Total Number of Graphs Possible (HELP!)

There was problem given to be in a hiring challenge which i was not able to solve completely so i am just curious to know what i missed.

Problem statement:

Given a number of vertices (n), determine the number of ways these vertices can form a graph, the graph can be disjoint so it is not necessary to connect all vertices. The answer can be very large so return modulo 10^9+7 ;

Constrains: 1 < n < 10^9

What i did is used the formula 2^(n*(n-1)/2) to get number of all possible graphs and used modular exponentiation to compute value, but i was not able to pass all test cases.

Can anyone tell me what i did wrong or is it some other concept to be used …??

I guess the answer should just be 2^n, as for each vertice you have two options: Select/ Not select

So for n vertices answer should be 22…2=2^n

Can the connected components be rearranged?

For example, is the graph 1 -> 2 -> 3 is different from 2 -> 1 -> 3?

I think all the vertices should be included, but the edges can be different.

Did you by any chance do
2^(n*(n-1)/2 % mod) ?
Your answer is correct if they mentioned distinct.


@dormordo For n=4 answer was 64

@crap_the_coder Yes i think because each will generate new graph and in example output answer for n = 4 was 64

@everule1 Yes i applied same formula and it passed 5 out of 13 test cases, is it related to modular exponentiation…?

Im sorry, but that is wrong…

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You can take %(mod -1)
I think you should read fermat’s little theorem

i know fermat’s little theorem but how is it useful in this case…??

Let 10^9 + 7 =p
Now say
You want 2^p+3
You take 2^3 =8 but 2^p+3 is 16 mod p
Because 2^p is 2 mod p.

But i have to take mod with overall value
let x = (n*(n-1))/2;

then what i was calculating is (2^x)%mod

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Please just post the code you submitted, as best as you can remember it :slight_smile:

using namespace std;

#define ll long long int
const ll MOD = 1e9 + 7;

ll powerLL(ll x, ll n)
ll ans = 1;
while (n) {
if (n & 1)
ans = ans * x % MOD;
n = n / 2;
x = x * x % MOD;
return ans;

ll powerStrings(string sa, string sb)

ll a = 0, b = 0; 

for (int i = 0; i < sa.length(); i++) 
    a = (a * 10 + (sa[i] - '0')) % MOD; 

for (int i = 0; i < sb.length(); i++) 
    b = (b * 10 + (sb[i] - '0')) % (MOD - 1); 

return powerLL(a, b); 


int main()

int n;  cin>>n;
ll tmp = (n*(n-1))/2;

string p = to_string(tmp); 

cout << powerStrings("2", p) << endl; 
return 0; 


1 Like

Since n is declared as an int, this will overflow for sufficiently large n: n=46342 should do the trick.

Lesson: Instead of trying to describe your code, just post it (preferably formatted for readability/ compilability) :slight_smile:


i had tried to declare it long long int also but it didn’t helped much, so what should i have done here then …??

Declaring n as long long int would eliminate one source of error, and I can’t see any others offhand.


Wait - why “- 1” here?

b = (b * 10 + (sb[i] - '0')) % (MOD - 1); 


Actually, you shouldn’t be taking the exponent modulo anything - just leave it alone.


Oh, I think I see. Hmmm … can’t find any further errors after you change n to be long long int.


Idk doesn’t give the right answer for 10^9 on codechef ide

What is the right answer? I get


but may have made a mistake.