# 2269. Find the K-Beauty of a Number

## Description

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:

• 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 (taking num as a string)

## Solutions

• class Solution {
public int divisorSubstrings(int num, int k) {
int ans = 0;
String s = "" + num;
for (int i = 0; i < s.length() - k + 1; ++i) {
int t = Integer.parseInt(s.substring(i, i + k));
if (t != 0 && num % t == 0) {
++ans;
}
}
return ans;
}
}

• class Solution {
public:
int divisorSubstrings(int num, int k) {
int ans = 0;
string s = to_string(num);
for (int i = 0; i < s.size() - k + 1; ++i) {
int t = stoi(s.substr(i, k));
ans += t && num % t == 0;
}
return ans;
}
};

• 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


• func divisorSubstrings(num int, k int) int {
ans := 0
s := strconv.Itoa(num)
for i := 0; i < len(s)-k+1; i++ {
t, _ := strconv.Atoi(s[i : i+k])
if t > 0 && num%t == 0 {
ans++
}
}
return ans
}

• function divisorSubstrings(num: number, k: number): number {
let ans = 0;
const s = num.toString();
for (let i = 0; i < s.length - k + 1; ++i) {
const t = parseInt(s.substring(i, i + k));
if (t !== 0 && num % t === 0) {
++ans;
}
}
return ans;
}