Helpful Maths - Implementation

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function helpful_maths(s)
digits := empty vector
for each character c in s
if c is a digit then
 add c to digits
 sort(digits.begin(), digits.end())
 print digits[0]
 for i from 1 to size of digits - 1
 print "+" followed by digits[i]
 print newline

The loop collects digits by checking if each character is between '0' and '9'. After sorting, you print the first digit without a plus sign, then loop through the rest adding plus signs before each digit. The time complexity is $O(n \log n)$ where $n$ is the number of digits.