One [topic category]

sliding window

Two [question difficulty]

Simple

Three [topic number]

2269. Find the K beauty value of a number

Four [title description]

The k beauty value of an integer num is defined as the number of substrings in num that satisfy the following conditions:
- The substring is of length k.
- The substring divides num evenly.
Given the integers num and k, please return the k value of num.
Notice:
- Prefix 0 is allowed.
- 0 is not divisible by any value.
A substring is a contiguous sequence of characters in a string.

Five [topic example]

Example 1:
- Input: num = 240, k = 2
- Output: 2
- Explanation: The following are substrings of length k in num:
  - "24" in "240": 24 can divide 240 evenly.
  - "40" in "240": 40 can divide 240 evenly.
- Therefore, the value of kbeauty is 2.
Example 2:
- Input: num = 430043, k = 2
- Output: 2
- Explanation: The following are substrings of length k in num:
  - "43" in "430043": 43 can divide 430043 evenly.
  - "30" in "430043": 30 is not divisible by 430043.
  - "00" in "430043": 0 is not divisible by 430043.
  - "04" in "430043": 4 is not divisible by 430043.
  - "43" in "430043": 43 can divide 430043 evenly.
- Therefore, the value of kbeauty is 2.

Six [topic prompt]

$1 <= num <= 10^9$
$1 <= k <= n u m . l e n g t h (treat n u m as a string)$

Seven [problem-solving ideas]

Using the idea of sliding windows
The fixed size of this window is the k given by the question, and it slides from left to right. If the number in this window is not 0 and can divide the given number, then it will be recorded in the result
You need to be careful not to cross the boundary. The idea is not difficult. If you don’t use library functions, the specific implementation is a little more complicated. You can see the following code for details, which is written in great detail.
Finally return the result

Eight 【Time Frequency】

Time complexity: $O (n)$ ， $n$ is the length of the incoming number
Space complexity: $O (1)$

Nine [code implementation]

Java language version

class Solution {
    
    
    public int divisorSubstrings(int num, int k) {
    
    
        int res = 0;
        int[] temp_num = new int[10];
        int i,count,temp;
        for(i = 0,temp = num;temp > 0;temp /= 10,i++){
    
    
            temp_num[i] = temp % 10;
        }
        count = i - k + 1;
        for(i = 0;i < count;i++){
    
    
            int sum = 0;
            for(int j = 0;j<k;j++){
    
    
                sum *= 10;
                sum += temp_num[i + k - j - 1];
            }
            if(sum != 0 && num % sum == 0){
    
    
                res++;
            }
        }
        return res;
    }
}

C language version

int divisorSubstrings(int num, int k)
{
    
    
    int res = 0;
    int* temp_num = (int*)malloc(sizeof(int) * 10);
    int i,count,temp;
    for(i = 0,temp = num;temp > 0;temp /= 10,i++)
    {
    
    
        temp_num[i] = temp % 10;
    }
    count = i - k + 1;
    for(i = 0;i < count;i++)
    {
    
    
        int sum = 0;
        for(int j = 0;j < k;j++)
        {
    
    
            sum *= 10;
            sum += temp_num[i + k - j - 1];
        }
        if(sum != 0 && num % sum == 0)
        {
    
    
            res++;
        }
    }
    return res;
}

Python language version

class Solution:
    def divisorSubstrings(self, num: int, k: int) -> int:
        res = 0
        num = str(num)
        for i in range(0,len(num) - k + 1):
            temp = int(num[i:i+k])
            if temp == 0:
                continue
            if int(num) % temp == 0:
                res += 1
        return res;

C++ language version

class Solution {
    
    
public:
    int divisorSubstrings(int num, int k) {
    
    
        int res = 0;
        int i,count,temp;
        int* temp_count = (int*)malloc(sizeof(int) * 10);
        for(i = 0,temp = num;temp > 0;temp /= 10,i++){
    
    
            temp_count[i] = temp % 10;          
        }
        count = i - k + 1;
        for(i = 0;i < count;i++){
    
    
            int sum = 0;
            for(int j = 0;j < k;j++){
    
    
                sum *= 10;
                sum += temp_count[i + k - j - 1];
            }
            if(sum != 0 && num % sum == 0){
    
    
                res++;
            }
        }
        return res;
    }
};