Given an array arr[], the task is to check whether the sum of primes in the array is divisible by any of the primes in the array. If it is then print YES, otherwise print NO.
Examples:
Input: arr[] = {2, 3}
Output: NO
Primes: 2, 3
Sum = 2 + 3 = 5 which is neither divisible by 2 nor 3Input: arr[] = {1, 2, 3, 4, 5}
Output: YES
2 + 3 + 5 = 10 is divisible by 2 as well as 5
Approach: The idea is to generate all primes upto maximum element from the array using Sieve of Eratosthenes.
- Traverse the array and check if the current element is prime or not. If it is prime then update sum = sum + arr[i].
- Traverse the array again and check if sum % arr[i] = 0 where arr[i] is prime. If it is then print YES. Otherwise print NO in the end.
Below is the implementation of the above approach:
C++
// C++ program to check if sum of primes from an array // is divisible by any of the primes from the same array #include <bits/stdc++.h> using namespace std; // Function to print "YES" if sum of primes from an array // is divisible by any of the primes from the same array void SumDivPrime( int A[], int n) { int max_val = *(std::max_element(A, A + n)) + 1; // USE SIEVE TO FIND ALL PRIME NUMBERS LESS // THAN OR EQUAL TO max_val // Create a boolean array "prime[0..n]". A // value in prime[i] will finally be false // if i is Not a prime, else true. vector< bool > prime(max_val + 1, true ); // Remaining part of SIEVE prime[0] = false ; prime[1] = false ; for ( int p = 2; p * p <= max_val; p++) { // If prime[p] is not changed, then // it is a prime if (prime[p] == true ) { // Update all multiples of p for ( int i = p * 2; i <= max_val; i += p) prime[i] = false ; } } int sum = 0; // Traverse through the array for ( int i = 0; i < n; ++i) { if (prime[A[i]]) sum += A[i]; } for ( int i = 0; i < n; ++i) { if (prime[A[i]] && sum % A[i] == 0) { cout << "YES" ; return ; } } cout << "NO" ; } // Driver program int main() { int A[] = { 1, 2, 3, 4, 5 }; int n = sizeof (A) / sizeof (A[0]); SumDivPrime(A, n); return 0; } |
Java
// Java program to check if sum of primes from an array // is divisible by any of the primes from the same array class Solution { //returns the maximum value static int max_element( int A[]) { int max=Integer.MIN_VALUE; for ( int i= 0 ;i<A.length;i++) if (max<A[i]) max=A[i]; return max; } // Function to print "YES" if sum of primes from an array // is divisible by any of the primes from the same array static void SumDivPrime( int A[], int n) { int max_val = (max_element(A)) + 1 ; // USE SIEVE TO FIND ALL PRIME NUMBERS LESS // THAN OR EQUAL TO max_val // Create a boolean array "prime[0..n]". A // value in prime[i] will finally be false // if i is Not a prime, else true. boolean prime[]= new boolean [max_val+ 1 ]; //initialize the array for ( int i= 0 ;i<=max_val;i++) prime[i]= true ; // Remaining part of SIEVE prime[ 0 ] = false ; prime[ 1 ] = false ; for ( int p = 2 ; p * p <= max_val; p++) { // If prime[p] is not changed, then // it is a prime if (prime[p] == true ) { // Update all multiples of p for ( int i = p * 2 ; i <= max_val; i += p) prime[i] = false ; } } int sum = 0 ; // Traverse through the array for ( int i = 0 ; i < n; ++i) { if (prime[A[i]]) sum += A[i]; } for ( int i = 0 ; i < n; ++i) { if (prime[A[i]] && sum % A[i] == 0 ) { System.out.println( "YES" ); return ; } } System.out.println( "NO" ); } // Driver program public static void main(String args[]) { int A[] = { 1 , 2 , 3 , 4 , 5 }; int n = A.length; SumDivPrime(A, n); } } //contributed by Arnab Kundu |
Python3
# Python3 program to check if sum of # primes from an array is divisible # by any of the primes from the same array import math # Function to print "YES" if sum of primes # from an array is divisible by any of the # primes from the same array def SumDivPrime(A, n): max_val = max (A) + 1 # USE SIEVE TO FIND ALL PRIME NUMBERS # LESS THAN OR EQUAL TO max_val # Create a boolean array "prime[0..n]". # A value in prime[i] will finally be # false if i is Not a prime, else true. prime = [ True ] * (max_val + 1 ) # Remaining part of SIEVE prime[ 0 ] = False prime[ 1 ] = False for p in range ( 2 , int (math.sqrt(max_val)) + 1 ): # If prime[p] is not changed, # then it is a prime if prime[p] = = True : # Update all multiples of p for i in range ( 2 * p, max_val + 1 , p): prime[i] = False sum = 0 # Traverse through the array for i in range ( 0 , n): if prime[A[i]]: sum + = A[i] for i in range ( 0 , n): if prime[A[i]] and sum % A[i] = = 0 : print ( "YES" ) return print ( "NO" ) # Driver Code A = [ 1 , 2 , 3 , 4 , 5 ] n = len (A) SumDivPrime(A, n) # This code is contributed # by saurabh_shukla |
C#
