Given a number, find the next higher number which has the exact same set of digits as the original number
You can do it in O(n) (where n is the number of digits) like this:
Starting from the right, you find the first pair-of-digits such that the left-digit is smaller than the right-digit. Let's refer to the left-digit by "digit-x". Find the smallest number larger than digit-x to the right of digit-x, and place it immediately left of digit-x. Finally, sort the remaining digits in ascending order - since they were already in descending order, all you need to do is reverse them (save for digit-x, which can be placed in the correct place in O(n)).
An example will make this more clear:
123456784987654321
start with a number
123456784 987654321
^the first place from the right where the left-digit is less than the right
Digit "x" is 4
123456784 987654321
^find the smallest digit larger than 4 to the right
123456785 4 98764321
^place it to the left of 4
123456785 4 12346789
123456785123446789
^sort the digits to the right of 5. Since all of them except
the '4' were already in descending order, all we need to do is
reverse their order, and find the correct place for the '4'
Proof of correctness:
Let's use capital letters to define digit-strings and lower-case for digits. The syntax AB means "the concatenation of strings A and B". < is lexicographical ordering, which is the same as integer ordering when the digit-strings are of equal length.
Our original number N is of the form AxB, where x is a single digit and B is sorted descending.
The number found by our algorithm is AyC, where y ∈ B is the smallest digit > x (it must exist due to the way x was chosen, see above), and C is sorted ascending.
Assume there is some number (using the same digits) N' such that AxB < N' < AyC. N' must begin with A or else it could not fall between them, so we can write it in the form AzD. Now our inequality is AxB < AzD < AyC, which is equivalent to xB < zD < yC where all three digit-strings contain the same digits.
In order for that to be true, we must have x <= z <= y. Since y is the smallest digit > x, z cannot be between them, so either z = x or z = y. Say z = x. Then our inequality is xB < xD < yC, which means B < D where both B and D have the same digits. However, B is sorted descending, so there is no string with those digits larger than it. Thus we cannot have B < D. Following the same steps, we see that if z = y, we cannot have D < C.
Therefore N' cannot exist, which means our algorithm correctly finds the next largest number.
An almost-identical problem appeared as a Code Jam problem and has a solution here:
http://code.google.com/codejam/contest/dashboard?c=186264#s=a&a=1
Here's a summary of the method using an example:
34722641
A. Split the sequence of digits in two, so that the right part is as long as possible while remaining in decreasing order:
34722 641
(If the entire number is in decreasing order, there's no bigger number to be made without adding digits.)
B.1. Select the last digit of the first sequence:
3472(2) 641
B.2. Find the smallest digit in the second sequence that is larger than it:
3472(2) 6(4)1
B.3. Swap them:
3472(2) 6(4)1
->
3472(4) 6(2)1
->
34724 621
C. Sort the second sequence into increasing order:
34724 126
D. Done!
34724126
Here's a compact (but partly brute force) solution in Python
def findnext(ii): return min(v for v in (int("".join(x)) for x in
itertools.permutations(str(ii))) if v>ii)
In C++ you could make the permutations like this: https://stackoverflow.com/a/9243091/1149664 (It's the same algorithm as the one in itertools)
Here's an implementation of the top answer described by Weeble and BlueRaja, (other answers). I doubt there's anything better.
def findnext(ii):
iis=list(map(int,str(ii)))
for i in reversed(range(len(iis))):
if i == 0: return ii
if iis[i] > iis[i-1] :
break
left,right=iis[:i],iis[i:]
for k in reversed(range(len(right))):
if right[k]>left[-1]:
right[k],left[-1]=left[-1],right[k]
break
return int("".join(map(str,(left+sorted(right)))))