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 <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; 
} 

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

<?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 
?> 

Output:

3 2 1

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

My Personal Notes

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

C++

Java

Python 3

C#

PHP

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