// C# program to check if sum of primes // from an array is divisible by any of // the primes from the same array class GFG { //returns the maximum value static int max_element( int [] A) { int max = System.Int32.MinValue; for ( int i = 0; i < A.Length; i++) if (max < A[i]) max = A[i]; return max; } // Function to print "YES" if sum of // primes from an array is divisible // by any of the primes from the same array static void SumDivPrime( int [] A, int n) { int max_val = (max_element(A)) + 1; // USE SIEVE TO FIND ALL PRIME NUMBERS LESS // THAN OR EQUAL TO max_val // Create a boolean array "prime[0..n]". A // value in prime[i] will finally be false // if i is Not a prime, else true. bool [] prime= new bool [max_val+1]; //initialize the array for ( int i = 0; i <= max_val; i++) prime[i] = true ; // Remaining part of SIEVE prime[0] = false ; prime[1] = false ; for ( int p = 2; p * p <= max_val; p++) { // If prime[p] is not changed, then // it is a prime if (prime[p] == true ) { // Update all multiples of p for ( int i = p * 2; i <= max_val; i += p) prime[i] = false ; } } int sum = 0; // Traverse through the array for ( int i = 0; i < n; ++i) { if (prime[A[i]]) sum += A[i]; } for ( int i = 0; i < n; ++i) { if (prime[A[i]] && sum % A[i] == 0) { System.Console.WriteLine( "YES" ); return ; } } System.Console.WriteLine( "NO" ); } // Driver code public static void Main() { int []A = { 1, 2, 3, 4, 5 }; int n = A.Length; SumDivPrime(A, n); } } // This code is contributed by mits |
PHP
<?php // PHP program to check if sum of primes // from an array is divisible by any of // the primes from the same array // Function to print "YES" if sum of primes // from an array is divisible by any of the // primes from the same array function SumDivPrime( $A , $n ) { $max_val = max( $A ); // USE SIEVE TO FIND ALL PRIME NUMBERS // LESS THAN OR EQUAL TO max_val // Create a boolean array "prime[0..n]". A // value in prime[i] will finally be false // if i is Not a prime, else true. $prime = array_fill (1, $max_val + 1, true); // Remaining part of SIEVE $prime [0] = false; $prime [1] = false; for ( $p = 2; $p * $p <= $max_val ; $p ++) { // If prime[p] is not changed, then // it is a prime if ( $prime [ $p ] == true) { // Update all multiples of p for ( $i = $p * 2; $i <= $max_val ; $i += $p ) $prime [ $i ] = false; } } $sum = 0; // Traverse through the array for ( $i = 0; $i < $n ; ++ $i ) { if ( $prime [ $A [ $i ]]) $sum += $A [ $i ]; } for ( $i = 0; $i < $n ; ++ $i ) { if ( $prime [ $A [ $i ]] && $sum % $A [ $i ] == 0) { echo "YES" ; return ; } } echo "NO" ; } // Driver Code $A = array ( 1, 2, 3, 4, 5 ); $n = sizeof( $A ) ; SumDivPrime( $A , $n ); // This code is contributed by Ryuga ?> |
Javascript
<script> // Javascript program to check if sum // of primes from an array is divisible // by any of the primes from the same array // Returns the maximum value function max_element(A) { var max = Number.MIN_VALUE; for ( var i = 0; i < A.length; i++) if (max < A[i]) max = A[i]; return max; } // Function to print "YES" if sum of primes // from an array is divisible by any of the // primes from the same array function SumDivPrime(A, n) { var max_val = (max_element(A)) + 1; // USE SIEVE TO FIND ALL PRIME NUMBERS LESS // THAN OR EQUAL TO max_val // Create a boolean array "prime[0..n]". A // value in prime[i] will finally be false // if i is Not a prime, else true. var prime = new Array(max_val + 1); // Initialize the array for ( var i = 0; i <= max_val; i++) prime[i] = true ; // Remaining part of SIEVE prime[0] = false ; prime[1] = false ; for ( var p = 2; p * p <= max_val; p++) { // If prime[p] is not changed, then // it is a prime if (prime[p] == true ) { // Update all multiples of p for ( var i = p * 2; i <= max_val; i += p) prime[i] = false ; } } var sum = 0; // Traverse through the array for ( var i = 0; i < n; ++i) { if (prime[A[i]]) sum += A[i]; } for ( var i = 0; i < n; ++i) { if (prime[A[i]] && sum % A[i] == 0) { document.write( "YES" ); return ; } } document.write( "NO" ); } // Driver code var A = [ 1, 2, 3, 4, 5 ]; var n = A.length; SumDivPrime(A, n); // This code is contributed by SoumikMondal </script> |
YES
Complexity Analysis:
- Time Complexity: O(max_val*log(log(max_val))), where max_val is the largest element of the array.
- Auxiliary Space: O(max_val)
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