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# Your Problem: # Given a non-negative int n, compute recursively (no loops) the count of the # occurrences of 8 as a digit, except that an 8 with another 8 immediately to # its left counts double, so 8818 yields 4. Note that mod (%) by 10 yields # the rightmost digit (126 % 10 is 6), while divide (/) by 10 removes the # rightmost digit (126 / 10 is 12). # # # count8(8) → 1 # count8(818) → 2 # count8(8818) → 4 # # Credit for this problem goes to Nick Parlante from CodingBat.com # Do not modify anything outside of the function or the function itself. # CODE START def getCount8(num:int) -> int: # CODE END # Press Ctrl+E (Command+E on mac) to test your code