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Find the K-Beauty of a Number - LeetCode 2269 Solution

Find the K-Beauty of a Number - Complete Solution Guide

Find the K-Beauty of a Number is LeetCode problem 2269, 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

The k-beauty of an integer num is defined as the number of substrings of num when it is read as a string that meet the following conditions: It has a length of k . It is a divisor of num . Given integers num and k , return the k-beauty of num . Note: Leading zeros are allowed. 0 is not a divisor of any value. A substring is a contiguous sequence of characters in a string. Example 1: Input: num = 240, k = 2 Output: 2 Explanation: The following are the substrings of num of length k: - "24" from "

Detailed Explanation

The problem asks you to find the 'k-beauty' of a number. The k-beauty is the count of substrings of length `k` within the number (treated as a string) that are also divisors of the original number. For example, if `num = 240` and `k = 2`, the substrings of length 2 are "24" and "40". Both are divisors of 240, so the k-beauty is 2. The problem explicitly states that leading zeros are allowed in substrings, but 0 is not considered a divisor.

Solution Approach

The solution iterates through all possible substrings of length `k` within the number represented as a string. For each substring, it converts it back to an integer, checks if it's non-zero, and then checks if the original number is divisible by the substring. If both conditions are true, the counter (`count`) is incremented. Finally, the function returns the `count`.

Step-by-Step Algorithm

  1. Convert the input integer `num` into a string `s`.
  2. Iterate through the string `s` using a sliding window of size `k`. The loop iterates from `i = 0` to `n - k`, where `n` is the length of the string.
  3. In each iteration, extract the substring `s[i:i+k]` and convert it to an integer `sub`.
  4. Check if `sub` is not equal to 0. If it is 0, skip to the next iteration.
  5. Check if `num` is divisible by `sub` using the modulo operator (`%`).
  6. If `sub` is not 0 and `num` is divisible by `sub`, increment the `count`.
  7. After iterating through all substrings, return the final `count`.

Key Insights

  • The problem elegantly combines string manipulation and divisibility checks. We need to iterate through all substrings of length `k`.
  • Converting the number to a string allows easy substring extraction. Integer division (`%`) is used to check for divisibility.
  • Handling the case where a substring is "0" is crucial because 0 is not a valid divisor. This needs to be explicitly checked.

Complexity Analysis

Time Complexity: O(n*k)

Space Complexity: O(1)

Topics

This problem involves: Math, String, Sliding Window.

Companies

Asked at: Postmates, Quora.