What is the time complexity of Perfect Number?

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

What data structures are used to solve Perfect Number?

This problem uses Math. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Perfect Number?

Insight 1: A naive approach of iterating through all numbers from 1 to `num-1` to find divisors would be too slow. We can optimize this by only iterating up to the square root of `num`. Insight 2: When we find a divisor `i` of `num`, then `num/i` is also a divisor. We can add both to the sum of divisors.

Which companies ask LeetCode 507. Perfect Number in interviews?

This problem is frequently asked at Accenture, Grammarly.

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Perfect Number - LeetCode 507 Solution

Q: How to solve Perfect Number?

The provided solution calculates the sum of the proper divisors of the input number `num`. It initializes the sum to 1 because 1 is always a divisor. It then iterates from 2 up to the square root of `num`. For each number `i` in this range, it checks if `i` is a divisor of `num`. If it is, `i` and `num/i` are added to the sum (unless `i * i == num`, in which case only `i` is added to avoid double-counting). Finally, it checks if the sum of the proper divisors is equal to `num`. If it is, then `num` is a perfect number, and the function returns `true`. Otherwise, it returns `false`.

Perfect Number - Complete Solution Guide

Perfect Number is LeetCode problem 507, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

A perfect number is a positive integer that is equal to the sum of its positive divisors , excluding the number itself. A divisor of an integer x is an integer that can divide x evenly. Given an integer n , return true if n is a perfect number, otherwise return false . Example 1: Input: num = 28 Output: true Explanation: 28 = 1 + 2 + 4 + 7 + 14 1, 2, 4, 7, and 14 are all divisors of 28. Example 2: Input: num = 7 Output: false Constraints: 1 <= num <= 10 8

Detailed Explanation

The problem asks us to determine if a given positive integer, `num`, is a perfect number. A perfect number is defined as a number that is equal to the sum of its proper divisors (all positive divisors excluding the number itself). The input is a single integer `num`, and the output should be a boolean value: `true` if `num` is a perfect number, and `false` otherwise. The constraint is that `1 <= num <= 10^8`.

Solution Approach

The provided solution calculates the sum of the proper divisors of the input number `num`. It initializes the sum to 1 because 1 is always a divisor. It then iterates from 2 up to the square root of `num`. For each number `i` in this range, it checks if `i` is a divisor of `num`. If it is, `i` and `num/i` are added to the sum (unless `i * i == num`, in which case only `i` is added to avoid double-counting). Finally, it checks if the sum of the proper divisors is equal to `num`. If it is, then `num` is a perfect number, and the function returns `true`. Otherwise, it returns `false`.

Step-by-Step Algorithm

Step 1: Handle the base cases: if `num` is less than or equal to 1, return `false` because perfect numbers are greater than 1.
Step 2: Initialize `sum_div` to 1, as 1 is always a divisor.
Step 3: Iterate from `i = 2` to the square root of `num` (inclusive).
Step 4: In each iteration, check if `i` divides `num` without any remainder (i.e., `num % i == 0`).
Step 5: If `i` is a divisor, add `i` to `sum_div`.
Step 6: If `i * i != num` (meaning `i` is not the square root of `num`), also add `num // i` to `sum_div`.
Step 7: After the loop finishes, compare `sum_div` with `num`. If they are equal, return `true`; otherwise, return `false`.

Key Insights

Insight 1: A naive approach of iterating through all numbers from 1 to `num-1` to find divisors would be too slow. We can optimize this by only iterating up to the square root of `num`.
Insight 2: When we find a divisor `i` of `num`, then `num/i` is also a divisor. We can add both to the sum of divisors.
Insight 3: Be careful not to double-count the square root if `num` is a perfect square (i.e., `i * i == num`).

Complexity Analysis

Time Complexity: O(sqrt(n))

Space Complexity: O(1)

Topics

This problem involves: Math.

Companies

Asked at: Accenture, Grammarly.

Perfect Number | LeetCode 507

Easy

Math

Time: O(sqrt(n))Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean checkPerfectNumber(int num) {
        if (num <= 1) return false;
        int sum = 1;
        for (int i = 2; i * i <= num; i++) {
            if (num % i == 0) {
                sum += i;
                if (i * i != num) {
                    sum += num / i;
                }
            }
        }
        return sum == num;
    }
}

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Problem Info

Description

Problem 507: Perfect Number - Detailed explanation and solution approaches.

Topics

Math

Companies

AccentureGrammarly

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:507

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

7 steps

1
Step 1: Handle the base cases: if `num` is less than or equal to 1, return `false` because perfect numbers are greater than 1.
2
Step 2: Initialize `sum_div` to 1, as 1 is always a divisor.
3
Step 3: Iterate from `i = 2` to the square root of `num` (inclusive).
4
Step 4: In each iteration, check if `i` divides `num` without any remainder (i.e., `num % i == 0`).
5
Step 5: If `i` is a divisor, add `i` to `sum_div`.
6
Step 6: If `i * i != num` (meaning `i` is not the square root of `num`), also add `num // i` to `sum_div`.
7
Step 7: After the loop finishes, compare `sum_div` with `num`. If they are equal, return `true`; otherwise, return `false`.

💡

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(sqrt(n))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(sqrt(n)) because the loop iterates up to the square root of `num`. The space complexity is O(1) because the algorithm uses a constant amount of extra space regardless of the size of the input.

📖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

Precomputed Perfect Numbers

Since the range of `num` is relatively small (1 to 10^8), we can precompute a list of perfect numbers within this range. Then, we can check if the given `num` is in this list. This approach would have O(1) time complexity for each query, but requires O(k) space to store the precomputed list (where k is the number of perfect numbers less than 10^8).

Time: O(1) for lookup, O(k) to precompute (k is the number of perfect numbers <= 10^8)Space: O(k) where k is number of perfect numbers <= 10^8.

Trade-offs: If we need to check multiple numbers, precomputing might be better. Otherwise, the original solution is suitable. There are only a few perfect numbers below 10^8: 6, 28, 496, 8128, 33550336. Therefore k is small. The space overhead would be minimal and performance for multiple queries would be improved significantly.

✨

Key Insights

1Insight 1: A naive approach of iterating through all numbers from 1 to `num-1` to find divisors would be too slow. We can optimize this by only iterating up to the square root of `num`.
2Insight 2: When we find a divisor `i` of `num`, then `num/i` is also a divisor. We can add both to the sum of divisors.
3Insight 3: Be careful not to double-count the square root if `num` is a perfect square (i.e., `i * i == num`).

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: `num` is 1. The problem constraints state `1 <= num <= 10^8`. The provided code handles this case by immediately returning `false` if `num <= 1`, as 1 is not a perfect number.
Edge case 2: `num` is a perfect square. The code handles this correctly by checking `if i * i != num` before adding the other divisor. This prevents double-counting the square root when it is a divisor.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Iterating up to `num` instead of `sqrt(num)`. This results in a time complexity of O(n), which can lead to Time Limit Exceeded (TLE) errors for large inputs.

✗Common mistake 2: Forgetting to handle the case where `num` is a perfect square. Double-counting the square root divisor will result in an incorrect sum.

✗Common mistake 3: Not handling the base case where num <= 1.

✅

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: Error detection in data transmission. Properties of perfect numbers, although not directly used, relate to number theory concepts used in checksums and error-correcting codes.

🏦Real-world scenario 2: Cryptography. While not directly used, the properties of divisors and prime factorization, explored when investigating perfect numbers, form the foundation of many cryptographic algorithms.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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 26, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 29

❓

Frequently Asked Questions

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Perfect Number"?

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

🤖 Need help with this problem?