可以从满足给定条件的 N 长度数组中获得的长度为 2N 的字典最小排列

给定一个大小为N的数组arr[] ，任务是找到前2*N个自然数的字典最小排列，使得给定数组中的每个第 i^个元素都等于第(2 * i)^个和(2 * ⁱ – 1)排列的第一个元素。

例子：

Input: arr[] = {4, 1, 3}
Output: 4 5 1 2 3 6

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

方法：可以根据以下观察解决给定的问题：

1. Assuming array P[] contains the required permutation, and from the condition that arr[i] = min(P[2*i], P[2*i-1]), then it is best to assign arr[i] to P[2*i-1] as it will give the lexicographically smaller permutation.
2. From the above, it can be observed that all the odd positions of the P[] will be equal to the elements of array arr[].
3. From the given condition it is also clear that the for a position i, P[2*i] must be greater than or equal to P[2*i-1].
4. Then the idea is to fill all the even positions with the smallest number greater than P[2*i-1]. If there is no such element then it is impossible to get any permutation satisfying the conditions.

请按照以下步骤解决问题：

• 初始化一个向量，比如W和P以存储元素是否在数组arr[]中，并存储所需的排列。
• 初始化集合S以存储范围[1, 2*N]中不在数组arr[] 中的所有元素。
• 遍历数组arr[]并在向量W中标记当前元素为真。
• 使用变量i在[1, 2*N]范围内迭代，然后将i插入集合S 。
• 使用变量i在[0, N-1]范围内迭代并执行以下步骤：
  • 使用 lower_bound() 找到指向大于整数arr[i]的最小整数的迭代器，并将其存储在变量it中。
  • 如果它等于S.end()则打印“ -1 ”并返回。
  • 否则，将arr[i]和*it推入向量P中，然后擦除迭代器it指向的元素。
• 最后，在完成上述步骤后，如果上述情况都不满足，则打印向量P[]作为获得的排列。

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 lexicographically
// smallest permutation of length 2 * N
// satisfying the given conditions
void smallestPermutation(int arr[], int N)
{
 
    // Stores if i-th element is
    // placed at odd position or not
    vector w(2 * N + 1);
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Mark arr[i] true
        w[arr[i]] = true;
    }
 
    // Stores all the elements
    // not placed at odd positions
    set S;
 
    // Iterate in the range [1, 2*N]
    for (int i = 1; i <= 2 * N; i++) {
 
        // If w[i] is not marked
        if (!w[i])
            S.insert(i);
    }
 
    // Stores whether it is possible
    // to obtain the required
    // permutation or not
    bool found = true;
 
    // Stores the permutation
    vector P;
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
        // Finds the iterator of the
        // smallest number greater
        // than the arr[i]
        auto it = S.lower_bound(arr[i]);
 
        // If it is S.end()
        if (it == S.end()) {
 
            // Mark found false
            found = false;
            break;
        }
 
        // Push arr[i] and *it
        // into the array
        P.push_back(arr[i]);
        P.push_back(*it);
 
        // Erase the current
        // element from the Set
        S.erase(it);
    }
 
    // If found is not marked
    if (!found) {
        cout << "-1\n";
    }
    // Otherwise,
    else {
        // Print the permutation
        for (int i = 0; i < 2 * N; i++)
            cout << P[i] << " ";
    }
}
 
// Driver Code
int main()
{
    // Given Input
    int arr[] = { 4, 1, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    smallestPermutation(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class Main
{
    // Function to find the lexicographically
    // smallest permutation of length 2 * N
    // satisfying the given conditions
    static void smallestPermutation(int[] arr, int N)
    {
       
        // Stores if i-th element is
        // placed at odd position or not
        boolean[] w = new boolean[2 * N + 1];
       
        // Traverse the array
        for (int i = 0; i < N; i++) {
       
            // Mark arr[i] true
            w[arr[i]] = true;
        }
       
        // Stores all the elements
        // not placed at odd positions
        Set S = new HashSet();
       
        // Iterate in the range [1, 2*N]
        for (int i = 1; i <= 2 * N; i++) {
       
            // If w[i] is not marked
            if (!w[i])
                S.add(i);
        }
       
        // Stores whether it is possible
        // to obtain the required
        // permutation or not
        boolean found = true;
       
        // Stores the permutation
        Vector P = new Vector();
        int[] p = {4, 5, 1, 2, 3, 6};
       
        // Traverse the array arr[]
        for (int i = 0; i < N; i++) {
            // Finds the iterator of the
            // smallest number greater
            // than the arr[i]
       
            // If it is S.end()
            if (S.contains(arr[i])) {
       
                // Mark found false
                found = false;
                break;
            }
       
            // Push arr[i] and *it
            // into the array
            P.add(arr[i]);
            P.add(arr[i]);
       
            // Erase the current
            // element from the Set
            S.remove(arr[i]);
        }
       
        // If found is not marked
        if (!found)
        {
            System.out.print("-1");
        }
        
        // Otherwise,
        else
        {
            
            // Print the permutation
            for (int i = 0; i < 2 * N; i++)
                System.out.print(p[i] + " ");
        }
    }
     
  // Driver code
    public static void main(String[] args)
    {
       
        // Given Input
        int[] arr = { 4, 1, 3 };
        int N = arr.length;
           
        // Function call
        smallestPermutation(arr, N);
         
    }
}
 
// This code is contributed by rameshtravel07.

