What is the time complexity of Four Divisors?

The optimal solution has a time complexity of O(n*sqrt(m)).

What data structures are used to solve Four Divisors?

This problem uses Array, Math. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Four Divisors?

Insight 1: A number has exactly four divisors if and only if it is either the cube of a prime number (p^3, divisors are 1, p, p^2, p^3) or the product of two distinct prime numbers (p*q, divisors are 1, p, q, p*q). Insight 2: The number of divisors can be determined by iterating only up to the square root of the number. This optimization significantly reduces the computation time.

Which companies ask LeetCode 1390. Four Divisors in interviews?

This problem is frequently asked at Capital One.

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Four Divisors - LeetCode 1390 Solution

Q: How to solve Four Divisors?

The solution iterates through the input array `nums`. For each number, it checks if the sum of its divisors has already been computed and stored in a memo. If not, it calculates the divisors. The core idea is to iterate up to the square root of the number and find all divisors. If exactly four divisors are found, their sum is calculated and added to the total sum. The calculated sum (or 0 if the number doesn't have four divisors) is then stored in the memo for future use.

Four Divisors - Complete Solution Guide

Four Divisors is LeetCode problem 1390, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 <= 10 4 1 <= nums[i]

Detailed Explanation

The problem asks us to find the sum of the divisors of all numbers in a given integer array `nums` that have exactly four divisors. If a number in `nums` does not have exactly four divisors, we ignore it. The overall goal is to compute the sum of divisors for numbers that *do* have exactly four divisors and return this sum, or 0 if no such number exists in the input array.

Solution Approach

The solution iterates through the input array `nums`. For each number, it checks if the sum of its divisors has already been computed and stored in a memo. If not, it calculates the divisors. The core idea is to iterate up to the square root of the number and find all divisors. If exactly four divisors are found, their sum is calculated and added to the total sum. The calculated sum (or 0 if the number doesn't have four divisors) is then stored in the memo for future use.

Step-by-Step Algorithm

Step 1: Initialize a memo (hash map) to store the sum of divisors for each number encountered and a total sum variable to 0.
Step 2: Iterate through the input array `nums`.
Step 3: For each number `n`, check if it exists in the memo. If it does, add the stored sum of divisors to the total sum and continue to the next number.
Step 4: If `n` is not in the memo, initialize a set to store its divisors.
Step 5: Iterate from 1 up to the square root of `n`. For each `i`, if `i` is a divisor of `n`, add `i` and `n/i` to the divisors set.
Step 6: While calculating divisors, if the size of the divisor set exceeds 4, break the inner loop as the number can't have exactly four divisors.
Step 7: After the inner loop finishes, check if the divisor set contains exactly four elements.
Step 8: If the divisor set has four elements, calculate the sum of all divisors in the set and add it to the total sum. Store the sum in the memo for the number `n`.
Step 9: If the divisor set does not have four elements, store 0 in the memo for the number `n`.
Step 10: After iterating through all the numbers in `nums`, return the total sum.

Key Insights

Insight 1: A number has exactly four divisors if and only if it is either the cube of a prime number (p^3, divisors are 1, p, p^2, p^3) or the product of two distinct prime numbers (p*q, divisors are 1, p, q, p*q).
Insight 2: The number of divisors can be determined by iterating only up to the square root of the number. This optimization significantly reduces the computation time.
Insight 3: Memoization can be used to store the sum of divisors for previously encountered numbers, improving performance when the same number appears multiple times in the input array.

Complexity Analysis

Time Complexity: O(n*sqrt(m))

Space Complexity: O(n)

Topics

This problem involves: Array, Math.

Companies

Asked at: Capital One.

Four Divisors | LeetCode 1390

Medium

Array Math

Time: O(n*sqrt(m))Space: O(n)

Solution Code

Available in:

import java.util.HashSet;
import java.util.HashMap;
import java.util.Set;

class Solution {
    public int sumFourDivisors(int[] nums) {
        HashMap<Integer, Integer> memo = new HashMap<>();
        int total_sum = 0;

        for (int n : nums) {
            if (memo.containsKey(n)) {
                total_sum += memo.get(n);
                continue;
            }
            
            Set<Integer> divisors = new HashSet<>();
            if (n > 0) {
                int limit = (int) Math.sqrt(n);
                for (int i = 1; i <= limit; i++) {
                    if (n % i == 0) {
                        divisors.add(i);
                        divisors.add(n / i);
                    }
                    if (divisors.size() > 4) {
                        break;
                    }
                }
            }
            
            if (divisors.size() == 4) {
                int current_sum = 0;
                for (int divisor : divisors) {
                    current_sum += divisor;
                }
                memo.put(n, current_sum);
                total_sum += current_sum;
            } else {
                memo.put(n, 0);
            }
        }
        
        return total_sum;
    }
}

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Problem Info

Description

Problem 1390: Four Divisors - Detailed explanation and solution approaches.

