Given N, we have to find the sum of products of all combinations taken 1 to N at a time. In simple words, we have to find the sum of products of all combinations taken 1 at a 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 many 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 combinationint 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 timevoid 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 << "\n"; sum = 0; // free from heap area free(combi); }}// Driver's codeint 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 forceimport 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 + "\n"); 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 = 0n = 5a = [0] * n # storing numbers from 1-N in arrayfor 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 forceusing 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 + "\n"); 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) |
Javascript
<script>// JavaScript program to find sum of all combination taken// (1 to N) at a time using brute force // to store sum of // every combination let sum = 0; function Combination(a, combi, n, r, depth, index) { // if we have reached // sufficient depth if (index == r) { // find the product // of combination let product = 1; for (let i = 0; i < r; i++) product = product * combi[i]; // add the product into sum sum += product; return; } // recursion to produce // different combination for (let 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 function allCombination(a, n) { for (let i = 1; i <= n; i++) { // creating temporary array // for storing combination let combi = []; // call combination with // r = i for combination // taken i at a time Combination(a, combi, n, i, 0, 0); // displaying sum document.write("f(" + i + ") --> " + sum + "<br/>"); sum = 0; } } // Driver code let n = 5; let a = []; // storing numbers // from 1-N in array for (let i = 0; i < n; i++) a[i] = i + 1; // calling allCombination allCombination(a, n);</script> |
Output
f(1) --> 15 f(2) --> 85 f(3) --> 225 f(4) --> 274 f(5) --> 120
Time complexity of the 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 the 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 taken// (1 to N) at a time using dynamic programming#include <iostream>using namespace std;// find the postfix sum arrayvoid 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 beforevoid 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 timevoid 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 << "\n"; // 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 << "\n"; // modify the array for overlapping problem modify(a, n); }}// Driver's Codeint 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;} |
Java
// Java Program to find sum of all combination taken// (1 to N) at a time using dynamic programmingimport java.util.*;class GFG { // find the postfix sum array static 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 static 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 static 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; } System.out.println("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]); } System.out.println("f(" + (i + 1) + ") --> " + sum); // modify the array for overlapping problem modify(a, n); } } // Driver's Code public static void main(String[] args) { 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); }}// This code is contributed by 29AjayKumar |
Python3
# Python3 Program to find # sum of all combination taken# (1 to N) at a time using # dynamic programming# Find the postfix sum arraydef postfix(a, n): for i in range (n - 1, 1, -1): a[i - 1] = a[i - 1] + a[i]# Modify the array such # that we don't have to# compute the products # which are obtained beforedef modify(a, n): for i in range (1, n): a[i - 1] = i * a[i];# Finding sum of all combination # taken 1 to N at a timedef allCombination(a, n): sum = 0 # sum taken 1 at time is # simply sum of 1 - N for i in range (1, n + 1): sum += i print ("f(1) --> ", sum ) # for sum of products for # all combination for i in range (1, n): # finding postfix array postfix(a, n - i + 1) # sum of products taken # i+1 at a time sum = 0 for j in range(1, n - i + 1): sum += (j * a[j]) print ("f(", i + 1, ") --> ", sum) # modify the array for # overlapping problem modify(a, n)# Driver's Codeif __name__ == "__main__": n = 5 a = [0] * n # storing numbers # from 1 to N for i in range(n): a[i] = i + 1 # calling allCombination allCombination(a, n)# This code is contributed by Chitranayal |
C#
// C# Program to find sum of all combination taken// (1 to N) at a time using dynamic programmingusing System; class GFG { // find the postfix sum array static 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 static 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 static 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; } Console.WriteLine("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]); } Console.WriteLine("f(" + (i + 1) + ") --> " + sum); // modify the array for overlapping problem modify(a, n); } } // Driver's Code public static void Main(String[] args) { 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); }}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript Program to find sum of all combination taken// (1 to N) at a time using dynamic programming // find the postfix sum array function postfix(a,n) { for (let 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 function modify(a,n) { for (let i = 1; i < n; i++) { a[i - 1] = i * a[i]; } } // finding sum of all combination // taken 1 to N at a time function allCombination(a,n) { let sum = 0; // sum taken 1 at time is simply sum of 1 - N for (let i = 1; i <= n; i++) { sum += i; } document.write("f(1) --> " + sum+"<br>"); // for sum of products for all combination for (let 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 (let j = 1; j <= n - i; j++) { sum += (j * a[j]); } document.write("f(" + (i + 1) + ") --> " + sum+"<br>"); // modify the array for overlapping problem modify(a, n); } } // Driver's Code let n = 5; let a = new Array(n); // storing numbers from 1 to N for (let i = 0; i < n; i++) { a[i] = i + 1; } // calling allCombination allCombination(a, n); // This code is contributed by avanitrachhadiya2155</script> |
Output
f(1) --> 15 f(2) --> 85 f(3) --> 225 f(4) --> 274 f(5) --> 120
Time Complexity: O(n^2)
Auxiliary Space: O(1)
You can also find the execution time of both the method for a large value of N and can see the difference for yourself.
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