Four Divisors

Four Divisors - Problem

Array Math Medium

Given an integer array nums, return the sum of divisors of the integers in that array that have exactly four divisors.

If there is no such integer in the array, return 0.

Note: A number has exactly four divisors when it is either:

The cube of a prime number (p³)
The product of two distinct prime numbers (p × q)

Input & Output

Example 1 — Mixed Numbers

$ Input: nums = [21,4,7]

› Output: 32

💡 Note: 21 has divisors [1,3,7,21] (4 divisors, sum=32). 4 has divisors [1,2,4] (3 divisors). 7 has divisors [1,7] (2 divisors). Only 21 has exactly 4 divisors.

Example 2 — Multiple Valid Numbers

$ Input: nums = [21,21]

› Output: 64

💡 Note: Both instances of 21 have exactly 4 divisors [1,3,7,21] with sum 32. Total: 32 + 32 = 64.

Example 3 — No Valid Numbers

$ Input: nums = [1,2,3,4,5]

› Output: 0

💡 Note: 1 has 1 divisor, 2 has 2 divisors, 3 has 2 divisors, 4 has 3 divisors, 5 has 2 divisors. None have exactly 4 divisors.

Constraints

1 ≤ nums.length ≤ 10⁴
1 ≤ nums[i] ≤ 10⁵

Visualization

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Asked in

G Google 12 f Facebook 8 a Amazon 6

The key insight is that only numbers of the form p³ (prime cubed) or p×q (product of two distinct primes) have exactly 4 divisors. Best approach uses mathematical pattern recognition to identify these cases directly. Time: O(n × √max(nums)), Space: O(1)

Common Approaches

✓ Bit Manipulation

⏱️ Time: N/A Space: N/A

Brute Force Divisor Counting

⏱️ Time: O(n × max(nums)) Space: O(1)

Iterate through each number in the array, then for each number check all possible divisors from 1 to n. If exactly 4 divisors are found, calculate their sum and add to result.

Mathematical Pattern Recognition

⏱️ Time: O(n × sqrt(max(nums))) Space: O(1)

A number has exactly 4 divisors if and only if it's either a cube of a prime (p³) or a product of two distinct primes (p×q). Check these patterns directly instead of counting all divisors.

Optimized Divisor Counting

⏱️ Time: O(n × sqrt(max(nums))) Space: O(1)

For each number, find divisors by checking only numbers from 1 to sqrt(n). For each divisor i found, both i and n/i are divisors (except when i = sqrt(n)).

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAX_NUMS 1000

int countDivisors(int n) {
    int count = 0;
    for (int i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            if (i * i == n) {
                count++;
            } else {
                count += 2;
            }
        }
    }
    return count;
}

int sumDivisors(int n) {
    int sum = 0;
    for (int i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            sum += i;
            if (i * i != n) {
                sum += n / i;
            }
        }
    }
    return sum;
}

int solution(int* nums, int numsSize) {
    int totalSum = 0;
    for (int i = 0; i < numsSize; i++) {
        if (countDivisors(nums[i]) == 4) {
            totalSum += sumDivisors(nums[i]);
        }
    }
    return totalSum;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[MAX_NUMS];
    int numsSize = 0;
    
    if (strstr(line, "[]")) {
        printf("0\n");
        return 0;
    }
    
    char* p = line;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        nums[numsSize++] = (int)strtol(p, &p, 10);
    }
    
    int result = solution(nums, numsSize);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

428 Likes

Ln 1, Col 1

Smart Actions

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