Given an array arr[], the task is to find the sum of the maximum and the minimum prime factor of every number in the given array.
Examples:
Input: arr[] = {15}
Output: 8
The maximum and the minimum prime factors
of 15 are 5 and 3 respectively.
Input: arr[] = {5, 10, 15, 20, 25, 30}
Output: 10 7 8 7 10 7
Approach: The idea is to use Sieve of Eratosthenes to precompute all the minimum and maximum prime factors of every number and store it in two arrays. After this precomputation, the sum of the minimum and the maximum prime factor can be found in constant time.
Below is the implementation of the above approach:
CPP
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;const int MAX = 100000;// max_prime[i] represent maximum prime// number that divides the number iint max_prime[MAX];// min_prime[i] represent minimum prime// number that divides the number iint min_prime[MAX];// Function to store the minimum prime factor// and the maximum prime factor in two arraysvoid sieve(int n){ for (int i = 2; i <= n; ++i) { // Check for prime number // if min_prime[i]>0, // then it is not a prime number if (min_prime[i] > 0) { continue; } // if i is a prime number // min_prime number that divide prime number // and max_prime number that divide prime number // is the number itself. min_prime[i] = i; max_prime[i] = i; int j = i + i; while (j <= n) { if (min_prime[j] == 0) { // If this number is being visited // for first time then this divisor // must be the smallest prime number // that divides this number min_prime[j] = i; } // Update prime number till // last prime number that divides this number // The last prime number that // divides this number will be maximum. max_prime[j] = i; j += i; } }}// Function to find the sum of the minimum// and the maximum prime factors of every// number from the given arrayvoid findSum(int arr[], int n){ // Pre-calculation sieve(MAX); // For every element of the given array for (int i = 0; i < n; i++) { // The sum of its smallest // and largest prime factor int sum = min_prime[arr[i]] + max_prime[arr[i]]; cout << sum << " "; }}// Driver codeint main(){ int arr[] = { 5, 10, 15, 20, 25, 30 }; int n = sizeof(arr) / sizeof(int); findSum(arr, n); return 0;} |
Java
// Java implementation of the approach class GFG{ static int MAX = 100000; // max_prime[i] represent maximum prime // number that divides the number i static int max_prime[] = new int[MAX + 1]; // min_prime[i] represent minimum prime // number that divides the number i static int min_prime[] = new int[MAX + 1]; // Function to store the minimum prime factor // and the maximum prime factor in two arrays static void sieve(int n) { for (int i = 2; i <= n; ++i) { // Check for prime number // if min_prime[i] > 0, // then it is not a prime number if (min_prime[i] > 0) { continue; } // if i is a prime number // min_prime number that divide prime number // and max_prime number that divide prime number // is the number itself. min_prime[i] = i; max_prime[i] = i; int j = i + i; while (j <= n) { if (min_prime[j] == 0) { // If this number is being visited // for first time then this divisor // must be the smallest prime number // that divides this number min_prime[j] = i; } // Update prime number till // last prime number that divides this number // The last prime number that // divides this number will be maximum. max_prime[j] = i; j += i; } } } // Function to find the sum of the minimum // and the maximum prime factors of every // number from the given array static void findSum(int arr[], int n) { // Pre-calculation sieve(MAX); // For every element of the given array for (int i = 0; i < n; i++) { // The sum of its smallest // and largest prime factor int sum = min_prime[arr[i]] + max_prime[arr[i]]; System.out.print(sum + " "); } } // Driver code public static void main (String[] args) { int arr[] = { 5, 10, 15, 20, 25, 30 }; int n = arr.length ; findSum(arr, n); } }// This code is contributed by AnkitRai01 |
Python
