具有给定数组作为子序列的字典上最小的排列数，直到 K

给定一个整数K和一个数组arr[]在[1, K]范围内有N个成对不同的整数，任务是找到前K个正整数的字典序最小排列，使得给定数组arr[]是一个子序列的排列。

例子：

Input: arr[] = {1, 3, 5, 7}, K = 8
Output: 1 2 3 4 5 6 7 8
Explanation: {1, 2, 3, 4, 5, 6, 7, 8} is the lexicographically smallest permutation of the first 8 positive integers such that the given array {1, 3, 5, 7} is also a subsequence of the permutation.

Input: arr[] = {6, 4, 2, 1}, K=7
Output: 3 5 6 4 2 1 7

编程需要懂一点英语

方法：给定的问题可以通过使用贪心方法来解决。以下是要遵循的步骤：

使用本文讨论的方法创建一个向量missing[] ，以升序存储范围[1, K]中不存在于给定数组arr[]中的整数。
创建两个指针p1和p2 ，它们分别将当前索引存储在arr[]和missing[]中。最初，两者都等于0 。
贪婪地将arr[p1]和缺少的 [p2]的最小值存储到一个向量中，比如ans[]并将相应的指针递增到下一个位置，直到存储的整数的计数小于K 。
为了使事情变得更容易，在数组missing[]和arr[]的末尾附加 INT_MAX 可以避免越界。
完成上述步骤后， ans[]中存储的所有值都是需要的结果。

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 lexicographically
// smallest permutation such that the
// given array is a subsequence of it
void findPermutation(int K, vector arr)
{
    // Stores the missing elements in
    // arr in the range [1, K]
    vector missing;
 
    // Stores if the ith element is
    // present in arr or not
    vector visited(K + 1, 0);
 
    // Loop to mark all integers present
    // in the array as visited
    for (int i = 0; i < arr.size(); i++) {
        visited[arr[i]] = 1;
    }
 
    // Loop to insert all the integers
    // not visited into missing
    for (int i = 1; i <= K; i++) {
        if (!visited[i]) {
            missing.push_back(i);
        }
    }
    // Append INT_MAX at end in order
    // to prevent going out of bounds
    arr.push_back(INT_MAX);
    missing.push_back(INT_MAX);
 
    // Pointer to the current element
    int p1 = 0;
 
    // Pointer to the missing element
    int p2 = 0;
 
    // Stores the required permutation
    vector ans;
 
    // Loop to construct the permutation
    // using greedy approach
    while (ans.size() < K) {
 
        // If missing element is smaller
        // that the current element insert
        // missing element
        if (arr[p1] < missing[p2]) {
            ans.push_back(arr[p1]);
            p1++;
        }
 
        // Insert current element
        else {
            ans.push_back(missing[p2]);
            p2++;
        }
    }
 
    // Print the required Permutation
    for (int i = 0; i < K; i++) {
        cout << ans[i] << " ";
    }
}
 
// Driver Code
int main()
{
    int K = 7;
    vector arr = { 6, 4, 2, 1 };
 
    // Function Call
    findPermutation(K, arr);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find the lexicographically
// smallest permutation such that the
// given array is a subsequence of it
static void findPermutation(int K, Vector arr)
{
    // Stores the missing elements in
    // arr in the range [1, K]
    Vector missing = new Vector();
 
    // Stores if the ith element is
    // present in arr or not
    boolean visited[] = new boolean[K + 1];
 
    // Loop to mark all integers present
    // in the array as visited
    for (int i = 0; i < arr.size(); i++) {
        visited[arr.get(i)] = true;
    }
 
    // Loop to insert all the integers
    // not visited into missing
    for (int i = 1; i <= K; i++) {
        if (!visited[i]) {
            missing.add(i);
        }
    }
    // Append Integer.MAX_VALUE at end in order
    // to prevent going out of bounds
    arr.add(Integer.MAX_VALUE);
    missing.add(Integer.MAX_VALUE);
 
    // Pointer to the current element
    int p1 = 0;
 
    // Pointer to the missing element
    int p2 = 0;
 
    // Stores the required permutation
    Vector ans = new Vector();
 
    // Loop to construct the permutation
    // using greedy approach
    while (ans.size() < K) {
 
