Given an array arr[] consisting of N positive integers, the task is to find the sum of factors of product of all array elements. Since the output can be very large, print it modulo 109 + 7.
Examples:
Input: arr[] = { 1, 2, 3, 4, 5 }
Output: 360
Explanation:
The product of all array elements = 1 * 2 * 3 * 4 * 5 = 120
All the factors of 120 are { 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 }
Therefore, the sum of factors is 360.Input: arr[] = { 1, 2 }
Output: 3
Explanation:
The product of all array elements = 1 * 2 = 2
All the factors of 2 are { 1, 2 }
Therefore, the sum of factors is 3.
Naive Approach: The simplest approach to solve this problem is to traverse the array and calculate the product of all elements of the array and calculate the sum of all the factors of the obtained product. But the problem with this approach is that, if the array elements are large, then the product may go out of bounds of the integer storing capacity and it will lead to wrong output.
Time Complexity: O(max(N, sqrt(product of array elements)))
Auxiliary Space: O(N)
Efficient approach: The above approach can be optimized based on the following observations:
If the product of array elements(P) =
Then, the sum of factors of P =
Follow the steps below to solve the problem:
- Initialize an integer, say ans, to store the sum of all the factors of the product of the array.
- Initialize an array of integer, say count[], where count[i] stores the frequency of prime factors i, in product of the array elements.
- Traverse the array count[], and check if count[i] is greater than zero or not. If found to be true, then multiply ans by (i(count[i] + 1)) – 1 and multiplicative inverse of (i -1)
- Finally, print the result obtained in ans
Below is the implementation of the above approach:
C
// C program to implement// the above approach#include <stdio.h>#define size 1000100#define inverse(a) power(a, mod - 2)typedef long long int ll;const ll mod = ((ll)(1e9 + 7));// Stores minimum prime// factorization of a numberint spf[size] = { 0 };// Function to add two numbersstatic inline ll add(ll a, ll b){ return (a % mod + b % mod) % mod;}// Function to subtract two numbersstatic inline ll sub(ll a, ll b){ return add(mod, a - b) % mod;}// Function to multiply two numbersstatic inline ll mul(ll a, ll b){ return (a % mod * b % mod) % mod;}// Function to calculate// x to the power yll power(ll x, ll y){ // Stores x ^ y ll res = 1; for (res = 1; y > 0; x = (x * x) % mod, y >>= 1) { // If y is odd if (y & 1) { // Update result res = (res * x) % mod; } } return res;}// Function to find the smallest prime factor// of numbers in the range [1, 1000100]void sieve(){ // Update the smallest prime factor of // all the numbers which is divisible by 2 for (int i = 2; i < size; i += 2) { // Update spf[i] spf[i] = 2; } for (int i = 3; i < size; i += 2) spf[i] = i; // Calculate the smallest prime factor // of all the numbers in the range [3, 1000100] for (int i = 3; i * i < size; i += 2) if (spf[i] == i) for (int j = i * i; j < size; j += i) spf[j] = i;}// Function to find the sum of factors of// product of all array elementslong long int sumof_factors(int a[], int n){ // Stores the sum of factors of // product of all array elements ll ans = 1; // count[i]: Stores frequency of // prime factor i in product of // all the array elements ll count[size] = { 0 }; // Traverse the array for (int i = 0; i < n; i++) { // Calculate the prime factor // of a[i] while (a[i] > 1) { // Update frequency of // prime factor spf[a[i]] count[spf[a[i]]]++; // Update a[i] a[i] /= spf[a[i]]; } } // Traverse the array, count[] for (ll i = 0; i < size; i++) // If frequency of prime factor i in // product of array elements // greater than 0 if (count[i] > 0) { // Calculate (i^(count[i]+1))-1 and // multiplicative inverse of (i -1) ll num1 = sub(power(i, count[i] + 1), 1); ll num2 = inverse(i - 1); ans = mul(ans, mul(num1, num2)); } return ans;}// Driver Codeint main(){ sieve(); int arr[] = { 1, 3, 2, 5, 4 }; int N = sizeof(arr) / sizeof(arr[0]); ll res = sumof_factors(arr, N); printf("%lld\n", res); return 0;} |
