打印给定排序算法失败的情况

给定整数N，任务是查找未通过以下排序算法的N个元素。如果N个元素均未失败，则打印-1。

loop i from 1 to n-1
   loop j from i to n-1 
      if a[j]>a[i+1]  
         swap(a[i], a[j+1])

例子：

Input: N = 10 
Output: 10 9 8 7 6 5 4 3 2 1 

Input: N = 2
Output: -1

方法：在解决各种情况时，我们可以观察到仅对于n <= 2 ，给定的算法是无效的。大于N的任何值将在给定算法上失败。使用此给定算法无法对由N个数字(1、2、3 …. N)组成的排序数组进行反向排序。

下面是上述方法的实现：

C++

// C++ program to find a case where the
// given algorithm fails
  
#include 
using namespace std;
  
// Function to print a case
// where the given sorting algorithm fails
void printCase(int n)
{
    // only case where it fails
    if (n <= 2) {
        cout << -1;
        return;
    }
  
    for (int i = n; i >= 1; i--)
        cout << i << " ";
}
  
// Driver Code
int main()
{
    int n = 3;
  
    printCase(n);
  
    return 0;
}

Java

// Java program to find a case where the
// given algorithm fails
  
import java.io.*;
  
class GFG {
    // Function to print a case where
// the given sorting algorithm fails
static void printCase(int n)
{
// only case where it fails
if (n <= 2) {
        System.out.print(-1);
    return; 
      
} 
for (int i = n; i >= 1; i--)
        System.out.print(i + " ");
}
  
// Driver Code
      
    public static void main (String[] args) {
        int n = 3;
        printCase(n);
  
  
//This code is contributed by akt_mit
    }
}

Python 3

# Python 3 program to find a case 
# where the given algorithm fails
  
# Function to print a case where
# the given sorting algorithm fails
def printCase(n):
  
    # only case where it fails
    if (n <= 2) :
        print("-1")
        return
  
    for i in range(n, 0, -1):
        print(i, end = " ")
  
# Driver Code
if __name__ == "__main__":
      
    n = 3
  
    printCase(n)
  
# This code is contributed 
# by ChitraNayal

C#

// C# program to find a case where the
// given algorithm fails
using System;
  
class GFG
{
// Function to print a case where
// the given sorting algorithm fails
static void printCase(int n)
{
    // only case where it fails
    if (n <= 2) 
    {
        Console.Write(-1);
        return;
    }
  
    for (int i = n; i >= 1; i--)
        Console.Write(i + " ");
}
  
// Driver Code
public static void Main()
{
    int n = 3;
  
    printCase(n);
}
}
  
// This code is contributed 
// by Akanksha Rai

PHP

= 1; $i--)
    {
        echo ($i);
        echo(" ");
    }
}
  
// Driver Code
$n = 3;
  
printCase($n);
  
// This code is contributed 
// by Shivi_Aggarwal
?>

输出：

3 2 1

时间复杂度：O(N)
辅助空间：O(1)