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 $O(\log num + k)$.

Python3JavaC++GoTypeScript

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;
}

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 $O(1)$.

Python3JavaC++GoTypeScript

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

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;
    }
};

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;
}

Comments