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;#define size 1000100#define inverse(a) power(a, mod - 2)typedef long long int ll;const ll mod = ((ll)(1e9 + 7));// Stores minimum prime// factorization of a numberint spf[size] = { 0 };// Function to add two numbersstatic inline ll add(ll a, ll b){ return (a % mod + b % mod) % mod;}// Function to subtract two numbersstatic inline ll sub(ll a, ll b){ return add(mod, a - b) % mod;}// Function to multiply two numbersstatic inline ll mul(ll a, ll b){ return (a % mod * b % mod) % mod;}// Function to calculate// x to the power yll power(ll x, ll y){ // Stores x ^ y ll res = 1; for (res = 1; y > 0; x = (x * x) % mod, y >>= 1) { // If y is odd if (y & 1) { // Update result res = (res * x) % mod; } } return res;}// Function to find the smallest prime factor// of numbers in the range [1, 1000100]void sieve(){ // Update the smallest prime factor of // all the numbers which is divisible by 2 for (int i = 2; i < size; i += 2) { // Update spf[i] spf[i] = 2; } for (int i = 3; i < size; i += 2) spf[i] = i; // Calculate the smallest prime factor // of all the numbers in the range [3, 1000100] for (int i = 3; i * i < size; i += 2) if (spf[i] == i) for (int j = i * i; j < size; j += i) spf[j] = i;}// Function to calculate sum of factors// of product of the given arraylong long int sumof_factors(int a[], int n){ // Stores the sum of factors of // product of all array elements ll ans = 1; // count[i]: Stores frequency of // prime factor i in product of // all the array elements ll count[size] = { 0 }; // Traverse the array for (int i = 0; i < n; i++) { // Calculate the prime factor // of a[i] while (a[i] > 1) { // Update frequency of // prime factor spf[a[i]] count[spf[a[i]]]++; // Update a[i] a[i] /= spf[a[i]]; } } // Traverse the array, count[] for (ll i = 0; i < size; i++) // If frequency of prime factor i in // product of array elements // greater than 0 if (count[i] > 0) { // Calculate (i^(count[i]+1))-1 and // multiplicative inverse of (i -1) ll num1 = sub(power(i, count[i] + 1), 1); ll num2 = inverse(i - 1); ans = mul(ans, mul(num1, num2)); } return ans;}// Driver Codeint main(){ sieve(); int arr[] = { 1, 3, 2, 5, 4 }; int N = sizeof(arr) / sizeof(arr[0]); ll res = sumof_factors(arr, N); cout << res; return 0;} |
Java
// Java program to implement// the above approachimport java.util.HashMap;import java.util.Map;class GFG { static final long mod = (int)(1e9 + 7); static final int size = (int)(1e6 + 100); // Function to subtract two numbers static final long sub(long a, long b) { return (mod + a % mod - b % mod) % mod; } // Function to multiply two numbers static final long mul(long a, long b) { return (a % mod * b % mod) % mod; } // Function to calculate // x to the power y static long power(long x, long y) { // Stores x ^ y long res = 1; for (res = 1; y > 0; x = (x * x) % mod, y >>= 1) { // If y is odd if ((y & 1) == 1) { // Update result res = (res * x) % mod; } } return res; } // Function to find inverse // of a mod 1e9 + 7 static long inverse(long a) { return power(a, mod - 2); } // Stores minimum prime // factorization of a number static int spf[] = new int[size]; // Function to find the smallest prime factor // of numbers in the range [1, 1000100] static void sieve() { for (int i = 1; i < size; i += 2) spf[i] = i; for (int i = 2; i < size; i += 2) spf[i] = 2; for (int i = 3; i * i < size; i += 2) if (spf[i] == i) for (int j = i * i; j < size; j += i) spf[j] = i; } // Function to calculate sum of factors // of product of the given array static long sumof_factors(int a[], int n) { // Traverse the array for (int i = 0; i < n; i++) if (a[i] == 0) return 0; // Stores the sum of factors of // product of all array elements long ans = 1; // count[i]: Stores frequency of // prime factor i in product of // all the array elements Map<Integer, Integer> count = new HashMap<Integer, Integer>(); // Traverse the array for (int num : a) { // Calculate the prime factor // of a[i] while (num > 1) { int temp = 0; try { temp = count.get(spf[num]); } catch (Exception e) { temp = 0; } // Update frequency of // prime factor spf[a[i]] count.put(spf[num], temp + 1); // Update num num /= spf[num]; } } for (Map.Entry<Integer, Integer> i : count.entrySet()) { // Calculate (i^(count[i]+1))-1 and // multiplicative inverse of (i -1) long num1 = sub( power(i.getKey(), i.getValue() + 1), 1); long num2 = inverse(i.getKey() - 1); ans = mul(ans, mul(num1, num2)); } return ans; } // Driver Code public static void main(String[] args) { sieve(); int n = 5; int a[] = { 1, 3, 2, 5, 4 }; System.out.println(sumof_factors(a, n)); }} |
Python3
# Python program to implement# the above approachfrom collections import defaultdictfrom math import sqrt# Function to find the smallest prime factor# of numbers in the range [1, 1000100]def computeSPF(size): # Stores smallest prime # factorization of a number spf = [i for i in range(size)] # Update the smallest prime factor of # all the numbers which is divisible by 2 for i in range(2, size, 2): spf[i] = 2 # Calculate the smallest prime factor # of all the numbers in the range [3, 1000100] for i in range(3, int(sqrt(size))+1, 2): if spf[i] == i: for j in range(i * i, size, i): spf[j] = i return spf# Function to calculate sum of factors# of product of the given arraydef sumof_factors(a, n, spf, mod): # Traverse the array if 0 in a: return 0 count = defaultdict(int) # Stores the sum of factors of # product of all array elements ans = 1 # Traverse the array for num in a: # Calculate the prime factor # of a[i] while num > 1: # Update frequency of # prime factor spf[a[i]] count[spf[num]] += 1 num //= spf[num] # Traverse the array, count[] for i in count: num1 = pow(i, count[i]+1, mod) - 1 num2 = pow(i-1, mod-2, mod) ans = (ans * num1 * num2) % mod return ans# Driver Codedef main(): spf = computeSPF(10**6) mod = 10**9 + 7 n = 4 a = [1, 3, 2, 5] ans = sumof_factors(a, n, spf, mod) print(ans)main() |
C#
