[Solved] Given a positive integer n. Find whether a number is amazing or not

Given a positive integer n. Find whether a number is amazing or not. Print True If number is amazing else False. An amazing number is a natural number that is a product of two prime numbers.

Input Format:

The first line of the input contains a single integer T, denoting the number of test cases.

The T test case follows, a single line of the input containing a positive integer N.

Sample Input:

2
6
8

Constraints:

1<=T<=100

1<=N<=100000

Output Format:

Print ‘True’ if it is amazing, otherwise print ‘False’

Sample Output:

True

False

Explanation:

6 is a semiprime number as it is a
product of two prime numbers 2 and 3.

8 is not a semiprime number.

Solution

Time complexity: O(root n)

Auxiliary space: O(1)

Reference: https://en.wikipedia.org/wiki/Semiprime

import java.util.*;
public class Sauravhathi {

    public static void main(String[] args){

        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-- > 0){
            int n = sc.nextInt();
            if (isAmazing(n) != 0)
            //sauravhathi
                System.out.println("True");
            else
                System.out.println("False");
        }
    }

    public static int isAmazing(int n) {
        int count = 0;

        for (int i = 2; count < 2 && i * i <= n; ++i){
                // sauravhathi
            while (n % i == 0) {
                n /= i;
                ++count;
            }
        }

        if (n > 1){
            ++count;
        }
        return count == 2 ? 1 : 0;
    }
}

def isAmazing(n):
    count = 0
    for i in range(2, n + 1):
        while n % i == 0:
            n /= i
            count += 1
            #sauravhathi
    return count == 2


N = int(input())
for i in range(N):
    n = int(input())
    if isAmazing(n):
        print("True")
    else:
        print("False")

#include <bits/stdc++.h>
using namespace std;

int isAmazing(int n) {
    int count = 0;
    for (int i = 2; count < 2 && i * i <= n; ++i) {
        while (n % i == 0) {
            n /= i;
             // sauravhathi
            ++count;
        }
    }
    if (n > 1) {
        ++count;
    }
    return count == 2 ? 1 : 0;
}

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        if (isAmazing(n))
            cout << "True" << endl;
        else
            cout << "False" << endl;
    }
    return 0;
}

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