LeetCode 2269: Find the K-Beauty of a Number

Explanation

To solve this problem, we can iterate through all possible substrings of length k of the given number num. For each substring, we check if it is a divisor of num. If it is a divisor, we increment a counter. Finally, we return the counter as the k-beauty of the number.

Time Complexity

The time complexity of this approach is O(n * k), where n is the length of the number num and k is the given length of the substrings.

Space Complexity

The space complexity is O(1) as we are not using any extra space that grows with the input size.

class Solution {
    public int beauty(int num, int k) {
        String numStr = String.valueOf(num);
        int count = 0;
        
        for (int i = 0; i <= numStr.length() - k; i++) {
            String sub = numStr.substring(i, i + k);
            int subNum = Integer.parseInt(sub);
            if (subNum != 0 && num % subNum == 0) {
                count++;
            }
        }
        
        return count;
    }
}