1390. Four Divisors LeetCode Solution

In this guide, you will get 1390. Four Divisors LeetCode Solution with the best time and space complexity. The solution to Four Divisors problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Four Divisors solution in C++
4. Four Divisors solution in Java
5. Four Divisors solution in Python
6. Additional Resources

Problem Statement of Four Divisors

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.

Example 1:

Input: nums = [21,4,7]
Output: 32
Explanation:
21 has 4 divisors: 1, 3, 7, 21
4 has 3 divisors: 1, 2, 4
7 has 2 divisors: 1, 7
The answer is the sum of divisors of 21 only.

Example 2:

Input: nums = [21,21]
Output: 64

Example 3:

Input: nums = [1,2,3,4,5]
Output: 0

Constraints:

1 <= nums.length <= 104
1 <= nums[i] <= 105

Complexity Analysis

• Time Complexity:
• Space Complexity:

1390. Four Divisors LeetCode Solution in C++

class Solution {
 public:
  int sumFourDivisors(vector<int>& nums) {
    int ans = 0;

    for (int num : nums) {
      int divisor = 0;
      for (int i = 2; i * i <= num; ++i)
        if (num % i == 0) {
          if (divisor == 0)
            divisor = i;
          else {
            divisor = 0;
            break;
          }
        }
      if (divisor > 0 && divisor * divisor < num)
        ans += 1 + num + divisor + num / divisor;
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

1390. Four Divisors LeetCode Solution in Java

class Solution {
  public int sumFourDivisors(int[] nums) {
    int ans = 0;

    for (int num : nums) {
      int divisor = 0;
      for (int i = 2; i * i <= num; ++i)
        if (num % i == 0) {
          if (divisor == 0)
            divisor = i;
          else {
            divisor = 0;
            break;
          }
        }
      if (divisor > 0 && divisor * divisor < num)
        ans += 1 + num + divisor + num / divisor;
    }

    return ans;
  }
}
// code provided by PROGIEZ

1390. Four Divisors LeetCode Solution in Python

class Solution:
  def sumFourDivisors(self, nums: list[int]) -> int:
    ans = 0

    for num in nums:
      divisor = 0
      for i in range(2, math.isqrt(num) + 1):
        if num % i == 0:
          if divisor == 0:
            divisor = i
          else:
            divisor = 0
            break
      if divisor > 0 and divisor * divisor < num:
        ans += 1 + num + divisor + num // divisor

    return ans
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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