        // If missing element is smaller
        // that the current element insert
        // missing element
        if (arr.get(p1) < missing.get(p2)) {
            ans.add(arr.get(p1));
            p1++;
        }
 
        // Insert current element
        else {
            ans.add(missing.get(p2));
            p2++;
        }
    }
 
    // Print the required Permutation
    for (int i = 0; i < K; i++) {
        System.out.print(ans.get(i)+ " ");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int K = 7;
    Integer []a = {6, 4, 2, 1};
    Vector arr = new Vector<>(Arrays.asList(a));
 
    // Function Call
    findPermutation(K, arr);
 
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python program for the above approach
 
# Function to find the lexicographically
# smallest permutation such that the
#  given array is a subsequence of it
def findPermutation(K, arr):
   
    # Stores the missing elements in
        # arr in the range [1, K]
    missing = []
 
    # Stores if the ith element is
    # present in arr or not
    visited = [0]*(K+1)
 
    # Loop to mark all integers present
    # in the array as visited
    for i in range(4):
        visited[arr[i]] = 1
 
    # Loop to insert all the integers
    # not visited into missing
    for i in range(1, K+1):
        if(not visited[i]):
            missing.append(i)
 
    # Append INT_MAX at end in order
        # to prevent going out of bounds
    INT_MAX = 2147483647
    arr.append(INT_MAX)
    missing.append(INT_MAX)
 
    # Pointer to the current element
    p1 = 0
 
    # Pointer to the missing element
    p2 = 0
 
    # Stores the required permutation
    ans = []
 
    # Loop to construct the permutation
    # using greedy approach
 
    while (len(ans) < K):
        # If missing element is smaller
                # that the current element insert
        # missing element
        if (arr[p1] < missing[p2]):
            ans.append(arr[p1])
            p1 = p1 + 1
 
        # Insert current element
        else:
            ans.append(missing[p2])
            p2 = p2 + 1
 
    # Print the required Permutation
    for i in range(0, K):
        print(ans[i], end=" ")
 
# Driver Code
if __name__ == "__main__":
    K = 7
    arr = [6, 4, 2, 1]
 
    # Function Call
    findPermutation(K, arr)
     
    # This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
using System.Collections;
class GFG{
 
// Function to find the lexicographically
// smallest permutation such that the
// given array is a subsequence of it
static void findPermutation(int K, ArrayList arr)
{
   
    // Stores the missing elements in
    // arr in the range [1, K]
    ArrayList missing = new ArrayList();
 
    // Stores if the ith element is
    // present in arr or not
    bool [] visited = new bool[K + 1];
 
    // Loop to mark all integers present
    // in the array as visited
    for (int i = 0; i < arr.Count; i++) {
        visited[(int)arr[i]] = true;
    }
 
    // Loop to insert all the integers
    // not visited into missing
    for (int i = 1; i <= K; i++) {
        if (!visited[i]) {
            missing.Add(i);
        }
    }
    // Append Int32.MaxValue at end in order
    // to prevent going out of bounds
    arr.Add(Int32.MaxValue);
    missing.Add(Int32.MaxValue);
 
    // Pointer to the current element
    int p1 = 0;
 
    // Pointer to the missing element
    int p2 = 0;
 
    // Stores the required permutation
    ArrayList ans = new ArrayList();
 
    // Loop to construct the permutation
    // using greedy approach
    while (ans.Count < K) {
 
        // If missing element is smaller
        // that the current element insert
        // missing element
        if ((int)arr[p1] < (int)missing[p2]) {
            ans.Add(arr[p1]);
            p1++;
        }
 
        // Insert current element
        else {
            ans.Add(missing[p2]);
            p2++;
        }
    }
 
    // Print the required Permutation
    for (int i = 0; i < K; i++) {
        Console.Write(ans[i]+ " ");
    }
}
 
// Driver Code
public static void Main()
{
    int K = 7;
    int [] a = {6, 4, 2, 1};
    ArrayList arr = new ArrayList(a);
 
    // Function Call
    findPermutation(K, arr);
}
}
 
// This code is contributed by ihritik.

Javascript

输出：

3 5 6 4 2 1 7

时间复杂度： O(N)
辅助空间： O(N)