Question

2269. Find the K-Beauty of a Number

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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:

• 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 (taking num as a string)

Solutions

Solution 1

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

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;
  }
};

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;
}