1952. Three Divisors LeetCode Solution

Last updated: February 25, 2025

In this guide, you will get 1952. Three Divisors LeetCode Solution with the best time and space complexity. The solution to Three Divisors problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Three Divisors solution in C++
Three Divisors solution in Java
Three Divisors solution in Python
Additional Resources

1952. Three Divisors LeetCode Solution image

Problem Statement of Three Divisors

Given an integer n, return true if n has exactly three positive divisors. Otherwise, return false.
An integer m is a divisor of n if there exists an integer k such that n = k * m.

Example 1:

Input: n = 2
Output: false
Explantion: 2 has only two divisors: 1 and 2.

Example 2:

Input: n = 4
Output: true
Explantion: 4 has three divisors: 1, 2, and 4.

Constraints:

1 <= n <= 104

Complexity Analysis

Time Complexity: O(\sqrt[4]{n})
Space Complexity: O(1)

1952. Three Divisors LeetCode Solution in C++

class Solution {
 public:
  bool isThree(int n) {
    if (n == 1)
      return false;
    // The numbers with exactly three divisors are perfect squares of a prime
    // number.
    const int root = sqrt(n);
    return root * root == n && isPrime(root);
  }

 private:
  bool isPrime(int num) {
    for (int i = 2; i <= sqrt(num); ++i)
      if (num % i == 0)
        return false;
    return true;
  }
};
/* code provided by PROGIEZ */

1952. Three Divisors LeetCode Solution in Java

class Solution {
  public boolean isThree(int n) {
    if (n == 1)
      return false;
    // The numbers with exactly three divisors are perfect squares of a prime
    // number.
    final int root = (int) Math.sqrt(n);
    return root * root == n && isPrime(root);
  }

  private boolean isPrime(int num) {
    for (int i = 2; i <= Math.sqrt(num); ++i)
      if (num % i == 0)
        return false;
    return true;
  }
}
// code provided by PROGIEZ

1952. Three Divisors LeetCode Solution in Python

class Solution:
  def isThree(self, n: int) -> bool:
    if n == 1:
      return False
    # The numbers with exactly three divisors are perfect squares of a prime
    # number.
    root = math.isqrt(n)
    return (root**2 == n and
            all(root % i != 0
                for i in range(2, math.isqrt(root) + 1)))
 # code by PROGIEZ

Additional Resources

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