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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 where is the number of digits.