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Formatted question description: https://leetcode.ca/all/2269.html
2269. Find the KBeauty of a Number
 Difficulty: Easy.
 Related Topics: Math, String, Sliding Window.
 Similar Questions: .
Problem
The kbeauty 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 kbeauty 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 "240": 24 is a divisor of 240.
 "40" from "240": 40 is a divisor of 240.
Therefore, the kbeauty is 2.
Example 2:
Input: num = 430043, k = 2
Output: 2
Explanation: The following are the substrings of num of length k:
 "43" from "430043": 43 is a divisor of 430043.
 "30" from "430043": 30 is not a divisor of 430043.
 "00" from "430043": 0 is not a divisor of 430043.
 "04" from "430043": 4 is not a divisor of 430043.
 "43" from "430043": 43 is a divisor of 430043.
Therefore, the kbeauty is 2.
Constraints:

1 <= num <= 109

1 <= k <= num.length
(takingnum
as a string)
Solution (Java, C++, Python)

class Solution { public int divisorSubstrings(int num, int k) { int i = 0; int j = 0; int count = 0; String s = String.valueOf(num); StringBuilder sb = new StringBuilder(); while (i < s.length() && j < s.length()) { sb.append(s.charAt(j)  '0'); int val = Integer.parseInt(sb.toString()); if (j  i + 1 == k) { if (val != 0 && num % val == 0) { count++; } sb.deleteCharAt(0); i++; j++; } else { j++; } } return count; } }

Todo

class Solution: def divisorSubstrings(self, num: int, k: int) > int: ans = 0 s = str(num) for i in range(len(s)  k + 1): t = int(s[i : i + k]) if t and num % t == 0: ans += 1 return ans ############ # 2269. Find the KBeauty of a Number # https://leetcode.com/problems/findthekbeautyofanumber/ class Solution: def divisorSubstrings(self, nums: int, k: int) > int: num = str(nums) n = len(num) res = 0 for i in range(n  k + 1): x = int(num[i : i + k]) if x != 0 and nums % x == 0: res += 1 return res
Explain:
nope.
Complexity:
 Time complexity : O(n).
 Space complexity : O(n).