Here is a summary of the problem:
Example | Input String | Answer |
0 | "596" | "569" |
1 | "93561" | "13569" |
2 | "5491727514" | "1491727554" |
3 | "10234" | "10234" |
4 | "93218910471211292416" | "13218910471211292496" |
Enumerating all combinations of i and j takes roughly n^{2} operations, so this is not a good way to solve the problem in general. However, the topcoder constraints limit the string to 50 characters, and 50*50 is a pretty small number, so it works easily.
In tests.sh, you can do the first 58 problems in under a second. The 59th, and last problem is roughly 9600 characters, and took my program roughly 9 seconds to do. You can test your solution on the first 58 problems with:
UNIX> head -n 58 tests.sh | sh
What are you going to swap it with? Well, you want to maximize the effect of replacing the first digit, so you should swap it with the smallest digit that's not zero (because we are not allowed to start the number with a zero). What if there are multiple of these? Again, let's look at example 2: "5491727514". We want to swap the '5' with a '1' and there are two '1' digits. The two potential swaps yield "1495727514" and "1491727554". The second of these is smaller -- we want to swap with the rightmost '1'.
Of course, that was just proof-by-Cosmo. Can you prove to yourself that you want the rightmost, minimum non-zero digit? It is correct.
As a second example, look at example 4: "93218910471211292416". '9' is definitely bigger than some digits to its right, so you are going to swap it. There are a bunch of ones, so you will swap it with a one, because that is the smallest non-zero digit. Which '1'? The rightmost one. That's how you get the answer "13218910471211292496".
This gives us a strategy for solving the problem, but if we program it in the most natural way, the program still runs in n^{2} operations. The most "natural" way is to work from the definition: start i at zero and have it go to the end of the string. For each value of i, you look at each value to the right of i and find the minimum, rightmost value (when i equals zero, you exclude zero). If this value is less than the digit in i, then you swap those digits and return. If the value is greater than or equal to the digit in i, then you increment i and try again.
If you think about it, when your input is a non-decreasing sequence of digits, this technique still uses n^{2} operations. Why? Because at iteration i, you look at all of the digits greater than i, and there are n-i of those (where the number has n digits).
How can we fix this? By thinking clearly and organizing our code so that we are not doing unnecessary, nested for loops. Interestingly, I solved it in one way, and Allen McBride (who TA's the class many years ago) solved it another. I'll describe both. Allen's solution is better.
i | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
v[i] | -1 | 4 | -1 | 1 | -1 | 2 | 3 | -1 | -1 | 0 |
Now, you run through the string as before, but finding the minimum value to the right of digit i simply requires you to run through the vector, whose size is limited to ten elements. Instead of requiring n^{2} operations in the worst case, it is linear!
When I ran tests.sh on this solution, the entire thing ran in 0.16 seconds.
Let's give an example that is similar to example 3: "10423". You should be able to see that the answer will be "10243". Here are the variables as you run through the string from right to left:
i | min_digit | min_nonzero | lpos |
Start | '0'+10 | '0'+10 | -1 |
4 | '3' | '3' | -1 |
3 | '2' | '2' | -1 |
2 | '2' | '2' | 2 |
1 | '0' | '2' | 2 |
0 | '0' | '1' | 2 |
In that last iteration, we don't set lpos equal to zero because the digit must be less then min_nonzero when lpos equals zero.
Now, when you're done, you are going to swap the digit at index lpos. Obviously, if lpos equals -1, you simply return the original string. If lpos equals zero, you want to find the rightmost minimum, non-zero digit, and swap with that. Otherwise, you want to find the rightmost, minimum digit that is to the right of lpos, and swap with that. In the example above, we start at index 2 (whose digit is '4'), and find the rightmost, minimum digit to the right of it -- that's the '2', which is at index 3. Swap the '2' and the '4', and you have your answer.
This solution is better than the Plank solution, because it doesn't require you to traverse the entire vector, if the problem structure is good, and it doesn't depend on a vector of digits like the Plank solution. Very nice, Allen!
When I programmed this up, it also ran tests.sh in 0.16 seconds.
The Plank solution: I traverse the entire vector to create my digit vector v, whose size is d. So my solution is O(n + d). Why the distinction? Well, suppose I'm working in base 100, but my string has two digits. Then, the d term dominates.
The McBride solution: Determining min_digit and min_nonzero is O(m+z) -- that's the number of elements that you look at to find min_digit or min_nonzero, and as always, the addition operator in the big-O equation stands for "either-or, depending on which is bigger". Finding lpos is O(l). So the running time of the overall solution is O(m+z+l). You'll note, this can be O(n) in the worst case, but if I were to generate the input randomly, and n is large, then it would be O(d). Why? Because on average we would find the rightmost 0 and 1 within the last d characters, and lpos would be a very small number.