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Formatted question description: https://leetcode.ca/all/2269.html
2269. Find the K-Beauty of a Number
- Difficulty: Easy.
- Related Topics: Math, String, Sliding Window.
- Similar Questions: .
Problem
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 "240": 24 is a divisor of 240.
- "40" from "240": 40 is a divisor of 240.
Therefore, the k-beauty 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 k-beauty 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 K-Beauty of a Number # https://leetcode.com/problems/find-the-k-beauty-of-a-number/ 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).