I'm confused. My solution followed the definition of frugal numbers from https://en.wikipedia.org/wiki/Frugal_number, which includes counting the number of digits in the exponents, and also checks frugality in other bases (not just base 10). But most solutions don't follow this. Can you clarify this?
It seems that the problem is not considering the exponent 1 as a valid digit, For instance, 115248 (6 digits) is called a frugal number in this problem since 115248 (6 digits) > 2^4*3*7^4 (5 digits). If we considered 3^1 as 2 digits instead of the 1-digit 3, 115248 should not be frugal. And the problem should have probably declared this restriction and that the considered base is 10.
PS: The OEIS also imposes this same restriction https://oeis.org/A046759, since 3645 is considered frugal: 3^6*5 instead of 3^6*5^1.
I recommend including the number 414720 to the test suite: 3^4*2^10*5.
Back to basics 9 - Indexed References
Matrix with different incremental runs
Sum the numbers on the main diagonal
Binary code (array)
generate nth pentatope number
find nth even fibonacci number
ZigZag - 01
Last non-zero digit
Hanging cable - 02
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