// C# program to implement// the above approachusing System;using System.Collections.Generic;class GFG { static long mod = (int)(1e9 + 7); static int size = (int)(1e6 + 100); // Function to subtract two numbers static long sub(long a, long b) { return (mod + a % mod - b % mod) % mod; } // Function to multiply two numbers static long mul(long a, long b) { return (a % mod * b % mod) % mod; } // Function to calculate // x to the power y static long power(long x, long y) { // Stores x ^ y long res = 1; for (res = 1; y > 0; x = (x * x) % mod, y >>= 1) { // If y is odd if ((y & 1) == 1) { // Update result res = (res * x) % mod; } } return res; } // Function to find inverse // of a mod 1e9 + 7 static long inverse(long a) { return power(a, mod - 2); } // Stores minimum prime // factorization of a number static int[] spf = new int[size]; // Function to find the smallest prime factor // of numbers in the range [1, 1000100] static void sieve() { for (int i = 1; i < size; i += 2) spf[i] = i; for (int i = 2; i < size; i += 2) spf[i] = 2; for (int i = 3; i * i < size; i += 2) if (spf[i] == i) for (int j = i * i; j < size; j += i) spf[j] = i; } // Function to calculate sum of factors // of product of the given array static long sumof_factors(int[] a, int n) { // Traverse the array for (int i = 0; i < n; i++) if (a[i] == 0) return 0; // Stores the sum of factors of // product of all array elements long ans = 1; // count[i]: Stores frequency of // prime factor i in product of // all the array elements Dictionary<int, int> count = new Dictionary<int, int>(); // Traverse the array for (int i = 0; i < a.Length; i++) { // Calculate the prime factor // of a[i] while (a[i] > 1) { int temp = 0; if (count.ContainsKey(spf[a[i]])) temp = count[spf[a[i]]]; // Update frequency of // prime factor spf[a[i]] count[spf[a[i]]] = temp + 1; // Update num a[i] /= spf[a[i]]; } } foreach(KeyValuePair<int, int> i in count) { // Calculate (i^(count[i]+1))-1 and // multiplicative inverse of (i -1) long num1 = sub(power(i.Key, i.Value + 1), 1); long num2 = inverse(i.Key - 1); ans = mul(ans, mul(num1, num2)); } return ans; } // Driver Code public static void Main(string[] args) { sieve(); int n = 5; int[] a = { 1, 3, 2, 5, 4 }; Console.WriteLine(sumof_factors(a, n)); }}// This code is contributed by ukasp. |
Javascript
// Function to calculate mod inversefunction expmod( base, exp, mod ){ if (exp == 0) return 1; if (exp % 2 == 0){ return Math.pow( expmod( base, (exp / 2), mod), 2) % mod; } else { return (base * expmod( base, (exp - 1), mod)) % mod; }}// Function to find the smallest prime factor// of numbers in the range [1, 1000100]function computeSPF(size) { // Stores smallest prime // factorization of a number const spf = Array.from(Array(size).keys()); // Update the smallest prime factor of // all the numbers which is divisible by 2 for (let i = 2; i < size; i += 2) { spf[i] = 2; } // Calculate the smallest prime factor // of all the numbers in the range [1, sqrt(size)] for (let i = 3; i <= Math.sqrt(size); i += 2) { if (spf[i] == i) { for (let j = i * i; j < size; j += i) { spf[j] = i; } } } return spf;}// Function to calculate sum of factors// of product of the given arrayfunction sumofFactors(a, n, spf, mod) { // Traverse the array if (a.includes(0)) { return 0; } let count = {}; // Stores the sum of factors of // product of all array elements let ans = 1; // Traverse the array for (let num of a) { // Calculate the prime factorization of num let primeFactors = {}; while (num > 1) { // Update frequency of prime factor spf[num // Update frequency of // prime factor spf[a[i]] if (!count.hasOwnProperty(spf[num])) { count[spf[num]] = 1; } else { count[spf[num]]++; } num = Math.floor(num / spf[num]); } } // Traverse the array, count[] for (let i of Object.keys(count)) { i = parseInt(i) let num1 = expmod(i, count[i] + 1, mod) - 1; let num2 = expmod(i - 1, mod - 2, mod); ans = (ans * num1 * num2) % mod; } return ans;}// Driver Codefunction main() { const spf = computeSPF(10 ** 6); const mod = 10 ** 9 + 7; const n = 4; const a = [1, 3, 2, 5]; const ans = sumofFactors(a, n, spf, mod); console.log(ans);}main(); |
Output:
360
Time Complexity: O(N * log(log(N)))
Auxiliary Space: O(N)
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