Given an integer N, the task is to find the sum of N and it’s maximum prime factor.
Examples:
Input: 19
Output: 38
Maximum prime factor of 19 is 19.
Hence, 19 + 19 = 38
Input: 8
Output: 10
8 + 2 = 10
Approach: Find the largest prime factor of the number and store it in maxPrimeFact then print the value of N + maxPrimeFact.
Below is the implementation of the above approach:
C++
// C++ program to find sum of n and // it's largest prime factor#include <cmath>#include <iostream>using namespace std;// Function to return the sum of n and // it's largest prime factorint maxPrimeFactors(int n){ int num = n; // Initialise maxPrime to -1. int maxPrime = -1; while (n % 2 == 0) { maxPrime = 2; n /= 2; } // n must be odd at this point, thus skip // the even numbers and iterate only odd numbers for (int i = 3; i <= sqrt(n); i += 2) { while (n % i == 0) { maxPrime = i; n = n / i; } } // This condition is to handle the case // when n is a prime number greater than 2 if (n > 2) maxPrime = n; // finally return the sum. int sum = maxPrime + num; return sum;}// Driver Program to check the above function.int main(){ int n = 19; cout << maxPrimeFactors(n); return 0;} |
Java
// Java program to find sum of n and// it's largest prime factorimport java.io.*;class GFG{// Function to return the sum of n// and it's largest prime factorstatic int maxPrimeFactors(int n){int num = n;// Initialise maxPrime to -1.int maxPrime = -1;while (n % 2 == 0){maxPrime = 2;n /= 2;}// n must be odd at this point,// thus skip the even numbers and// iterate only odd numbersfor (int i = 3; i <= Math.sqrt(n); i += 2) { while (n % i == 0) { maxPrime = i; n = n / i; } } // This condition is to handle the case // when n is a prime number greater than 2 if (n > 2) { maxPrime = n; }// finally return the sum.int sum = maxPrime + num;return sum;}// Driver Codepublic static void main (String[] args){int n = 19;System.out.println(maxPrimeFactors(n));}}// This code is contributed by anuj_67 |
Python3
# Python 3 program to find sum of n and # it's largest prime factorfrom math import sqrt# Function to return the sum of n and # it's largest prime factordef maxPrimeFactors(n): num = n # Initialise maxPrime to -1. maxPrime = -1; while (n % 2 == 0): maxPrime = 2 n = n / 2 # n must be odd at this point, thus skip # the even numbers and iterate only odd numbers p = int(sqrt(n) + 1) for i in range(3, p, 2): while (n % i == 0): maxPrime = i n = n / i # This condition is to handle the case # when n is a prime number greater than 2 if (n > 2): maxPrime = n # finally return the sum. sum = maxPrime + num return sum# Driver Codeif __name__ == '__main__': n = 19 print(maxPrimeFactors(n))# This code is contributed by # Surendra_Gangwar |
C#
// C# program to find sum of n and// it's largest prime factorusing System;class GFG{// Function to return the sum of n// and it's largest prime factorstatic int maxPrimeFactors(int n){int num = n;// Initialise maxPrime to -1.int maxPrime = -1;while (n % 2 == 0){ maxPrime = 2; n /= 2;}// n must be odd at this point,// thus skip the even numbers and// iterate only odd numbersfor (int i = 3; i <= Math.Sqrt(n); i += 2) { while (n % i == 0) { maxPrime = i; n = n / i; } } // This condition is to handle the case // when n is a prime number greater than 2 if (n > 2) { maxPrime = n;}// finally return the sum.int sum = maxPrime + num;return sum;}// Driver Codestatic void Main (){ int n = 19; Console.WriteLine(maxPrimeFactors(n));}}// This code is contributed by Ryuga |
PHP
<?php// PHP program to find sum of n and // it's largest prime factor// Function to return the sum of n // and it's largest prime factorfunction maxPrimeFactors($n){ $num = $n; // Initialise maxPrime to -1. $maxPrime = -1; while ($n % 2 == 0) { $maxPrime = 2; $n /= 2; } // n must be odd at this point, // thus skip the even numbers // and iterate only odd numbers for ($i = 3; $i <= sqrt($n); $i += 2) { while ($n % $i == 0) { $maxPrime = $i; $n = $n / $i; } } // This condition is to handle the case // when n is a prime number greater than 2 if ($n > 2) $maxPrime = $n; // finally return the sum. $sum = $maxPrime + $num; return $sum;}// Driver Code$n = 19;echo maxPrimeFactors($n);// This code is contributed // by inder_verma?> |
Javascript
<script>// Javascript program to find sum of n and// it's largest prime factor// Function to return the sum of n// and it's largest prime factorfunction maxPrimeFactors(n){ var num = n; // Initialise maxPrime to -1. var maxPrime = -1; while (n % 2 == 0) { maxPrime = 2; n = parseInt(n/2); } // n must be odd at this point, // thus skip the even numbers and // iterate only odd numbers for (var i = 3; i <= parseInt(Math.sqrt(n)); i += 2) { while (n % i == 0) { maxPrime = i; n = parseInt(n / i); } } // This condition is to handle the case // when n is a prime number greater than 2 if (n > 2) { maxPrime = n; } // finally return the sum. var sum = maxPrime + num; return sum;}// Driver Codevar n = 19;document.write(maxPrimeFactors(n));// This code contributed by shikhasingrajput </script> |
Output:
38
Time Complexity: O(sqrtn*logn)
Auxiliary Space: O(1)
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