# Print a case where the given sorting algorithm fails

Given an integer N, the task is to find N elements which fail the below-sorting algorithm. If none of the N elements fail, then print -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])
```

Examples:

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

Input: N = 2
Output: -1
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: On solving for various cases, we can observe that only for n<=2, the given algorithm is invalid. Any value of N above that will fail on the given algorithm. A sorted array consisiting of N numbers(1, 2, 3 . . N) in reverse order cannot be sorted using this given algorithm.

Below is the implementation of the above approach:

## 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 ` `?> `

Output:

```3 2 1
```

Time Complexity: O(N)
Auxiliary Space: O(1)

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