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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...

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).

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