Given an array A[] consisting of N integers, the task is to check if the sum of numbers of digits in each array element is a prime number or not.
Examples:
Input: A[] = {1, 11, 12}
Output: Yes
Explanation: Count of digits of A[0], A[1] and A[2] are 1, 2, 2 respectively. Therefore, total sum of count of digits = 1 + 2 + 2 = 5, which is prime.Input: A[] = {1, 11, 123}
Output: No
Approach: Follow the steps below to solve the problem:
- Initialize a variable sum, to store the sum of the number of digits of array elements.
- Traverse the array and convert each array element to its equivalent string
- Add the length of every string to sum.
- Check if the value of sum after complete traversal of the array, is prime or not.
- Print Yes if found to be true. Otherwise, print No.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to check whether // a number is prime or not bool isPrime( int n)
{ // Corner cases
if (n <= 1)
return false ;
if (n <= 3)
return true ;
// If given number is a
// multiple of 2 or 3
if (n % 2 == 0 || n % 3 == 0)
return false ;
for ( int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false ;
return true ;
} // Function to check if sum // of count of digits of all // array elements is prime or not void CheckSumPrime( int A[], int N)
{ // Initialize sum with 0
int sum = 0;
// Traverse over the array
for ( int i = 0; i < N; i++) {
// Convert array element to string
string s = to_string(A[i]);
// Add the count of
// digits to sum
sum += s.length();
}
// Print the result
if (isPrime(sum)) {
cout << "Yes" << endl;
}
else {
cout << "No" << endl;
}
} // Drive Code int main()
{ int A[] = { 1, 11, 12 };
int N = sizeof (A) / sizeof (A[0]);
// Function call
CheckSumPrime(A, N);
return 0;
} |
Java
// Java program for the above approach import java.io.*;
import java.util.*;
class GFG{
// Function to check whether // a number is prime or not static boolean isPrime( int n)
{ // Corner cases
if (n <= 1 )
return false ;
if (n <= 3 )
return true ;
// If given number is a
// multiple of 2 or 3
if (n % 2 == 0 || n % 3 == 0 )
return false ;
for ( int i = 5 ; i * i <= n; i = i + 6 )
if (n % i == 0 || n % (i + 2 ) == 0 )
return false ;
return true ;
} // Function to check if sum // of count of digits of all // array elements is prime or not static void CheckSumPrime( int [] A, int N)
{ // Initialize sum with 0
int sum = 0 ;
// Traverse over the array
for ( int i = 0 ; i < N; i++)
{
// Convert array element to string
String s = Integer.toString(A[i]);
// Add the count of
// digits to sum
sum += s.length();
}
// Print the result
if (isPrime(sum) == true )
{
System.out.println( "Yes" );
}
else
{
System.out.println( "No" );
}
} // Drive Code public static void main(String[] args)
{ int [] A = { 1 , 11 , 12 };
int N = A.length;
// Function call
CheckSumPrime(A, N);
} } // This code is contributed by akhilsaini |
Python3
# Python3 program for the above approach import math
# Function to check whether # a number is prime or not def isPrime(n):
# Corner cases
if (n < = 1 ):
return False
if (n < = 3 ):
return True
# If given number is a
# multiple of 2 or 3
if (n % 2 = = 0 or n % 3 = = 0 ):
return False
for i in range ( 5 , int (math.sqrt(n) + 1 ), 6 ):
if (n % i = = 0 or n % (i + 2 ) = = 0 ):
return False
return True
# Function to check if sum # of count of digits of all # array elements is prime or not def CheckSumPrime(A, N):
# Initialize sum with 0
sum = 0
# Traverse over the array
for i in range ( 0 , N):
# Convert array element to string
s = str (A[i])
# Add the count of
# digits to sum
sum + = len (s)
# Print the result
if (isPrime( sum ) = = True ):
print ( "Yes" )
else :
print ( "No" )
# Drive Code if __name__ = = '__main__' :
A = [ 1 , 11 , 12 ]
N = len (A)
# Function call
CheckSumPrime(A, N)
# This code is contributed by akhilsaini |
C#
// C# program for the above approach using System;
class GFG{
// Function to check whether // a number is prime or not static bool isPrime( int n)
{ // Corner cases
if (n <= 1)
return false ;
if (n <= 3)
return true ;
// If given number is a
// multiple of 2 or 3
if (n % 2 == 0 || n % 3 == 0)
return false ;
for ( int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false ;
return true ;
} // Function to check if sum // of count of digits of all // array elements is prime or not static void CheckSumPrime( int [] A, int N)
{ // Initialize sum with 0
int sum = 0;
// Traverse over the array
for ( int i = 0; i < N; i++)
{
// Convert array element to string
String s = A[i].ToString();
// Add the count of
// digits to sum
sum += s.Length;
}
// Print the result
if (isPrime(sum) == true )
{
Console.WriteLine( "Yes" );
}
else
{
Console.WriteLine( "No" );
}
} // Drive Code public static void Main()
{ int [] A = { 1, 11, 12 };
int N = A.Length;
// Function call
CheckSumPrime(A, N);
} } // This code is contributed by akhilsaini |
Javascript
<script> // Javascript program for the above approach // Function to check whether // a number is prime or not function isPrime(n)
{ // Corner cases
if (n <= 1)
return false ;
if (n <= 3)
return true ;
// If given number is a
// multiple of 2 or 3
if (n % 2 == 0 || n % 3 == 0)
return false ;
for (let i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false ;
return true ;
} // Function to check if sum // of count of digits of all // array elements is prime or not function CheckSumPrime(A, N)
{ // Initialize sum with 0
let sum = 0;
// Traverse over the array
for (let i = 0; i < N; i++) {
// Convert array element to string
let s = new String(A[i]);
// Add the count of
// digits to sum
sum += s.length;
}
// Print the result
if (isPrime(sum)) {
document.write( "Yes" + "<br>" );
}
else {
document.write( "No" + "<br>" );
}
} // Drive Code let A = [ 1, 11, 12 ];
let N = A.length
// Function call
CheckSumPrime(A, N);
// This code is contributed by gfgking
</script> |
Output:
Yes
Time Complexity: O(N3/2)
Auxiliary Space: O(1)