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 consisting 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 failsvoid printCase(int n){ // only case where it fails if (n <= 2) { cout << -1; return; } for (int i = n; i >= 1; i--) cout << i << " ";}// Driver Codeint main(){ int n = 3; printCase(n); return 0;} |
Java
// Java program to find a case where the// given algorithm failsimport java.io.*;class GFG { // Function to print a case where// the given sorting algorithm failsstatic void printCase(int n){// only case where it failsif (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 failsdef printCase(n): # only case where it fails if (n <= 2) : print("-1") return for i in range(n, 0, -1): print(i, end = " ")# Driver Codeif __name__ == "__main__": n = 3 printCase(n)# This code is contributed # by ChitraNayal |
C#
// C# program to find a case where the// given algorithm failsusing System;class GFG{// Function to print a case where// the given sorting algorithm failsstatic 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 Codepublic 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 failsfunction 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?> |
Javascript
<script> // Javascript 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) { document.write(-1); return; } for (let i = n; i >= 1; i--) document.write(i + " "); } let n = 3; printCase(n); </script> |
Output:
3 2 1
Time Complexity: O(N)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
