# Solving Leetcode Interviews in Seconds with AI: Three Divisors


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1952" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	Here's the breakdown:

*   **Key Idea:** A number has exactly three divisors if and only if it is the square of a prime number. This is because the divisors would be 1, the prime number itself, and its square.

*   **Prime Check Optimization:** Efficiently determine if a number is prime by only checking divisibility up to the square root of the number.

*   **Square Root Check:** Compute the integer square root of the input `n` and verify if squaring it yields the original number. If it does, then check if the square root is a prime number.

*   **Complexity:**
    *   Runtime: O(sqrt(n))
    *   Space: O(1)

	
	# Code
	```python
	import math

def isThree(n: int) -> bool:
    """
    Given an integer n, return true if n has exactly three positive divisors.
    Otherwise, return false.
    """
    if n < 4:
        return False

    root = int(math.sqrt(n))
    if root * root != n:
        return False

    if root <= 1:
      return False

    for i in range(2, int(math.sqrt(root)) + 1):
        if root % i == 0:
            return False

    return True
	```