Getiing TLE in 1 testcase

#1
#2

@aiman.mumtaz
You have to modify your code in such a way, that you don’t end producing any duplicates.
This trick helps in achieving that

if (a[i] != a[j] or i == j) {
		swap(a[i], a[j]);
		permute(a, i + 1, n, s);
	}
#3

why is the i==j condition used??

#5

to consider the case when a[i]==a[j] only once.

#6

What will be the case when i == j?

#7

to take into the account the first occurence.

#8

Is using vector better than using set?

#9

If yes, then how? Please Explain

#10

Vector and set both are different and have different uses. There is no clear cut answer, which one is better.

#11

So I can use both for this problem?

#12

Yes, but using set would be an overkill, since we are not producing duplicates

#13

actually we are producing duplicates with 1 1 3 as input

#14

Please refer to the code I mentioned earlier, it is not producing any duplicates… Please check