Three Divisors

Three Divisors - Problem

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.

Input & Output

Example 1 — Perfect Square of Prime

$ Input: n = 4

› Output: true

💡 Note: 4 = 2², and 2 is prime. The divisors of 4 are: 1, 2, 4 (exactly 3 divisors)

Example 2 — Not a Square of Prime

$ Input: n = 6

› Output: false

💡 Note: The divisors of 6 are: 1, 2, 3, 6 (4 divisors, not 3)

Example 3 — Another Square of Prime

$ Input: n = 9

› Output: true

💡 Note: 9 = 3², and 3 is prime. The divisors of 9 are: 1, 3, 9 (exactly 3 divisors)

Constraints

1 ≤ n ≤ 10⁴

Visualization

Tap to expand

Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is that only perfect squares of prime numbers have exactly 3 divisors. Best approach is the mathematical solution checking if n = p² where p is prime. Time: O(√n), Space: O(1)

Common Approaches

✓ Brute Force - Count All Divisors

⏱️ Time: O(n) Space: O(1)

Iterate through all numbers from 1 to n and count how many divide n evenly. Return true if exactly 3 divisors are found.

Mathematical Insight

⏱️ Time: O(√n) Space: O(1)

A number has exactly 3 divisors if and only if it's the square of a prime number. The divisors are: 1, the prime, and the square itself.

Optimized - Check Up to √n

⏱️ Time: O(√n) Space: O(1)

Instead of checking all numbers up to n, only check up to √n. For each divisor i found, we also get n/i as a divisor (except when i = √n).

Brute Force - Count All Divisors — Algorithm Steps

Initialize counter to 0
For each number i from 1 to n, check if n % i == 0
If divisible, increment counter
Return true if counter equals 3

Visualization

Tap to expand

Step-by-Step Walkthrough

Initialize

Start with count = 0

Check Each Number

Test if each number from 1 to n divides evenly

Count Result

Return true if exactly 3 divisors found

Code -

solution.c — C

#include <stdio.h>
#include <stdbool.h>

bool solution(int n) {
    int count = 0;
    for (int i = 1; i <= n; i++) {
        if (n % i == 0) {
            count++;
        }
    }
    return count == 3;
}

int main() {
    int n;
    scanf("%d", &n);
    bool result = solution(n);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single loop from 1 to n checking each number

✓ Linear Growth

Space Complexity

O(1)

Only using a counter variable

✓ Linear Space

25.0K Views

Medium Frequency

~15 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Count All Divisors — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler