How do you solve Four Divisors on LeetCode?

Four Divisors (LeetCode #1390) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n√m) time complexity and O(1) space complexity. Key insight: A number has exactly four divisors if it is the product of two distinct primes or a cube of a prime.

What is the brute force solution for Four Divisors?

The brute force approach for Four Divisors is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking each number in the array and finding all its divisors. If a number has exactly four divisors, we sum them up. This method is straightforward but can be inefficient for larger numbers. The optimal Optimal Solution approach improves this to O(n√m).

What algorithm or data structure is used to solve Four Divisors?

Four Divisors uses the following concepts: Array, Math. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Four Divisors?

Always consider edge cases, such as small numbers or primes. Explain your thought process clearly while coding; it helps the interviewer understand your reasoning. Practice identifying patterns in problems, such as divisor counting, to improve your problem-solving speed.

What are common mistakes when solving Four Divisors?

Not checking for the case where the divisor is the square root and counting it twice. Failing to recognize that only certain forms of numbers can have exactly four divisors.

#1390

Four Divisors

Medium

ArrayMathHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n√m)
Space	O(1)	O(1)

💡

Intuition

Time O(n√m)Space O(1)

The optimal approach leverages the fact that a number can only have exactly four divisors in specific cases: either it is the product of two distinct primes or a cube of a prime. This allows us to check divisors only up to the square root of each number, significantly reducing the number of checks needed.

⚙️

Algorithm

3 steps

1Step 1: Initialize a variable to hold the total sum of divisors.
2Step 2: For each number in the input array, check for divisors only up to the square root of the number.
3Step 3: Count the divisors. If exactly four are found, sum them up.

solution.py17 lines

1# Full working Python code
2
3def sum_of_four_divisors(nums):
4    total_sum = 0
5    for num in nums:
6        divisors = []
7        for i in range(1, int(num**0.5) + 1):
8            if num % i == 0:
9                divisors.append(i)
10                if i != num // i:
11                    divisors.append(num // i)
12        if len(divisors) == 4:
13            total_sum += sum(divisors)
14    return total_sum
15
16# Example usage
17print(sum_of_four_divisors([21, 4, 7]))  # Output: 32

ℹ

Complexity note: The time complexity is O(n√m) where n is the number of elements in the array and m is the maximum value in the array. We only check up to the square root of each number for divisors, which reduces the number of checks significantly. The space complexity is O(1) since we are using a constant amount of space.

1A number has exactly four divisors if it is the product of two distinct primes or a cube of a prime.
2Finding divisors only up to the square root of a number significantly reduces the number of iterations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.