Sum of sum of all subsets of a set formed by first N natural numbers
Last Updated :
21 Sep, 2022
Given N, and ff(N) = f(1) + f(2) + ...... + f(N), where f(k) is the sum of all subsets of a set formed by first k natural numbers. The task is to find ff(N) modulo 1000000007.
Examples:
Input: 2
Output: 7
f(1) + f(2)
f(1) = 1 = 1
f(2) = 1 + 2 + {1 + 2} = 6
Input: 3
Output: 31
f(1) + f(2) + f(3)
f(1) = 1 = 1
f(2) = 1 + 2 + {1 + 2} = 6
f(3) = 1 + 2 + 3 + {1 + 2} + {2 + 3} + {1 + 3} + {1 + 2 + 3} = 24
Approach: Find a pattern of the sequence that will form. The values of f(1), f(2), f(3) are 1, 6 and 31 respectively. Let's find f(4).
f(4) = 1 + 2 + 3 + 4 + {1 + 2} + {1 + 3} + {1 + 4}
+ {2 + 3} + {2 + 4} + {3 + 4} + {1 + 2 + 3} + {1 + 2 + 4}
+ {1 + 3 + 4} + {2 + 3 + 4} + {1 + 2 + 3 + 4} = 80.
Hence ff(N) will be
ff(1) = f(1) = 1
ff(2) = f(1) + f(2) = 7
ff(3) = f(1) + f(2) + f(3) = 31
ff(4) = f(1) + f(2) + f(3) + f(4) = 111
.
.
.
The series formed is 1, 7, 31, 111... There exists a formula for it which is 2^n*(n^2 + n + 2) - 1. where, N is starting from zero.
Below is the implementation of the above approach.
C++
// C++ program to find Sum of all
// subsets of a set formed by
// first N natural numbers | Set-2
#include <bits/stdc++.h>
using namespace std;
// modulo value
#define mod (int)(1e9 + 7)
// Iterative Function to calculate (x^y)%p in O(log y)
int power(int x, int y, int p)
{
int res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0) {
// If y is odd, multiply x with the result
if (y & 1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// function to find ff(n)
int check(int n)
{
// In formula n is starting from zero
n--;
// calculate answer using
// formula 2^n*(n^2 + n + 2) - 1
int ans = n * n;
// whenever answer is greater than
// or equals to mod then modulo it.
if (ans >= mod)
ans %= mod;
ans += n + 2;
if (ans >= mod)
ans %= mod;
ans = (power(2, n, mod) % mod * ans % mod) % mod;
// adding modulo while subtraction is very necessary
// otherwise it will cause wrong answer
ans = (ans - 1 + mod) % mod;
return ans;
}
// Driver code
int main()
{
int n = 4;
// function call
cout << check(n) << endl;
return 0;
}
// C program to find Sum of all
// subsets of a set formed by
// first N natural numbers | Set-2
#include <stdio.h>
// modulo value
#define mod (int)(1e9 + 7)
// Iterative Function to calculate (x^y)%p in O(log y)
int power(int x, int y, int p)
{
int res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0) {
// If y is odd, multiply x with the result
if (y & 1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// function to find ff(n)
int check(int n)
{
// In formula n is starting from zero
n--;
// calculate answer using
// formula 2^n*(n^2 + n + 2) - 1
int ans = n * n;
// whenever answer is greater than
// or equals to mod then modulo it.
if (ans >= mod)
ans %= mod;
ans += n + 2;
if (ans >= mod)
ans %= mod;
ans = (power(2, n, mod) % mod * ans % mod) % mod;
// adding modulo while subtraction is very necessary
// otherwise it will cause wrong answer
ans = (ans - 1 + mod) % mod;
return ans;
}
// Driver code
int main()
{
int n = 4;
// function call
printf("%d\n",check(n));
return 0;
}
// This code is contributed by kothavvsaakash.
// Java program to find Sum of all
// subsets of a set formed by
// first N natural numbers | Set-2
class Geeks {
// Iterative Function to calculate
// (x^y)%p in O(log y)
static int power(int x, int y, int p)
{
// Initialize result
int res = 1;
// Update x if it is more
// than or equal to p
x = x % p;
while (y > 0) {
// If y is odd, multiply x
// with the result
if (y != 0)
res = (res * x) % p;
// y must be even now
// y = y / 2
y = y >> 1;
x = (x * x) % p;
}
return res;
}
// function to find ff(n)
static int check(int n)
{
// modulo value
int mod = (int)(1e9 + 7);
// In formula n is
// starting from zero
n--;
// calculate answer using
// formula 2^n*(n^2 + n + 2) - 1
int ans = n * n;
// whenever answer is greater than
// or equals to mod then modulo it.
