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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

Solution 1: Enumeration

We can convert $num$ to a string $s$, then enumerate all substrings of $s$ with length $k$, convert them to an integer $t$, and check if $t$ is divisible by $num$. If it is, we increment the answer.

The time complexity is $O(\log num \times k)$, and the space complexity is $O(\log num + k)$.

Solution 2: Sliding Window

We can maintain a sliding window of length $k$. Initially, the window contains the lowest $k$ digits of $num$. Then, for each iteration, we move the window one digit to the right, update the number in the window, and check if the number in the window is divisible by $num$. If it is, we increment the answer.

The time complexity is $O(\log num)$, and the space complexity is $O(1)$.

• Java
• C++
• Python
• Go
• TypeScript
• Java 2
• C++ 2
• Python 2
• Go 2
• TypeScript 2

• 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;
      }
  }
  
  
  // Solution 2
  class Solution {
      public int divisorSubstrings(int num, int k) {
          int x = 0, p = 1;
          int t = num;
          for (; k > 0; --k) {
              int v = t % 10;
              t /= 10;
              x = p * v + x;
              p *= 10;
          }
          int ans = x != 0 && num % x == 0 ? 1 : 0;
          for (p /= 10; t > 0; t /= 10) {
              x /= 10;
              int v = t % 10;
              x = p * v + x;
              ans += (x != 0 && num % x == 0 ? 1 : 0);
          }
          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;
      }
  };
  
  
  // Solution 2
  class Solution {
  public:
      int divisorSubstrings(int num, int k) {
          int x = 0;
          long long p = 1;
          int t = num;
          for (; k > 0; --k) {
              int v = t % 10;
              t /= 10;
              x = p * v + x;
              p *= 10;
          }
          int ans = x != 0 && num % x == 0 ? 1 : 0;
          for (p /= 10; t > 0; t /= 10) {
              x /= 10;
              int v = t % 10;
              x = p * v + x;
              ans += (x != 0 && num % x == 0 ? 1 : 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
  
  
  # Solution 2
  class Solution:
      def divisorSubstrings(self, num: int, k: int) -> int:
          x, p = 0, 1
          t = num
          for _ in range(k):
              t, v = divmod(t, 10)
              x = p * v + x
              p *= 10
          ans = int(x != 0 and num % x == 0)
          p //= 10
          while t:
              x //= 10
              t, v = divmod(t, 10)
              x = p * v + x
              ans += int(x != 0 and num % x == 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
  }
  
  
  // Solution 2
  func divisorSubstrings(num int, k int) (ans int) {
  	x, p, t := 0, 1, num
  	for ; k > 0; k-- {
  		v := t % 10
  		t /= 10
  		x = p*v + x
  		p *= 10
  	}
  	if x != 0 && num%x == 0 {
  		ans++
  	}
  	for p /= 10; t > 0; t /= 10 {
  		x /= 10
  		v := t % 10
  		x = p*v + x
  		if x != 0 && num%x == 0 {
  			ans++
  		}
  	}
  	return
  }
  
  

• 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;
  }
  
  
  // Solution 2
  function divisorSubstrings(num: number, k: number): number {
      let [x, p, t] = [0, 1, num];
      for (; k > 0; k--) {
          const v = t % 10;
          t = Math.floor(t / 10);
          x = p * v + x;
          p *= 10;
      }
      let ans = x !== 0 && num % x === 0 ? 1 : 0;
      for (p = Math.floor(p / 10); t > 0; t = Math.floor(t / 10)) {
          x = Math.floor(x / 10);
          x = p * (t % 10) + x;
          ans += x !== 0 && num % x === 0 ? 1 : 0;
      }
      return ans;
  }
  
  

• class Solution {
      public int divisorSubstrings(int num, int k) {
          int x = 0, p = 1;
          int t = num;
          for (; k > 0; --k) {
              int v = t % 10;
              t /= 10;
              x = p * v + x;
              p *= 10;
          }
          int ans = x != 0 && num % x == 0 ? 1 : 0;
          for (p /= 10; t > 0; t /= 10) {
              x /= 10;
              int v = t % 10;
              x = p * v + x;
              ans += (x != 0 && num % x == 0 ? 1 : 0);
          }
          return ans;
      }
  }
  

• class Solution {
  public:
      int divisorSubstrings(int num, int k) {
          int x = 0;
          long long p = 1;
          int t = num;
          for (; k > 0; --k) {
              int v = t % 10;
              t /= 10;
              x = p * v + x;
              p *= 10;
          }
          int ans = x != 0 && num % x == 0 ? 1 : 0;
          for (p /= 10; t > 0; t /= 10) {
              x /= 10;
              int v = t % 10;
              x = p * v + x;
              ans += (x != 0 && num % x == 0 ? 1 : 0);
          }
          return ans;
      }
  };
  

• class Solution:
      def divisorSubstrings(self, num: int, k: int) -> int:
          x, p = 0, 1
          t = num
          for _ in range(k):
              t, v = divmod(t, 10)
              x = p * v + x
              p *= 10
          ans = int(x != 0 and num % x == 0)
          p //= 10
          while t:
              x //= 10
              t, v = divmod(t, 10)
              x = p * v + x
              ans += int(x != 0 and num % x == 0)
          return ans
  
  

• func divisorSubstrings(num int, k int) (ans int) {
  	x, p, t := 0, 1, num
  	for ; k > 0; k-- {
  		v := t % 10
  		t /= 10
  		x = p*v + x
  		p *= 10
  	}
  	if x != 0 && num%x == 0 {
  		ans++
  	}
  	for p /= 10; t > 0; t /= 10 {
  		x /= 10
  		v := t % 10
  		x = p*v + x
  		if x != 0 && num%x == 0 {
  			ans++
  		}
  	}
  	return
  }
  

• function divisorSubstrings(num: number, k: number): number {
      let [x, p, t] = [0, 1, num];
      for (; k > 0; k--) {
          const v = t % 10;
          t = Math.floor(t / 10);
          x = p * v + x;
          p *= 10;
      }
      let ans = x !== 0 && num % x === 0 ? 1 : 0;
      for (p = Math.floor(p / 10); t > 0; t = Math.floor(t / 10)) {
          x = Math.floor(x / 10);
          x = p * (t % 10) + x;
          ans += x !== 0 && num % x === 0 ? 1 : 0;
      }
      return ans;
  }
  
  

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