Find the KBeauty of a Number
Implementation


Explanation
 For Input:
num = 430043, k = 2
 Substrings of Length 2: “43”, “30”, “00”, “04”, “43”
 Valid Divisors: “43”, “43” (only these two are nonzero and divide 430043 evenly)
 Output:
2
By counting each valid divisor individually, we address the issue with the original implementation and obtain the expected result. The complexity of the code remains the same, with a time complexity of ( O(nk) ) and a space complexity of ( O(1) ), as we are only using a constant amount of additional space.