if (ans >= mod)
ans %= mod;
ans += n + 2;
if (ans >= mod)
ans %= mod;
ans = (power(2, n, mod) % mod *
ans % mod) % mod;
// adding modulo while subtraction
// is very necessary otherwise it
// will cause wrong answer
ans = (ans - 1 + mod) % mod;
return ans;
}
// Driver Code
public static void main(String args[])
{
int n = 4;
// function call
System.out.println(check(n));
}
}
// This code is contributed by ankita_saini
#Python3 program to find Sum of all
# subsets of a set formed by
# first N natural numbers | Set-2
# modulo value
mod = (int)(1e9 + 7)
# Iterative Function to calculate (x^y)%p in O(log y)
def power(x,y,p):
res = 1 # Initialize result
x = x % p # Update x if it is more than or
# equal to p
while (y > 0):
# If y is odd, multiply x with the result
if (y & 1):
res = (res * x) % p
# y must be even now
y = y >> 1 # y = y/2
x = (x * x) % p
return res
# function to find ff(n)
def check(n):
# In formula n is starting from zero
n=n-1
# calculate answer using
# formula 2^n*(n^2 + n + 2) - 1
ans = n * n
# whenever answer is greater than
# or equals to mod then modulo it.
if (ans >= mod):
ans %= mod
ans += n + 2
if (ans >= mod):
ans %= mod
ans = (pow(2, n, mod) % mod * ans % mod) % mod
# adding modulo while subtraction is very necessary
# otherwise it will cause wrong answer
ans = (ans - 1 + mod) % mod
return ans
#Driver code
if __name__=='__main__':
n = 4
# function call
print(check(n))
# This code is contributed by ash264
// C# program to find Sum
// of all subsets of a set
// formed by first N natural
// numbers | Set-2
using System;
class GFG
{
// Iterative Function
// to calculate (x^y)%p
// in O(log y)
static int power(int x, int y,
int p)
{
// Initialize result
int res = 1;
// Update x if it is more
// than or equal to p
x = x % p;
while (y > 0)
{
// If y is odd, multiply
// x with the result
if (y != 0)
res = (res * x) % p;
// y must be even
// now y = y / 2
y = y >> 1;
x = (x * x) % p;
}
return res;
}
// function to find ff(n)
static int check(int n)
{
// modulo value
int mod = (int)(1e9 + 7);
// In formula n is
// starting from zero
n--;
// calculate answer
// using formula
// 2^n*(n^2 + n + 2) - 1
int ans = n * n;
// whenever answer is
// greater than or equals
// to mod then modulo it.
if (ans >= mod)
ans %= mod;
ans += n + 2;
if (ans >= mod)
ans %= mod;
ans = (power(2, n, mod) % mod *
ans % mod) % mod;
// adding modulo while
// subtraction is very
// necessary otherwise it
// will cause wrong answer
ans = (ans - 1 + mod) % mod;
return ans;
}
// Driver Code
public static void Main(String []args)
{
int n = 4;
// function call
Console.WriteLine(check(n));
}
}
// This code is contributed
// by ankita_saini
<?php
// PHP program to find Sum
// of all subsets of a set
// formed by first N natural
// numbers | Set-2
// modulo value
// Iterative Function to
// calculate (x^y)%p in O(log y)
function power($x, $y, $p)
{
$res = 1; // Initialize result
$x = $x % $p; // Update x if it
// is more than or
// equal to p
while ($y > 0)
{
// If y is odd, multiply
// x with the result
if ($y & 1)
$res = ($res * $x) % $p;
// y must be even now
$y = $y >> 1; // y = y/2
$x = ($x * $x) % $p;
}
return $res;
}
// function to find ff(n)
function check($n)
{
$mod = 1e9+7;
// In formula n is
// starting from zero
$n--;
// calculate answer using
// formula 2^n*(n^2 + n + 2) - 1
$ans = $n * $n;
// whenever answer is greater
// than or equals to mod then
// modulo it.
if ($ans >= $mod)
$ans %= $mod;
$ans += $n + 2;
if ($ans >= $mod)
$ans %= $mod;
$ans = (power(2, $n, $mod) %
$mod * $ans %
$mod) % $mod;
// adding modulo while subtraction
// is very necessary otherwise it
// will cause wrong answer
$ans = ($ans - 1 + $mod) % $mod;
return $ans;
}
// Driver code
$n = 4;
// function call
echo check($n) ."\n";
// This code is contributed
// by Akanksha Rai(Abby_akku)
?>
<script>
// javascript program to find Sum of all
// subsets of a set formed by
// first N natural numbers | Set-2
// modulo value
const mod = (1e9 + 7);
// Iterative Function to calculate (x^y)%p in O(log y)
function power( x, y, p)
{
let res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with the result
if (y & 1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// function to find ff(n)
function check( n)
{
// In formula n is starting from zero
n--;
// calculate answer using
// formula 2^n*(n^2 + n + 2) - 1
let ans = n * n;
// whenever answer is greater than
// or equals to mod then modulo it.
if (ans >= mod)
ans %= mod;
ans += n + 2;
if (ans >= mod)
ans %= mod;
ans = (power(2, n, mod) % mod * ans % mod) % mod;
// adding modulo while subtraction is very necessary
// otherwise it will cause wrong answer
ans = (ans - 1 + mod) % mod;
return ans;
}
// Driver code
let n = 4;
// function call
document.write(check(n));
// This code is contributed by aashish1995
</script>
Time complexity: O(log n).
Auxiliary Space: O(1)
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