Minimum Good Permutation - MINPERM CodeChef Easy Question

Welcome back !

Here is the link to today's PROBLEM

So , the question is simple enough , we just need to find a perumtation of 1 to n where pi != i and is smallest .

So , what we do is we make two cases.

If the length is even we just swap two adjacent elements in pairs , like 1,2 to 2,1 1,2,3,4 to 2,1,4,3 and so on..

If the lenght is odd , we swap adjacent elements amd in the end swap nth and n-1th pair . Eg. 1,2,3 to 2,1,3 and then finally 2,3,1

1,2,3,4,5 to 2,1,4,3,5 and then 2,1,4,5,3.

Here is the Code for that :-

Code :-

#include <iostream>

using namespace std;

int main() {

    int t;

    scanf("%d",&t);

    while(t--)

    {

        long n;

        scanf("%ld",&n);

        long arr[n];

        long i=0;

        if(n%2==0)

        {

            while(i!=n)

            {

                arr[i]=i+1+1;

                arr[i+1]=i+1;

                i=i+2;

            }

        }

        else

        {

            while(i!=n+1)

            {

                arr[i]=i+1+1;

                arr[i+1]=i+1;

                i=i+2;

            }

            arr[n-1]--;

            long k;

            k=arr[n-2];

            arr[n-2]=arr[n-1];

            arr[n-1]=k;

            

            

        }

        i=0;

        while(i!=n)

        {

            cout<<arr[i]<<" ";

            i++;

        }

        cout<<endl;

    }

// your code goes here

return 0;

}

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