Sum of element whose prime factors are present in array
Last Updated :
17 Aug, 2022
Given an array arr[] of non-negative integers where 2 ? arr[i] ? 106. The task is to find the sum of all those elements from the array whose prime factors are present in the same array. Examples:
Input: arr[] = {2, 3, 10} Output: 5 Factor of 2 is 2 which is present in the array Factor of 3 is 3, also present in the array Factors of 10 are 2 and 5, out of which only 2 is present in the array. So, sum = 2 + 3 = 5 Input: arr[] = {5, 11, 55, 25, 100} Output: 96
Approach: The idea is to first calculate the least prime factor till maximum element of the array with Prime Factorization using Sieve.
- Now, Iterate the array and for an element arr[i]
- If arr[i] != 1:
- If leastPrimeFactor(arr[i]) is present in the array then update arr[i] / leastPrimeFactor(arr[i]) and go to step 2.
- Else go to step 1.
- Else update sum = sum + originalVal(arr[i]).
- Print the sum in the end.
Below is the implementation of the above approach:
C++
// C++ program to find the sum of the elements of an array
// whose prime factors are present in the same array
#include <bits/stdc++.h>
using namespace std;
#define MAXN 1000001
// Stores smallest prime factor for every number
int spf[MAXN];
// Function to calculate SPF (Smallest Prime Factor)
// for every number till MAXN
void sieve()
{
spf[1] = 1;
for (int i = 2; i < MAXN; i++)
// Marking smallest prime factor for every
// number to be itself.
spf[i] = i;
// Separately marking spf for every even
// number as 2
for (int i = 4; i < MAXN; i += 2)
spf[i] = 2;
for (int i = 3; i * i < MAXN; i++) {
// If i is prime
if (spf[i] == i) {
// Marking SPF for all numbers divisible by i
for (int j = i * i; j < MAXN; j += i)
// Marking spf[j] if it is not
// previously marked
if (spf[j] == j)
spf[j] = i;
}
}
}
// Function to return the sum of the elements of an array
// whose prime factors are present in the same array
int sumFactors(int arr[], int n)
{
// Function call to calculate smallest prime factors of
// all the numbers upto MAXN
sieve();
// Create map for each element
std::map<int, int> map;
for (int i = 0; i < n; ++i)
map[arr[i]] = 1;
int sum = 0;
for (int i = 0; i < n; ++i) {
int num = arr[i];
// If smallest prime factor of num is present in array
while (num != 1 && map[spf[num]] == 1) {
num /= spf[num];
}
// Each factor of arr[i] is present in the array
if (num == 1)
sum += arr[i];
}
return sum;
}
// Driver program
int main()
{
int arr[] = { 5, 11, 55, 25, 100 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function call to print required answer
cout << sumFactors(arr, n);
return 0;
}
// Java program to find the sum of the elements of an array
// whose prime factors are present in the same array
import java.util.*;
public class GFG{
final static int MAXN = 1000001 ;
// Stores smallest prime factor for every number
static int spf[] = new int [MAXN];
// Function to calculate SPF (Smallest Prime Factor)
// for every number till MAXN
static void sieve()
{
spf[1] = 1;
for (int i = 2; i < MAXN; i++)
// Marking smallest prime factor for every
// number to be itself.
spf[i] = i;
// Separately marking spf for every even
// number as 2
for (int i = 4; i < MAXN; i += 2)
spf[i] = 2;
for (int i = 3; i * i < MAXN; i++) {
// If i is prime
if (spf[i] == i) {
// Marking SPF for all numbers divisible by i
for (int j = i * i; j < MAXN; j += i)
// Marking spf[j] if it is not
// previously marked
if (spf[j] == j)
spf[j] = i;
}
}
}
// Function to return the sum of the elements of an array
// whose prime factors are present in the same array
static int sumFactors(int arr[], int n)
{
// Function call to calculate smallest prime factors of
// all the numbers upto MAXN
sieve();
// Create map for each element
Map map=new HashMap();
for(int i = 0 ; i < MAXN ; ++i)
map.put(i,0) ;
for (int i = 0; i < n; ++i)
map.put(arr[i],1);
int sum = 0;
for (int i = 0; i < n; ++i) {
int num = arr[i];
// If smallest prime factor of num is present in array
while (num != 1 && (int)(map.get(spf[num])) == 1) {
num /= spf[num];
}
// Each factor of arr[i] is present in the array
if (num == 1)
sum += arr[i];
}
return sum;
}
// Driver program
public static void main(String []args)
{
int arr[] = { 5, 11, 55, 25, 100 };
int n = arr.length ;
// Function call to print required answer
System.out.println(sumFactors(arr, n)) ;
}
// This code is contributed by Ryuga
}
# Python program to find the sum of the
# elements of an array whose prime factors
# are present in the same array
from collections import defaultdict
MAXN = 1000001
MAXN_sqrt = int(MAXN ** (0.5))
# Stores smallest prime factor
# for every number
spf = [None] * (MAXN)
# Function to calculate SPF (Smallest
# Prime Factor) for every number till MAXN
def sieve():
spf[1] = 1
for i in range(2, MAXN):
# Marking smallest prime factor
# for every number to be itself.
spf[i] = i
# Separately marking spf for every
# even number as 2
for i in range(4, MAXN, 2):
spf[i] = 2
for i in range(3, MAXN_sqrt):
# If i is prime
if spf[i] == i:
# Marking SPF for all numbers
# divisible by i
for j in range(i * i, MAXN, i):
# Marking spf[j] if it is
# not previously marked
if spf[j] == j:
spf[j] = i
# Function to return the sum of the elements
# of an array whose prime factors are present
# in the same array
def sumFactors(arr, n):
# Function call to calculate smallest
# prime factors of all the numbers upto MAXN
sieve()
# Create map for each element
Map = defaultdict(lambda:0)
for i in range(0, n):
Map[arr[i]] = 1
Sum = 0
for i in range(0, n):
num = arr[i]
# If smallest prime factor of num
# is present in array
while num != 1 and Map[spf[num]] == 1:
num = num // spf[num]
# Each factor of arr[i] is present
# in the array
if num == 1:
Sum += arr[i]
return Sum
# Driver Code
if __name__ == "__main__":
arr = [5, 11, 55, 25, 100]
n = len(arr)
# Function call to print
# required answer
print(sumFactors(arr, n))
# This code is contributed by Rituraj Jain
// C# program to find the sum of the elements
// of an array whose prime factors are present
// in the same array
using System;
using System.Collections.Generic;
class GFG
{
static int MAXN = 1000001;
// Stores smallest prime factor for every number
static int []spf = new int [MAXN];
// Function to calculate SPF (Smallest Prime Factor)
// for every number till MAXN
static void sieve()
{
spf[1] = 1;
for (int i = 2; i < MAXN; i++)
// Marking smallest prime factor for
// every number to be itself.
spf[i] = i;
// Separately marking spf for every even
// number as 2
for (int i = 4; i < MAXN; i += 2)
spf[i] = 2;
for (int i = 3; i * i < MAXN; i++)
{
// If i is prime
if (spf[i] == i)
{
// Marking SPF for all numbers divisible by i
for (int j = i * i; j < MAXN; j += i)
// Marking spf[j] if it is not
// previously marked
if (spf[j] == j)
spf[j] = i;
}
}
}
// Function to return the sum of the elements
// of an array whose prime factors are present
// in the same array
static int sumFactors(int []arr, int n)
{
// Function call to calculate smallest
// prime factors of all the numbers upto MAXN
sieve();
// Create map for each element
Dictionary<int, int> map = new Dictionary<int, int>();
for(int i = 0 ; i < MAXN ; ++i)
map.Add(i, 0);
for (int i = 0; i < n; ++i)
{
if(map.ContainsKey(arr[i]))
{
map[arr[i]] = 1;
}
else
{
map.Add(arr[i], 1);
}
}
int sum = 0;
for (int i = 0; i < n; ++i)
{
int num = arr[i];
// If smallest prime factor of num
// is present in array
while (num != 1 &&
(int)(map[spf[num]]) == 1)
{
num /= spf[num];
}
// Each factor of arr[i] is present
// in the array
if (num == 1)
sum += arr[i];
}
return sum;
}
// Driver Code
public static void Main(String []args)
{
int []arr = { 5, 11, 55, 25, 100 };
int n = arr.Length;
// Function call to print required answer
Console.WriteLine(sumFactors(arr, n));
}
}
// This code is contributed by PrinciRaj1992
// JavaScript program to find the sum of the
// elements of an array whose prime factors
// are present in the same array
let MAXN = 1000001
let MAXN_sqrt = Math.floor(MAXN ** (0.5))
// Stores smallest prime factor
// for every number
let spf = new Array(MAXN)
// Function to calculate SPF (Smallest
// Prime Factor) for every number till MAXN
function sieve()
{
spf[1] = 1
for (var i = 2; i < MAXN; i++)
{
// Marking smallest prime factor
// for every number to be itself.
spf[i] = i
}
// Separately marking spf for every
// even number as 2
for (var i = 4; i < MAXN; i += 2)
spf[i] = 2
for (var i = 3; i < MAXN_sqrt; i++)
{
// If i is prime
if (spf[i] == i)
{
// Marking SPF for all numbers
// divisible by i
for (var j = i * i; j < MAXN; j += i)
{
// Marking spf[j] if it is
// not previously marked
if (spf[j] == j)
spf[j] = i
}
}
}
}
// Function to return the sum of the elements
// of an array whose prime factors are present
// in the same array
function sumFactors(arr, n)
{
// Function call to calculate smallest
// prime factors of all the numbers upto MAXN
sieve()
// Create map for each element
let Map = {}
for (var i = 0; i < n; i++)
Map[arr[i]] = 1
let Sum = 0
for (var i = 0; i < n; i++)
{
let num = arr[i]
// If smallest prime factor of num
// is present in array
while (num != 1 && Map.hasOwnProperty(spf[num]))
num = Math.floor(num / spf[num])
// Each factor of arr[i] is present
// in the array
if (num == 1)
Sum += arr[i]
}
return Sum
}
// Driver Code
let arr = [5, 11, 55, 25, 100]
let n = arr.length
// Function call to print
// required answer
console.log(sumFactors(arr, n))
// This code is contributed by phasing17
Time Complexity: O(MAXN*log(log(MAXN)))
Auxiliary Space: O(MAXN)
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