Get the next biggest number with the same digits using Python

The challenge

Create a function that takes a positive integer and returns the next bigger number that can be formed by rearranging its digits. For example:

12 ==> 21
513 ==> 531
2017 ==> 2071
nextBigger(num: 12)   # returns 21
nextBigger(num: 513)  # returns 531
nextBigger(num: 2017) # returns 2071

If the digits can’t be rearranged to form a bigger number, return -1 (or nil in Swift):

9 ==> -1
111 ==> -1
531 ==> -1
nextBigger(num: 9)   # returns nil
nextBigger(num: 111) # returns nil
nextBigger(num: 531) # returns nil

Test cases

Test.assert_equals(next_bigger(12),21)
Test.assert_equals(next_bigger(513),531)
Test.assert_equals(next_bigger(2017),2071)
Test.assert_equals(next_bigger(414),441)
Test.assert_equals(next_bigger(144),414)

The solution in Python

def next_bigger(n):
    # create a list representation of the input integer
    arr = list(str(n))
    
    # sort in reverse to get the largest possible number
    max_n = int("".join(sorted(arr, reverse=True)))
    
    # sort to get the minimum number
    min_n = sorted(arr)
    
    # copy the input to a new variable
    m = n
    
    # loop while less than or equal to the max
    while m <= max_n:
        # increment our number
        m += 1
        # if found in our min list
        if sorted(list(str(m))) == min_n:
            # return the new number
            return m
        
    # if all else fails, return -1
    return -1

An alternative

def next_bigger(n):
    # if the number is the same as the reverse
    if str(n) == ''.join(sorted(str(n))[::-1]):
        # return -1 as it doesn't change
        return -1

    # keep a new temp variable
    a = n

    # loop forever
    while True:
        # increment the number
        a += 1

        # we have a match!
        if sorted(str(a)) == sorted(str(n)):
            # return the answer!
            return a
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