Given N, we have to find the sum of products of all combination taken 1 to N at a time. In simple words, we have to find the sum of products of all combination taken 1 at time, then 2 at a time, then 3 at a time till N at a time.
If you think closely about the problem, a large value of N could result in producing a large number of combinations.
Examples:
Input : N = 3
Output : f(1) = 6
f(2) = 11
f(3) = 6
Explanation: f(x) is sum of products of all
combination taken x at a time
1 + 2 + 3 = 6
f(2) = (1*2) + (1*3) + (2*3) = 11
f(3) = (1*2*3)
Input : N = 4
Output : f(1) = 10
f(2) = 35
f(3) = 50
f(4) = 24
Explanation: f(1) = 1 + 2 + 3 + 4 = 10
f(2) = (1*2) + (1*3) + (1*4) +
(2*3) + (2*4) + (3*4)
= 35
f(3) = (1*2*3) + (1*2*4) +(1*3*4) +
(2*3*4)
= 50
f(4) = (1*2*3*4) = 24
A Brute force approach would be to produce all the combinations and then find their products and sum.
Recursion would do the trick to produce the combinations taken x at a time.
Example : N = 4 taken 3 at a time
C++
// Program to find SOP of all combination taken // (1 to N) at a time using brute force #include <iostream> using namespace std; // to store sum of every combination int sum = 0; void Combination(int a[], int combi[], int n, int r, int depth, int index) { // if we have reached sufficient depth if (index == r) { // find the product of combination int product = 1; for (int i = 0; i < r; i++) product = product * combi[i]; // add the product into sum sum += product; return; } // recursion to produce different combination for (int i = depth; i < n; i++) { combi[index] = a[i]; Combination(a, combi, n, r, i + 1, index + 1); } } // function to print sum of products of // all combination taken 1-N at a time void allCombination(int a[], int n) { for (int i = 1; i <= n; i++) { // creating temporary array for storing // combination int *combi = new int[i]; // call combination with r = i // for combination taken i at a time Combination(a, combi, n, i, 0, 0); // displaying sum cout << "f(" << i << ") --> " << sum << "
"; sum = 0; // free from heap area free(combi); } } // Driver's code int main() { int n = 5; int *a = new int[n]; // storing numbers from 1-N in array for (int i = 0; i < n; i++) a[i] = i + 1; // calling allCombination allCombination(a, n); return 0; } |
Java
// Program to find SOP of // all combination taken // (1 to N) at a time using // brute force import java.io.*; class GFG { // to store sum of // every combination static int sum = 0; static void Combination(int []a, int []combi, int n, int r, int depth, int index) { // if we have reached // sufficient depth if (index == r) { // find the product // of combination int product = 1; for (int i = 0; i < r; i++) product = product * combi[i]; // add the product into sum sum += product; return; } // recursion to produce // different combination for (int i = depth; i < n; i++) { combi[index] = a[i]; Combination(a, combi, n, r, i + 1, index + 1); } } // function to print sum of // products of all combination // taken 1-N at a time static void allCombination(int []a, int n) { for (int i = 1; i <= n; i++) { // creating temporary array // for storing combination int []combi = new int[i]; // call combination with // r = i for combination // taken i at a time Combination(a, combi, n, i, 0, 0); // displaying sum System.out.print("f(" + i + ") --> " + sum + "
"); sum = 0; } } // Driver code public static void main(String args[]) { int n = 5; int []a = new int[n]; // storing numbers // from 1-N in array for (int i = 0; i < n; i++) a[i] = i + 1; // calling allCombination allCombination(a, n); } } // This code is contributed by // Manish Shaw(manishshaw1) |
Python3
# Python3 Program to find SOP of all combination
# taken (1 to N) at a time using brute force
# to store sum of every combination
def Combination(a, combi, n, r, depth, index):
global Sum
# if we have reached sufficient depth
if index == r:
# find the product of combination
product = 1
for i in range(r):
product = product * combi[i]
