next_number.py

'''
    Quick and dirty code to see if I could figure out the next
    larger number with all the same digits of the number passed.
    
    Note: After coming up with the solution I found the following
    
    http://stackoverflow.com/questions/9368205/given-a-number-find-the-next-higher-number-which-has-the-exact-same-set-of-digi
'''
def get_next(the_num):
    the_num = str(the_num)
    last = -1
    for x in xrange(-1, -len(the_num)-1, -1):
        if last > the_num[x]:
            # x is the index of the number that needs to be moved
            left = the_num[:x]
            mid = the_num[x]
            right = sorted(the_num[x+1:])

            found = -1
            for x in xrange(len(right)):
                if int(right[x]) > int(mid):
                    found = x
                    break
            if found != -1:
                the_num = list(left) + [right[found]]
                right[found] = mid
                the_num += sorted(right)
            break
        last = the_num[x]

    return int(''.join(the_num))


val = 23468
last = -1
while(last != val):
    print val
    last = val
    val = get_next(val)
print val

# S/O Numbers:
# print 123456784987654321
# print get_next(123456784987654321)