具有 K 个完美索引的前 N 个自然数的字典序最小排列

给定两个正整数N和K ，任务是按字典顺序找到前N 个自然数的最小排列，使得恰好有K 个完美索引。

An index i in an array is said to be perfect if all the elements at indices smaller than i are smaller than the element at index i.
 

例子：

Input: N = 10, K = 3
Output: 8 9 10 1 2 3 4 5 6 7
Explanation: There are exactly 3 perfect indices 0, 1 and 2.

Input: N = 12, K = 4
Output: 9 10 11 12 1 2 3 4 5 6 7 8

朴素的方法：这个想法是生成前N 个自然数的所有可能排列，并打印出字典序最小且具有K 个完美索引的排列。
时间复杂度： O(N*N!)
辅助空间： O(1)

高效方法：为了优化上述方法，其思想是观察到最小排列将具有范围[1, N]的最后K 个元素按递增顺序排在前面。其余元素可以按递增顺序附加。请按照以下步骤解决问题：

• 初始化数组A[]以存储前N 个自然数的字典序最小排列。
• 使用变量（例如i ）迭代范围[0, K – 1] ，然后更新数组元素A[i]以存储(N – K + 1) + i 。
• 使用变量i迭代范围[K, N – 1]并将数组元素A[i]更新为(i – K + 1) 。
• 完成上述步骤后，打印数组A[] ，该数组包含按字典顺序排列的具有K 个完美索引的最小排列。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to print the lexicographically
// smallest permutation with K perfect indices
void findPerfectIndex(int N, int K)
{
    // Iterator to traverse the array
    int i = 0;
 
    // Traverse first K array indices
    for (; i < K; i++) {
        cout << (N - K + 1) + i << " ";
    }
 
    // Traverse remaining indices
    for (; i < N; i++) {
        cout << i - K + 1 << " ";
    }
}
 
// Driver Code
int main()
{
    int N = 10, K = 3;
    findPerfectIndex(N, K);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to print the lexicographically
// smallest permutation with K perfect indices
static void findPerfectIndex(int N, int K)
{
   
    // Iterator to traverse the array
    int i = 0;
 
    // Traverse first K array indices
    for (; i < K; i++)
    {
        System.out.print((N - K + 1) + i+ " ");
    }
 
    // Traverse remaining indices
    for (; i < N; i++)
    {
        System.out.print(i - K + 1+ " ");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 10, K = 3;
    findPerfectIndex(N, K);
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python program for the above approach
 
# Function to print the lexicographically
# smallest permutation with K perfect indices
def findPerfectIndex(N, K) :
     
    # Iterator to traverse the array
    i = 0
 
    # Traverse first K array indices
    for i in range(K):
        print((N - K + 1) + i, end = " ")
     
 
    # Traverse remaining indices
    for i in range(3, N):
        print( i - K + 1, end = " ")
     
# Driver Code
 
N = 10
K = 3
findPerfectIndex(N, K)
 
# This code is contributed by code_hunt.

C#

// C# program for the above approach
using System;
class GFG
{
 
// Function to print the lexicographically
// smallest permutation with K perfect indices
static void findPerfectIndex(int N, int K)
{
   
    // Iterator to traverse the array
    int i = 0;
 
    // Traverse first K array indices
    for (; i < K; i++)
    {
        Console.Write((N - K + 1) + i+ " ");
    }
 
    // Traverse remaining indices
    for (; i < N; i++)
    {
        Console.Write(i - K + 1+ " ");
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    int N = 10, K = 3;
    findPerfectIndex(N, K);
}
}
 
// This code is contributed by susmitakundugoaldanga.

Javascript

 
// javascript program for the above approach
// Function to prvar the lexicographically
// smallest permutation with K perfect indices
function findPerfectIndex(N , K)
{
   
    // Iterator to traverse the array
    var i = 0;
 
    // Traverse first K array indices
    for (; i < K; i++)
    {
        document.write((N - K + 1) + i+ " ");
    }
 
    // Traverse remaining indices
    for (; i < N; i++)
    {
        document.write(i - K + 1+ " ");
    }
}
 
// Driver Code
var N = 10, K = 3;
findPerfectIndex(N, K);
 
// This code is contributed by 29AjayKumar

输出：

8 9 10 1 2 3 4 5 6 7

时间复杂度： O(N)
辅助空间： O(1)