Three Divisors - Practice Coding Problems

Three Divisors - Problem

Three Divisors is a fascinating number theory problem that asks you to identify special numbers with exactly three positive divisors.

Given an integer n, return true if n has exactly three positive divisors, otherwise return false.

A divisor of n is any positive integer m where n = k × m for some integer k. For example, the divisors of 12 are: 1, 2, 3, 4, 6, and 12.

Key Insight: Numbers with exactly three divisors have a very special property - they are squares of prime numbers! This is because a prime squared (p²) has divisors: 1, p, and p².

Input & Output

example_1.py — Basic Case

$ Input: n = 9

› Output: true

💡 Note: The divisors of 9 are: 1, 3, and 9. Since there are exactly 3 divisors, return true. Note that 9 = 3², which is the square of prime 3.

example_2.py — Non-prime Square

$ Input: n = 12

› Output: false

💡 Note: The divisors of 12 are: 1, 2, 3, 4, 6, and 12. Since there are 6 divisors (not 3), return false.

example_3.py — Edge Case

$ Input: n = 1

› Output: false

💡 Note: The only divisor of 1 is 1 itself. Since there is only 1 divisor (not 3), return false.

Visualization

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Understanding the Visualization

Prime Property

A prime p has exactly 2 divisors: 1 and p

Square of Prime

p² has divisors: 1, p, and p² (exactly 3)

Non-prime Squares

If n = ab where a,b > 1, then n² has more divisors

Verification

Check if n is perfect square and √n is prime

Key Takeaway

🎯 Key Insight: Numbers with exactly three divisors are always squares of prime numbers (p²), which have divisors 1, p, and p².

Time & Space Complexity

Time Complexity

⏱️

O(√n)

We only check divisors up to the square root of n

✓ Linear Growth

Space Complexity

O(1)

Only using a counter variable and loop variables

✓ Linear Space

Constraints

1 ≤ n ≤ 10⁴
n is a positive integer
The solution should handle edge cases like n = 1

Asked in

G Google 15 a Amazon 12 ⊞ Microsoft 8 f Meta 6

The optimal approach recognizes that only squares of prime numbers have exactly three divisors. Instead of counting all divisors, check if n is a perfect square and if its square root is prime. This reduces complexity from O(n) to O(√n) and leverages mathematical insights for elegant problem solving.

Common Approaches

Approach	Time	Space	Notes
✓ Optimized: Check up to √n	O(√n)	O(1)	Count divisors by checking only up to square root, counting pairs
Brute Force (Count All Divisors)	O(n)	O(1)	Count all divisors by checking every number from 1 to n

Optimized: Check up to √n — Algorithm Steps

Initialize count = 0
Loop from i = 1 to √n
If n % i == 0, increment count
If i ≠ n/i, increment count again (for the paired divisor)
Return true if count equals 3

Visualization

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Step-by-Step Walkthrough

Find √n

Calculate square root to limit search space

Check Pairs

For each divisor i, also count n/i

Handle Squares

Don't double-count when i = n/i

Verify Count

Check if total divisors equals 3

Code -

solution.c — C

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

bool hasThreeDivisors(int n) {
    int count = 0;
    int sqrtN = (int) sqrt(n);
    
    for (int i = 1; i <= sqrtN; i++) {
        if (n % i == 0) {
            if (i * i == n) {
                count += 1;
            } else {
                count += 2;
            }
            
            if (count > 3) return false;
        }
    }
    
    return count == 3;
}

int main() {
    int n;
    scanf("%d", &n);
    printf("%s\n", hasThreeDivisors(n) ? "true" : "false");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(√n)

We only check divisors up to the square root of n

✓ Linear Growth

Space Complexity

O(1)

Only using a counter variable and loop variables

✓ Linear Space

Constraints

1 ≤ n ≤ 10⁴
n is a positive integer
The solution should handle edge cases like n = 1

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Optimized: Check up to √n — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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