# Python3 implementation of the approachMAX = 100000# max_prime[i] represent maximum prime# number that divides the number imax_prime = [0]*(MAX + 1)# min_prime[i] represent minimum prime# number that divides the number imin_prime = [0]*(MAX + 1)# Function to store the minimum prime factor# and the maximum prime factor in two arraysdef sieve(n): for i in range(2, n + 1): # Check for prime number # if min_prime[i]>0, # then it is not a prime number if (min_prime[i] > 0): continue # if i is a prime number # min_prime number that divide prime number # and max_prime number that divide prime number # is the number itself. min_prime[i] = i max_prime[i] = i j = i + i while (j <= n): if (min_prime[j] == 0): # If this number is being visited # for first time then this divisor # must be the smallest prime number # that divides this number min_prime[j] = i # Update prime number till # last prime number that divides this number # The last prime number that # divides this number will be maximum. max_prime[j] = i j += i# Function to find the sum of the minimum# and the maximum prime factors of every# number from the given arraydef findSum(arr, n): # Pre-calculation sieve(MAX) # For every element of the given array for i in range(n): # The sum of its smallest # and largest prime factor sum = min_prime[arr[i]] + max_prime[arr[i]] print(sum, end = " ")# Driver codearr = [5, 10, 15, 20, 25, 30]n = len(arr)findSum(arr, n)# This code is contributed by mohit kumar 29 |
C#
// C# implementation of the approach using System;class GFG{ static int MAX = 100000; // max_prime[i] represent maximum prime // number that divides the number i static int []max_prime = new int[MAX + 1]; // min_prime[i] represent minimum prime // number that divides the number i static int []min_prime = new int[MAX + 1]; // Function to store the minimum prime factor // and the maximum prime factor in two arrays static void sieve(int n) { for (int i = 2; i <= n; ++i) { // Check for prime number // if min_prime[i] > 0, // then it is not a prime number if (min_prime[i] > 0) { continue; } // if i is a prime number // min_prime number that divide prime number // and max_prime number that divide prime number // is the number itself. min_prime[i] = i; max_prime[i] = i; int j = i + i; while (j <= n) { if (min_prime[j] == 0) { // If this number is being visited // for first time then this divisor // must be the smallest prime number // that divides this number min_prime[j] = i; } // Update prime number till // last prime number that divides this number // The last prime number that // divides this number will be maximum. max_prime[j] = i; j += i; } } } // Function to find the sum of the minimum // and the maximum prime factors of every // number from the given array static void findSum(int []arr, int n) { // Pre-calculation sieve(MAX); // For every element of the given array for (int i = 0; i < n; i++) { // The sum of its smallest // and largest prime factor int sum = min_prime[arr[i]] + max_prime[arr[i]]; Console.Write(sum + " "); } } // Driver code public static void Main(String[] args) { int []arr = { 5, 10, 15, 20, 25, 30 }; int n = arr.Length ; findSum(arr, n); } }// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of the approach var MAX = 100000; // max_prime[i] represent maximum prime // number that divides the number i var max_prime = Array.from({length: MAX + 1}, (_, i) => 0);// min_prime[i] represent minimum prime // number that divides the number i var min_prime = Array.from({length: MAX + 1}, (_, i) => 0); // Function to store the minimum prime factor // and the maximum prime factor in two arrays function sieve(n) { for (var i = 2; i <= n; ++i) { // Check for prime number // if min_prime[i] > 0, // then it is not a prime number if (min_prime[i] > 0) { continue; } // if i is a prime number // min_prime number that divide prime number // and max_prime number that divide prime number // is the number itself. min_prime[i] = i; max_prime[i] = i; var j = i + i; while (j <= n) { if (min_prime[j] == 0) { // If this number is being visited // for first time then this divisor // must be the smallest prime number // that divides this number min_prime[j] = i; } // Update prime number till // last prime number that divides this number // The last prime number that // divides this number will be maximum. max_prime[j] = i; j += i; } } } // Function to find the sum of the minimum // and the maximum prime factors of every // number from the given array function findSum(arr , n) { // Pre-calculation sieve(MAX); // For every element of the given array for (i = 0; i < n; i++) { // The sum of its smallest // and largest prime factor var sum = min_prime[arr[i]] + max_prime[arr[i]]; document.write(sum + " "); } } // Driver code var arr = [ 5, 10, 15, 20, 25, 30 ]; var n = arr.length ;findSum(arr, n); // This code contributed by shikhasingrajput </script> |
Output:
10 7 8 7 10 7
Time Complexity: O(n2)
Auxiliary Space: O(100000)
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