Find the K-Beauty of a Number - Practice Coding Problems

Find the K-Beauty of a Number - 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.

Input & Output

Example 1 — Basic Case

$ Input: num = 240, k = 2

› Output: 2

💡 Note: Substrings of length 2: "24" and "40". Both 24 and 40 divide 240 evenly, so k-beauty is 2.

Example 2 — With Leading Zero

$ Input: num = 430, k = 2

› Output: 2

💡 Note: Substrings: "43", "30". 430 ÷ 43 = 10 and 430 ÷ 30 = 14.33... Only "43" divides evenly, but "30" also works: 430 ÷ 30 = 14.33 (not integer). Actually 43 divides 430, and 30 doesn't divide 430 evenly.

Example 3 — No Valid Divisors

$ Input: num = 123, k = 3

› Output: 1

💡 Note: Only substring is "123". Since 123 ÷ 123 = 1, it divides itself, so k-beauty is 1.

Constraints

1 ≤ num ≤ 10⁹
1 ≤ k ≤ num.length

Visualization

Tap to expand

Understanding the Visualization

Input

Number 240 with k=2

Extract Substrings

Get all 2-length substrings: "24", "40"

Count Divisors

Count how many substrings divide the original number

Key Takeaway

🎯 Key Insight: Convert to string, slide window of size k, and count valid divisors

Asked in

a Amazon 15 G Google 12

The key insight is to convert the number to a string and use a sliding window to extract all k-length substrings, then test each for divisibility. The sliding window approach provides clean code structure. Time: O(n), Space: O(n) where n is the number of digits.

Common Approaches

Approach	Time	Space	Notes
✓ Sliding Window Optimization	O(n)	O(n)	Use sliding window technique to efficiently process each k-length substring
Brute Force with String Conversion	O(n)	O(n)	Convert number to string and check every k-length substring for divisibility

Sliding Window Optimization — Algorithm Steps

Step 1: Convert number to string representation
Step 2: Use sliding window of size k to get each substring
Step 3: Convert each substring to integer and test divisibility
Step 4: Return count of valid divisors

Visualization

Tap to expand

Step-by-Step Walkthrough

Initialize Window

Start with first k characters

Slide Window

Move window one position at a time

Test Each Window

Check divisibility for each window position

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int num, int k) {
    char s[20];
    sprintf(s, "%d", num);
    int n = strlen(s);
    int count = 0;
    
    // Sliding window of size k
    for (int i = 0; i <= n - k; i++) {
        char window[20];
        strncpy(window, s + i, k);
        window[k] = '\0';
        int divisor = atoi(window);
        
        // Check if it's a valid divisor
        if (divisor != 0 && num % divisor == 0) {
            count++;
        }
    }
    
    return count;
}

int main() {
    int num, k;
    scanf("%d %d", &num, &k);
    printf("%d\n", solution(num, k));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass with sliding window over n digits

✓ Linear Growth

Space Complexity

O(n)

String representation and temporary substring storage

⚡ Linearithmic Space

12.5K Views

Medium Frequency

~15 min Avg. Time

342 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Sliding Window Optimization — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler