LeetCode 728 | Self Dividing Numbers | O(n log n) Time | O(k) Space | Easy Solution

Description: Self Dividing Numbers is a beginner-friendly LeetCode problem that helps improve your understanding of digit manipulation, modulo operations, and number properties. You are given two integers left and right. Return a list of all the self-dividing numbers in the range [left, right]. A self-dividing number is a number that: Is divisible by every digit it contains Does not contain the digit 0 For example, 128 is a self-dividing number because: 128 % 1 = 0 128 % 2 = 0 128 % 8 = 0 ✅ Approach The idea is simple and efficient: • Traverse every number from left to right • For each number, extract its digits one by one • If any digit is 0, the number is not self-dividing • Check whether the original number is divisible by each digit • If all digits divide the number evenly, add it to the result list • Continue until all numbers in the range are processed This approach efficiently verifies whether each number satisfies the self-dividing condition. ✅ Time Complexity O(n log n) → For each number, we process all of its digits ✅ Space Complexity O(k) → Where k is the number of self-dividing numbers stored in the result 🔗 LeetCode https://leetcode.com/u/TR0kHGhQN9/ 🔗 LinkedIn   / dhruvi-patel-bb2b9239a   📝 Notes • Great problem for practicing digit extraction • Uses modulo and division operations extensively • Helps build intuition for number property checks • No advanced data structures are required • Frequently asked in beginner coding interviews • Demonstrates how to validate conditions on individual digits • An excellent example of an O(n log n) time and O(k) space solution #leetcode #dsa #coding #programming #java #python #cplusplus #math #numbers #leetcodeeasy #codinginterview #algorithms #softwareengineer #computerscience #datastructures #selfdividingnumbers #digitmanipulation #numbertheory #interviewpreparation #codingpractice #leetcodeproblems