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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 (taking num 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;
    }
}

############

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:
    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


  • 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;
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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