Three Divisors - Practice Coding Problems

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

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Understanding the Visualization

Input

Given integer n to analyze

Count Divisors

Find all positive divisors of n

Check Count

Return true if exactly 3 divisors found

Key Takeaway

🎯 Key Insight: Only perfect squares of prime numbers have exactly 3 divisors: 1, the prime, and its square

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

Approach	Time	Space	Notes
✓ Optimized - Check Up to √n	O(√n)	O(1)	Count divisors by checking only up to square root of n
Mathematical Insight	O(√n)	O(1)	Recognize that only squares of primes have exactly 3 divisors
Brute Force - Count All Divisors	O(n)	O(1)	Check every number from 1 to n and count divisors

Optimized - Check Up to √n — Algorithm Steps

Initialize counter to 0
For each number i from 1 to √n
If n % i == 0, count both i and n/i as divisors
Handle perfect square case to avoid double counting
Return true if count equals 3

Visualization

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Step-by-Step Walkthrough

Check 1 to √n

Only test numbers up to square root

Count Pairs

For each divisor i, also count n/i

Handle Perfect Square

Count √n only once if n is perfect square

Code -

solution.c — C

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

bool solution(int n) {
    int count = 0;
    int sqrtN = sqrt(n);
    
    for (int i = 1; i <= sqrtN; i++) {
        if (n % i == 0) {
            if (i * i == n) {
                count += 1;
            } else {
                count += 2;
            }
        }
    }
    
    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)

Loop only goes up to square root of n

✓ Linear Growth

Space Complexity

O(1)

Only using a counter variable

✓ Linear Space

25.0K Views

Medium Frequency

~15 min Avg. Time

890 Likes

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Optimized - Check Up to √n — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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