从 P 前缀反转后获得的给定数组中找到原始数组

给定一个大小为N的数组arr[]和一个整数P (P < N)，任务是从通过P 个前缀反转获得的数组中找到原始数组，其中在第 i 个反转中，包含索引的数组的大小为 i 的前缀范围[0, i-1]被反转。

例子：

Input: arr[] = {4, 2, 1, 3, 5, 6}, P = 4.
Output:  1 2 3 4 5 6
Explanation: {1, 2, 3, 4, 5, 6} on prefix reversal P = 1 converts to {1, 2, 3, 4, 5, 6}.
{1, 2, 3, 4, 5, 6} on prefix reversal P = 2 converts to {2, 1, 3, 4, 5, 6}.
{2, 1, 3, 4, 5, 6} on prefix reversal P = 3 converts to {3, 1, 2, 4, 5, 6}
{3, 1, 2, 4, 5, 6} on prefix reversal P = 4 converts to  {4, 2, 1, 3, 5, 6} 
So answer is {1, 2, 3, 4, 5, 6}

Input: arr[] = {10, 9, 8, 3, 5, 6}, P = 3
Output: 9 8 10 3 5 6 

朴素方法：为了解决这个问题，在每个步骤中反转大小为i的前缀，对于范围[1，P]中的i ，从P大小的前缀开始，然后逐渐减小大小。

时间复杂度： O(N ² )
辅助空间： O(1)

高效方法：此解决方案基于两个指针方法。由于只有P个前缀反转，因此数组的前P个元素只会受到影响，其余的保持不变。因此，在P前缀反转之后，可以观察到原始数组和数组的模式。只应修改前P个元素。按照以下步骤解决上述问题：

• 初始化两个变量l = 0和r = P-1
• 初始化向量res以存储修改后的前缀和index = 0以跟踪奇数和偶数索引处的元素。
• 使用 while 循环遍历arr[] 的前缀。
  • 如果索引甚至将arr[l]推入向量 res 并增加l 。
  • 否则将arr[r]推入向量res并递减r 。
  • 也增加索引。
• 现在反转res并将修改后的前缀分配给 arr 的长度为 p 的前缀。
• 打印原始数组。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
void find_original_array(int arr[], int n, int p)
{
    // Initialize the r and l
    int r = p - 1;
    int l = 0;
 
    // Initialize index = 0
    // to track elements at
    // odd and even positions
    int index = 0;
    vector res;
 
    while (l <= r) {
 
        // If index is even
        if (index % 2 == 0) {
            res.push_back(arr[l++]);
        }
 
        // If index is odd
        else {
            res.push_back(arr[r--]);
        }
 
        // Increment index
        index = index + 1;
    }
 
    // Reverse the res
    reverse(res.begin(), res.end());
 
    // Assign the modified prefix to arr
    for (int i = 0; i < res.size(); i++) {
        arr[i] = res[i];
    }
 
    // Print the array arr
    // which is the original array
    // modified from the
    // prefix reversed array
    for (int i = 0; i < n; i++) {
        cout << arr[i] << " ";
    }
}
 
// Driver code
int main()
{
 
    int arr[] = { 4, 2, 1, 3, 5, 6 }, P = 4;
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    find_original_array(arr, n, P);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class GFG
{
  static void find_original_array(int arr[], int n, int p)
  {
 
    // Initialize the r and l
    int r = p - 1;
    int l = 0;
 
    // Initialize index = 0
    // to track elements at
    // odd and even positions
    int index = 0;
    ArrayList res = new ArrayList();
 
    while (l <= r) {
 
      // If index is even
      if (index % 2 == 0) {
        res.add(arr[l++]);
      }
 
      // If index is odd
      else {
        res.add(arr[r--]);
      }
 
      // Increment index
      index = index + 1;
    }
 
    // Reverse the res
    Collections.reverse(res);
 
    // Assign the modified prefix to arr
    for (int i = 0; i < res.size(); i++) {
      arr[i] = (int)res.get(i);
    }
 
    // Print the array arr
    // which is the original array
    // modified from the
    // prefix reversed array
    for (int i = 0; i < n; i++) {
      System.out.print(arr[i] + " ");
    }
  }
 
  // Driver code
  public static void main(String args[])
  {
 
    int arr[] = { 4, 2, 1, 3, 5, 6 }, P = 4;
    int n = arr.length;
 
    // Function call
    find_original_array(arr, n, P);
 
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Python program for the above approach
def find_original_array(arr, n, p):
 
    # Initialize the r and l
    r = p - 1;
    l = 0;
 
    # Initialize index = 0
    # to track elements at
    # odd and even positions
    index = 0;
    res = []
 
    while (l <= r):
 
        # If index is even
        if (index % 2 == 0):
            res.append(arr[l]);
            l += 1;
         
        # If index is odd
        else:
            res.append(arr[r]);
            r -= 1;
         
        # Increment index
        index = index + 1;
     
    # Reverse the res
    res.reverse();
 
    # Assign the modified prefix to arr
    for i in range(len(res)):
        arr[i] =  res[i];
     
    # Print array arr
    # which is the original array
    # modified from the
    # prefix reversed array
    for i in range(n):
        print(arr[i], end=" ");
     
# Driver code
if __name__ == '__main__':
 
    arr = [ 4, 2, 1, 3, 5, 6 ]
    P = 4;
    n = len(arr);
 
    # Function call
    find_original_array(arr, n, P);
 
# This code is contributed by gauravrajput1

C#

// C# program for the above approach
using System;
using System.Collections;
 
class GFG
{
  static void find_original_array(int []arr, int n, int p)
  {
 
    // Initialize the r and l
    int r = p - 1;
    int l = 0;
 
    // Initialize index = 0
    // to track elements at
    // odd and even positions
    int index = 0;
    ArrayList res = new ArrayList();
 
    while (l <= r) {
 
      // If index is even
      if (index % 2 == 0) {
        res.Add(arr[l++]);
      }
 
      // If index is odd
      else {
        res.Add(arr[r--]);
      }
 
      // Increment index
      index = index + 1;
    }
 
    // Reverse the res
    res.Reverse();
 
    // Assign the modified prefix to arr
    for (int i = 0; i < res.Count; i++) {
      arr[i] = (int)res[i];
    }
 
    // Print the array arr
    // which is the original array
    // modified from the
    // prefix reversed array
    for (int i = 0; i < n; i++) {
      Console.Write(arr[i] + " ");
    }
  }
 
  // Driver code
  public static void Main()
  {
 
    int []arr = { 4, 2, 1, 3, 5, 6 };
    int P = 4;
    int n = arr.Length;
 
    // Function call
    find_original_array(arr, n, P);
 
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

    // JavaScript program for the above approach
 
    const find_original_array = (arr, n, p) => {
     
        // Initialize the r and l
        let r = p - 1;
        let l = 0;
 
        // Initialize index = 0
        // to track elements at
        // odd and even positions
        let index = 0;
        let res = [];
 
        while (l <= r) {
 
            // If index is even
            if (index % 2 == 0) {
                res.push(arr[l++]);
            }
 
            // If index is odd
            else {
                res.push(arr[r--]);
            }
 
            // Increment index
            index = index + 1;
        }
 
        // Reverse the res
        res.reverse();
 
        // Assign the modified prefix to arr
        for (let i = 0; i < res.length; i++) {
            arr[i] = res[i];
        }
 
        // Print the array arr
        // which is the original array
        // modified from the
        // prefix reversed array
        for (let i = 0; i < n; i++) {
            document.write(`${arr[i]} `);
        }
    }
 
    // Driver code
    let arr = [4, 2, 1, 3, 5, 6], P = 4;
    let n = arr.length;
 
    // Function call
    find_original_array(arr, n, P);
 
// This code is contributed by rakeshsahni
 

输出

1 2 3 4 5 6 

时间复杂度： O(N)，其中 N 是数组的长度。
辅助空间： O(N)