Write a recursive function named mergeDigitPairs that accepts an integer parameter n and returns the integer formed by combining each pair of digits from n into a single digit that is their sum.
For example, if passed the number 1234, you should combine the digits 12 into 1+2 or 3, and combine the digits 34 into 3+4 or 7, leading to a returned result of 37.
If adding a given pair of digits produces a two-digit number, repeat the process until you have a single-digit number to replace the original pair.
For example, if passed the number 1168, the 11 becomes 1+1 or 2, but the 68 becomes 6+8 or 14, so we merge them again by saying that 14 is 1+4 or 5, so the pair 68 turns into 5, leading to an overall result of 25.
If passed a number with an odd number of digits, the first (most significant) digit is left untouched.
For example, the number 13372 becomes 169 because the 3+3 becomes 6 and the 7+2 becomes 9.
If passed a negative number, perform the same process as usual but return a negative result.
For example, when passed -1234, return -37.
If passed a single-digit number, simply return that number itself.
The following table shows several calls and their expected return values:
| Call |
Return Value |
mergeDigitPairs(1234) |
37 |
mergeDigitPairs(3186507) |
3927 |
mergeDigitPairs(-52874) |
-512 |
mergeDigitPairs(88888888) |
7777 |
mergeDigitPairs(20581974) |
2412 |
mergeDigitPairs(0) |
0 |
mergeDigitPairs(6) |
6 |
mergeDigitPairs(6) |
-5 |
Constraints:
- Do not use any loops; you must use recursion.
- Do not declare any global variables.
- Do not use arrays or a string to solve this problem (for example, do not convert
n
into a string).
- You may define other "helper" functions if you like; they are subject to these same constraints.
Note: Because JavaScript integers are number types, dividing two integers which do not
divide evenly returns the decimal value of the result instead of an integer (for example, 1 / 2 results in 0.5, not 0).
Use parseInt(a / b) to parse the result into an integer if needed to solve this problem.
