Given a number N, find the smallest prime divisor of N.
Examples:
Input: 25
Output: 5Input: 31
Output: 31
Approach:
- Check if the number is divisible by 2 or not.
- Iterate from i = 3 to sqrt(N) and making a jump of 2.
- If any of the numbers divide N then it is the smallest prime divisor.
- If none of them divide, then N is the answer.
Below is the implementation of the above algorithm:
// C++ program to count the number of // subarrays that having 1 #include <bits/stdc++.h> using namespace std;
// Function to find the smallest divisor int smallestDivisor( int n)
{ // if divisible by 2
if (n % 2 == 0)
return 2;
// iterate from 3 to sqrt(n)
for ( int i = 3; i * i <= n; i += 2) {
if (n % i == 0)
return i;
}
return n;
} // Driver Code int main()
{ int n = 31;
cout << smallestDivisor(n);
return 0;
} |
// Java program to count the number of // subarrays that having 1 import java.io.*;
class GFG {
// Function to find the smallest divisor static int smallestDivisor( int n)
{ // if divisible by 2
if (n % 2 == 0 )
return 2 ;
// iterate from 3 to sqrt(n)
for ( int i = 3 ; i * i <= n; i += 2 ) {
if (n % i == 0 )
return i;
}
return n;
} // Driver Code public static void main (String[] args) {
int n = 31 ;
System.out.println (smallestDivisor(n));
}
} |
# Python3 program to count the number # of subarrays that having 1 # Function to find the smallest divisor def smallestDivisor(n):
# if divisible by 2
if (n % 2 = = 0 ):
return 2 ;
# iterate from 3 to sqrt(n)
i = 3 ;
while (i * i < = n):
if (n % i = = 0 ):
return i;
i + = 2 ;
return n;
# Driver Code n = 31 ;
print (smallestDivisor(n));
# This code is contributed by mits |
// C# program to count the number // of subarrays that having 1 using System;
class GFG
{ // Function to find the // smallest divisor static int smallestDivisor( int n)
{ // if divisible by 2
if (n % 2 == 0)
return 2;
// iterate from 3 to sqrt(n)
for ( int i = 3;
i * i <= n; i += 2)
{
if (n % i == 0)
return i;
}
return n;
} // Driver Code static public void Main ()
{ int n = 31;
Console.WriteLine(smallestDivisor(n));
} } // This code is contributed // by Sach_Code |
<?php // PHP program to count the number // of subarrays that having 1 // Function to find the smallest divisor function smallestDivisor( $n )
{ // if divisible by 2
if ( $n % 2 == 0)
return 2;
// iterate from 3 to sqrt(n)
for ( $i = 3; $i * $i <= $n ; $i += 2)
{
if ( $n % $i == 0)
return $i ;
}
return $n ;
} // Driver Code $n = 31;
echo smallestDivisor( $n );
// This code is contributed by Sachin ?> |
<script> // javascript program to count the number of // subarrays that having 1 // Function to find the smallest divisor
function smallestDivisor(n) {
// if divisible by 2
if (n % 2 == 0)
return 2;
// iterate from 3 to sqrt(n)
for ( var i = 3; i * i <= n; i += 2) {
if (n % i == 0)
return i;
}
return n;
}
// Driver Code
var n = 31;
document.write(smallestDivisor(n));
// This code is contributed by todaysgaurav </script> |
31
How to efficiently find prime factors of all numbers till n?
Please refer Least prime factor of numbers till n
Time Complexity: O(sqrt(N)), as we are using a loop to traverse sqrt (N) times. As the condition is i*i<=N, on application of sqrt function on both the sides we get sqrt (i*i) <= sqrt(N), which is i<= sqrt(N), therefore the loop will traverse for sqrt(N) times.
Auxiliary Space: O(1), as we are not using any extra space.