按字典顺序排列的前N个自然数的最小排列，具有K个完美指数

给定两个正整数N和K ，任务是从字典上找到前N个自然数的最小排列，从而恰好有K个完美指数。

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

天真的方法：这个想法是生成所有第一个N个自然数的所有可能排列，并打印在字典上最小的，具有K个完美索引的该排列。
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，我们的想法是观察到最小的排列将以递增的顺序在前面排列范围为[1，N]的最后K个元素。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 []以存储按字典顺序排列的前N个自然数的最小排列。
• 使用变量i遍历[0，K – 1]范围，并更新数组元素A [i]以存储(N – K + 1)+ i 。
• 使用变量i在[K，N – 1]范围内进行迭代，并将数组元素A [i]更新为(i – K + 1) 。
• 完成上述步骤后，按字典顺序打印数组A [] ，该数组按字典顺序包含具有K个完善索引的最小排列。

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.

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)