might I suggest using Inclusion-Exclusion instead? much easier...

Enjoyed solving it!

int q;
cin>>q;
for(int k=1;k<=q;k++)
{
int n;
cin>>n;
int array[n];
for(int i=0;i<n;i++)
{
cin>>array[i];
}
int count=0;
sort(array,array+n);

do
{
int j;
for(j=0;j<(n-1);j++)
{
if(array[j]!=array[j+1])
{
continue;
}
else
{
break;
}
}
if(j==(n-1))
{
count+=1;
}
}while(next_permutation(array,array+n));
cout<<count<<endl;
}

can anyone suggest me why this program run successfully for testcase0 but for other test cases it show terminated due to timeout

As an alternative, you can try using a hash map. It's pretty unlikely that your solution is using all 9 million elements in the 2D array since that would mean your solution is too slow anyway. As social media usage worldwide continues to grow, so does the importance of incorporating social channels into the marketing plan for your startup with a social media marketing strategy. While many websites and companies outsource digital marketing. Among all spdload.com is reliable and suthentic service provider.

The number of beautiful permutations = number of total permutations - number of ugly permutations.

An ugly permutation is one that violates the condition fora beautiful permutation, i.e. it contains two identical numbers next to each other.

To calculate the number of ugly permutations: 1. Calculate the number of pairs of identical elements present in the input array. 2. Choose a particular pair. 3. Now place this pair in each of the positions in the array and permutate the rest of the array with the remaining elements. 4. Repeat steps 2&3. above for all possible pairs of identical elements. 5. Take care of repititions.