# add the product into sum
Sum += product
return
# recursion to produce different
# combination
for i in range(depth, n):
combi[index] = a[i]
Combination(a, combi, n, r,
i + 1, index + 1)
# function to print sum of products of
# all combination taken 1-N at a time
def allCombination(a, n):
global Sum
for i in range(1, n + 1):
# creating temporary array for
# storing combination
combi = [0] * i
# call combination with r = i
# for combination taken i at a time
Combination(a, combi, n, i, 0, 0)
# displaying sum
print(“f(“, i, “) –> “, Sum)
Sum = 0
# Driver Code
Sum = 0
n = 5
a = [0] * n
# storing numbers from 1-N in array
for i in range(n):
a[i] = i + 1
# calling allCombination
allCombination(a, n)
# This code is contributed by PranchalK
C#
// Program to find SOP of // all combination taken // (1 to N) at a time using // brute force using System; class GFG { // to store sum of // every combination static int sum = 0; static void Combination(int []a, int []combi, int n, int r, int depth, int index) { // if we have reached // sufficient depth if (index == r) { // find the product // of combination int product = 1; for (int i = 0; i < r; i++) product = product * combi[i]; // add the product into sum sum += product; return; } // recursion to produce // different combination for (int i = depth; i < n; i++) { combi[index] = a[i]; Combination(a, combi, n, r, i + 1, index + 1); } } // function to print sum of // products of all combination // taken 1-N at a time static void allCombination(int []a, int n) { for (int i = 1; i <= n; i++) { // creating temporary array // for storing combination int []combi = new int[i]; // call combination with // r = i for combination // taken i at a time Combination(a, combi, n, i, 0, 0); // displaying sum Console.Write("f(" + i + ") --> " + sum + "
"); sum = 0; } } // Driver code static void Main() { int n = 5; int []a = new int[n]; // storing numbers // from 1-N in array for (int i = 0; i < n; i++) a[i] = i + 1; // calling allCombination allCombination(a, n); } } // This code is contributed by // Manish Shaw(manishshaw1) |
Output:
f(1) --> 15 f(2) --> 85 f(3) --> 225 f(4) --> 274 f(5) --> 120
The Time complexity of above code is exponential when the value of N is large.
An Efficient Method is to use the concept of dynamic programming. We don’t have to find sum of products every time. We can make use of previous results.
Let’s take an example: N = 4
C++
// CPP Program to find sum of all combination takne // (1 to N) at a time using dynamic programming #include <iostream> using namespace std; // find the postfix sum array void postfix(int a[], int n) { for (int i = n - 1; i > 0; i--) a[i - 1] = a[i - 1] + a[i]; } // modify the array such that we don't have to // compute the products which are obtained before void modify(int a[], int n) { for (int i = 1; i < n; i++) a[i - 1] = i * a[i]; } // finding sum of all combination taken 1 to N at a time void allCombination(int a[], int n) { int sum = 0; // sum taken 1 at time is simply sum of 1 - N for (int i = 1; i <= n; i++) sum += i; cout << "f(1) --> " << sum << "
"; // for sum of products for all combination for (int i = 1; i < n; i++) { // finding postfix array postfix(a, n - i + 1); // sum of products taken i+1 at a time sum = 0; for (int j = 1; j <= n - i; j++) { sum += (j * a[j]); } cout << "f(" << i + 1 << ") --> " << sum << "
"; // modify the array for overlapping problem modify(a, n); } } // Driver's Code int main() { int n = 5; int *a = new int[n]; // storing numbers from 1 to N for (int i = 0; i < n; i++) a[i] = i + 1; // calling allCombination allCombination(a, n); return 0; } |
Output:
f(1) --> 15 f(2) --> 85 f(3) --> 225 f(4) --> 274 f(5) --> 120
The Time Complexity of above method is O(n^2) which far more better than the brute force method.
You can also find the execution time of both the method for large value of N and can see the difference for yourself.
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