Welcome to Subscribe On Youtube

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:

• 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

• Java
• C++
• Python
• Go
• TypeScript

• 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;
  }
  
  

All Problems

All Solutions