Smallest prime divisor of a number

Given a number N, find the smallest prime divisor of N. 

Examples:

Input: 25 
Output: 5

Input: 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++

// 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

// 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

# 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#

// 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
// 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
?>

Javascript

<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>

Output:

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.