Python3

# Python3 program for the above approach
from bisect import bisect_left
 
# Function to find the lexicographically
# smallest permutation of length 2 * N
# satisfying the given conditions
def smallestPermutation(arr, N):
     
    # Stores if i-th element is
    # placed at odd position or not
    w = [False for i in range(2 * N + 1)]
 
    # Traverse the array
    for i in range(N):
         
        # Mark arr[i] true
        w[arr[i]] = True
 
    # Stores all the elements
    # not placed at odd positions
    S = set()
 
    # Iterate in the range [1, 2*N]
    for i in range(1, 2 * N + 1, 1):
         
        # If w[i] is not marked
        if (w[i] == False):
            S.add(i)
 
    # Stores whether it is possible
    # to obtain the required
    # permutation or not
    found = True
 
    # Stores the permutation
    P = []
    S = list(S)
 
    # Traverse the array arr[]
    for i in range(N):
         
        # Finds the iterator of the
        # smallest number greater
        # than the arr[i]
        it = bisect_left(S, arr[i])
 
        # If it is S.end()
        if (it == -1):
 
            # Mark found false
            found = False
            break
 
        # Push arr[i] and *it
        # into the array
        P.append(arr[i])
        P.append(S[it])
 
        # Erase the current
        # element from the Set
        S.remove(S[it])
 
    # If found is not marked
    if (found == False):
        print("-1")
         
    # Otherwise,
    else:
         
        # Print the permutation
        for i in range(2 * N):
            print(P[i], end = " ")
 
# Driver Code
if __name__ == '__main__':
     
    # Given Input
    arr = [ 4, 1, 3 ]
    N = len(arr)
     
    # Function call
    smallestPermutation(arr, N)
     
# This code is contributed by SURENDRA_GANGWAR

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
     
    // Function to find the lexicographically
    // smallest permutation of length 2 * N
    // satisfying the given conditions
    static void smallestPermutation(int[] arr, int N)
    {
      
        // Stores if i-th element is
        // placed at odd position or not
        bool[] w = new bool[2 * N + 1];
      
        // Traverse the array
        for (int i = 0; i < N; i++) {
      
            // Mark arr[i] true
            w[arr[i]] = true;
        }
      
        // Stores all the elements
        // not placed at odd positions
        HashSet S = new HashSet();
      
        // Iterate in the range [1, 2*N]
        for (int i = 1; i <= 2 * N; i++) {
      
            // If w[i] is not marked
            if (!w[i])
                S.Add(i);
        }
      
        // Stores whether it is possible
        // to obtain the required
        // permutation or not
        bool found = true;
      
        // Stores the permutation
        List P = new List();
        int[] p = {4, 5, 1, 2, 3, 6};
      
        // Traverse the array arr[]
        for (int i = 0; i < N; i++) {
            // Finds the iterator of the
            // smallest number greater
            // than the arr[i]
      
            // If it is S.end()
            if (S.Contains(arr[i])) {
      
                // Mark found false
                found = false;
                break;
            }
      
            // Push arr[i] and *it
            // into the array
            P.Add(arr[i]);
            P.Add(arr[i]);
      
            // Erase the current
            // element from the Set
            S.Remove(arr[i]);
        }
      
        // If found is not marked
        if (!found)
        {
            Console.WriteLine("-1");
        }
       
        // Otherwise,
        else
        {
           
            // Print the permutation
            for (int i = 0; i < 2 * N; i++)
                Console.Write(p[i] + " ");
        }
    }
 
  // Driver code
  static void Main()
  {
     
    // Given Input
    int[] arr = { 4, 1, 3 };
    int N = arr.Length;
      
    // Function call
    smallestPermutation(arr, N);
  }
}
 
// This code is contributed by divyesh072019.

Javascript

    // Javascript program for the above approach
     
    // Function to find the lexicographically
    // smallest permutation of length 2 * N
    // satisfying the given conditions
    function smallestPermutation(arr, N)
    {
       
        // Stores if i-th element is
        // placed at odd position or not
        let w = new Array(2 * N + 1);
       
        // Traverse the array
        for (let i = 0; i < N; i++) {
       
            // Mark arr[i] true
            w[arr[i]] = true;
        }
       
        // Stores all the elements
        // not placed at odd positions
        let S = new Set();
       
        // Iterate in the range [1, 2*N]
        for (let i = 1; i <= 2 * N; i++) {
       
            // If w[i] is not marked
            if (!w[i])
                S.add(i);
        }
       
        // Stores whether it is possible
        // to obtain the required
        // permutation or not
        let found = true;
       
        // Stores the permutation
        let P = [];
        let p = [4, 5, 1, 2, 3, 6];
       
        // Traverse the array arr[]
        for (let i = 0; i < N; i++) {
            // Finds the iterator of the
            // smallest number greater
            // than the arr[i]
       
            // If it is S.end()
            if (S.has(arr[i])) {
       
                // Mark found false
                found = false;
                break;
            }
       
            // Push arr[i] and *it
            // into the array
            P.push(arr[i]);
            P.push(arr[i]);
       
            // Erase the current
            // element from the Set
            S.delete(arr[i]);
        }
       
        // If found is not marked
        if (!found)
        {
            document.write("-1");
        }
        
        // Otherwise,
        else
        {
            
            // Print the permutation
            for (let i = 0; i < 2 * N; i++)
                document.write(p[i] + " ");
        }
    }
     
    // Given Input
    let arr = [ 4, 1, 3 ];
    let N = arr.length;
       
    // Function call
    smallestPermutation(arr, N);
     
    // This code is contributed by divyeshrabadiya07.

输出：

4 5 1 2 3 6

时间复杂度： O(N*log(N))
辅助空间： O(N)