate until nextPower is at most N
    while (nextPower <= N)
        nextPower *= 2;
 
    // If X is equal to 1
    if (X == 1)
        return (int) nextPower - N;
 
    // Stores the power of 2 less than or
    // equal to N
    long prevPower = nextPower / 2;
 
    // Return the final value
    return 2 * (N - (int)prevPower) + 1;
}
 
// Driver Code
    public static void main (String[] args) {
        int arr[] = { 1, 2, 3, 4, 5 };
    int X = 1;
    int N =arr.length;
 
   System.out.println(rotate(arr, N, X));
 
    }
}
 
    // This code is contributed by Potta Lokesh

Python3

# Python3 program for the above approach
 
# Function to find the last element
# left after performing N - 1 queries
# of type X
def rotate(arr, N, X):
    # Stores the next power of 2
    nextPower = 1
 
    # Iterate until nextPower is at most N
    while (nextPower <= N):
        nextPower *= 2
 
    # If X is equal to 1
    if (X == 1):
        ans = nextPower - N
        return ans
 
    # Stores the power of 2 less than or
    # equal to N
    prevPower = nextPower // 2
 
    # Return the final value
    return 2 * (N - prevPower) + 1
 
 
# Driver Code
arr = [1, 2, 3, 4, 5]
X = 1
N = len(arr)
print(rotate(arr, N, X))
 
# This code is contributed by amreshkumar3.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the last element
// left after performing N - 1 queries
// of type X
static int rotate(int []arr, int N, int X)
{
     
    // Stores the next power of 2
    int nextPower = 1;
 
    // Iterate until nextPower is at most N
    while (nextPower <= N)
        nextPower *= 2;
 
    // If X is equal to 1
    if (X == 1)
        return nextPower - N;
 
    // Stores the power of 2 less than or
    // equal to N
    int prevPower = nextPower / 2;
 
    // Return the final value
    return 2 * (N - prevPower) + 1;
}
 
// Driver Code
public static void Main()
{
    int []arr = { 1, 2, 3, 4, 5 };
    int X = 1;
    int N = arr.Length;
 
    Console.Write(rotate(arr, N, X));
}
}
 
// This code is contributed by SURENDRA_GANGWAR

Javascript

       // JavaScript program for the above approach
 
       // Function to find the last element
       // left after performing N - 1 queries
       // of type X
       function rotate(arr, N, X) {
           // Stores the next power of 2
           let nextPower = 1;
 
           // Iterate until nextPower is at most N
           while (nextPower <= N)
               nextPower *= 2;
 
           // If X is equal to 1
           if (X == 1)
               return nextPower - N;
 
           // Stores the power of 2 less than or
           // equal to N
           let prevPower = nextPower / 2;
 
           // Return the final value
           return 2 * (N - prevPower) + 1;
       }
 
       // Driver Code
       let arr = [1, 2, 3, 4, 5];
       let X = 1;
       let N = arr.length;
 
       document.write(rotate(arr, N, X));
 
   // This code is contributed by Potta Lokesh
 
   

输出

3

时间复杂度： O(log N)
辅助空间： O(1)