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

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find a case where the
// given algorithm fails
  
#include <bits/stdc++.h>
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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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
    }
}

chevron_right


Python 3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to find a case where 
// the given algorithm fails
  
// Function to print a case where 
// the given sorting algorithm fails
function printCase($n)
{
    // only case where it fails
    if ($n <= 2) 
    {
        echo (-1);
        return;
    }
  
    for ($i = $n; $i >= 1; $i--)
    {
        echo ($i);
        echo(" ");
    }
}
  
// Driver Code
$n = 3;
  
printCase($n);
  
// This code is contributed 
// by Shivi_Aggarwal
?>

chevron_right


Output:

3 2 1

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



My Personal Notes arrow_drop_up

Just another competitive programmer

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.





Article Tags :
Practice Tags :


1


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.