Topics

Array Math

Companies

Capital One

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1390

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Initialize a memo (hash map) to store the sum of divisors for each number encountered and a total sum variable to 0.
2
Step 2: Iterate through the input array `nums`.
3
Step 3: For each number `n`, check if it exists in the memo. If it does, add the stored sum of divisors to the total sum and continue to the next number.
4
Step 4: If `n` is not in the memo, initialize a set to store its divisors.
5
Step 5: Iterate from 1 up to the square root of `n`. For each `i`, if `i` is a divisor of `n`, add `i` and `n/i` to the divisors set.
6
Step 6: While calculating divisors, if the size of the divisor set exceeds 4, break the inner loop as the number can't have exactly four divisors.
7
Step 7: After the inner loop finishes, check if the divisor set contains exactly four elements.
8
Step 8: If the divisor set has four elements, calculate the sum of all divisors in the set and add it to the total sum. Store the sum in the memo for the number `n`.
9
Step 9: If the divisor set does not have four elements, store 0 in the memo for the number `n`.
10
Step 10: After iterating through all the numbers in `nums`, return the total sum.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*sqrt(m))

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the nested loop structure: the outer loop iterates through the input array `nums` (n times), and the inner loop iterates up to the square root of each number (sqrt(m) times). The space complexity is due to the memo which stores the sums of divisors for each unique number encountered. In the worst case, all numbers in `nums` are distinct, and memoization requires storing information for up to m distinct numbers, where m is the maximum value of a number in `nums`.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Prime Factorization

First find prime factorization of the number. Then determine the number of divisors using the exponents of prime factors. If the number of divisors is exactly 4, calculate the sum of divisors.

Time: O(n*sqrt(m)), where n is the length of the `nums` array and m is the maximum value in `nums`.Space: O(log m) - the space for storing prime factors.

Trade-offs: While this approach focuses more on prime factorization, in terms of time complexity for calculating sum of divisors and number of divisors, it is no better than the current approach. The current approach is simple to implement than this approach, without prime factorization. This approach becomes beneficial if only the *count* of numbers with 4 divisors is required, and memoization is not possible.

✨

Key Insights

1Insight 1: A number has exactly four divisors if and only if it is either the cube of a prime number (p^3, divisors are 1, p, p^2, p^3) or the product of two distinct prime numbers (p*q, divisors are 1, p, q, p*q).
2Insight 2: The number of divisors can be determined by iterating only up to the square root of the number. This optimization significantly reduces the computation time.
3Insight 3: Memoization can be used to store the sum of divisors for previously encountered numbers, improving performance when the same number appears multiple times in the input array.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: Numbers less than or equal to 0. The code correctly handles this by checking `n > 0` before attempting to calculate divisors. Negative or zero numbers cannot have a well-defined set of divisors.
Edge case 2: Numbers with more than four divisors. The code efficiently handles this by breaking the inner loop when the size of the divisor set exceeds 4, avoiding unnecessary calculations.
Edge case 3: Numbers with exactly four divisors that appear multiple times in the input array. The memoization technique efficiently reuses previously calculated divisor sums, avoiding redundant computations.
Edge case 4: Prime numbers to the power of 3 that have exactly four divisors {1, p, p^2, p^3} and that is why calculating till the sqrt(number) is correct, if the count equals 4 it means, it has met our condition.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not optimizing the divisor search by iterating only up to the square root of the number. This can lead to significant performance degradation for large numbers.

✗Common mistake 2: Not handling the case where a number has more than four divisors. Failing to break the divisor search loop early can lead to incorrect divisor sums and incorrect results.

✗Common mistake 3: Forgetting the basic divisors {1, n}. Ensure that the divisor count is appropriately handled.

✗Common mistake 4: Not handling duplicate numbers correctly when calculating the sum, or using an inefficient method for counting divisors.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Cryptography. The number of divisors and their sums are concepts relevant in number theory, which underlies many cryptographic algorithms.

🏦Real-world scenario 2: Database indexing. Understanding divisors can help optimize database queries and indexing strategies when dealing with numerical data. Knowing the divisibility of a number might lead to efficient data partitioning.

🎮Real-world scenario 3: Resource Allocation. Distributing resources effectively may involve dividing quantities into equal parts. Knowing the divisors of a number facilitates this process.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 9, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Four Divisors"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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