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Print a case where the given sorting algorithm fails
  • Last Updated : 08 Jan, 2019

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++

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// 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;
}

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Java

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// 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
    }
}

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Python 3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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Output:

3 2 1

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

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