Let's assume we will consider binary numbers which has length 2n and n might be about 1000. We are looking for kth number (k is limited by 10^9) which has following properties:
Amount of 1's is equal to amount of 0's what can be described as following: #(1) = #(0)
Every prefix of this number has to contain atleast as much 0's as 1's. It might be easier to understand it after negating the sentence, which is: There is no prefix which would contain more 1's than 0's.
And basically that's it.
So to make it clear let's do some example:
n=2, k=2
we have to take binary number of length 2n:
0000
0001
0010
0011
0100
0101
0110
0111
1000
and so on...
And now we have to find 2nd number which fulfill those two requirements. So we see 0011 is the first one, and 0101 is second one.
If we change k=3, then answer doesn't exist since there are number which have same amount of opposite bits, but for 0110, there is prefix 011 so number doesn't fulfill second constraint and same would be with all numbers which has 1 as most significant bit.
So what I did so far to find algorithm?
Well my first idea was to generate all possible bits settings, and check whether it has those two properties, but generate them all would take O(2^(2n)) which is not an option for n=1000.
Additionally I realize there is no need to check all numbers which are smaller than 0011 for n=2, 000111 for n=3, and so on... frankly speaking those which half of most significant bits remains "untouched" because those numbers have no possibility to fulfill #(1) = #(0) condition. Using that I can reduce n by half, but it doesn't help much. Instead of 2 * forever I have forever running algorithm. It's still O(2^n) complexity, which is way too big.
Any idea for algorithm?
Conclusion
This text has been created as a result of my thoughts after reading Andy Jones post.
First of all I wouldn't post code I have used since it's point 6 in following document from Andy's post Kasa 2009. All you have to do is consider nr as that what I described as k. Unranking Dyck words algorithm, would help us find out answer much faster. However it has one bottleneck.
while (k >= C(n-i,j))
Considering that n <= 1000, Catalan number can be quite huge, even C(999,999). We can use some big number arithmetic, but on the other hand I came up with little trick to overpass it and use standard integer.
We don't want to know how big actually Catalan number is as long as it's bigger than k. So now we will create Catalan numbers caching partial sums in n x n table.
... ...
5 | 42 ...
4 | 14 42 ...
3 | 5 14 28 ...
2 | 2 5 9 14 ...
1 | 1 2 3 4 5 ...
0 | 1 1 1 1 1 1 ...
---------------------------------- ...
0 1 2 3 4 5 ...
To generate it is quite trivial:
C(x,0) = 1
C(x,y) = C(x,y-1) + C(x-1,y) where y > 0 && y < x
C(x,y) = C(x,y-1) where x == y
So what we can see only this:
C(x,y) = C(x,y-1) + C(x-1,y) where y > 0 && y < x
can cause overflow.
Let's stop at this point and provide definition.
k-flow - it's not real overflow of integer but rather information that value of C(x,y) is bigger than k.
My idea is to check after each running of above formula whether C(x,y) is grater than k or any of sum components is -1. If it is we put -1 instead, which would act as a marker, that k-flow has happened. I guess it quite obvious that if k-flow number is sum up with any positive number it's still be k-flowed in particular sum of 2 k-flowed numbers is k-flowed.
The last what we have to prove is that there is no possibility to create real overflow. Real overflow might only happen if we sum up a + b which non of them is k-flowed but as sum they generated the real overflow.
Of course it's impossible since maximum value can be described as a + b <= 2 * k <= 2*10^9 <= 2,147,483,647 where last value in this inequality is value of int with sign. I assume also that int has 32 bits, as in my case.
The numbers you are describing correspond to Dyck words. Pt 2 of Kasa 2009 gives a simple algorithm for enumerating them in lexicographic order. Its references should be helpful if you want to do any further reading.
As an aside (and be warned I'm half asleep as I write this, so it might be wrong), the wikipedia article notes that the number of Dyck words of length 2n is the n th Catalan number, C(n). You might want to find the smallest n such that C(n) is larger than the k you're looking for, and then enumerate Dyck words starting from X^n Y^n.
I'm sorry for misunderstood this problem last time, so I edit it and now I can promise the correction and you can test the code first, the complexity is O(n^2), the detail answer is follow
First, we can equal the problem to the next one
We are looking for kth largest number (k is limited by 10^9) which has following properties:
Amount of 1's is equal to amount of 0's what can be described as following: #(1) = #(0)
Every prefix of this number has to contain at least as much [[1's as 0's]], which means: There is no prefix which would contain more [[0's than 1's]].
Let's give an example to explain it: let n=3 and k=4, the amount of satisfied numbers is 5, and the picture below has explain what we should determine in previous problem and new problem:
| 000111 ------> 111000 ^
| 001011 ------> 110100 |
| 001101 ------> 110010 |
| previous 4th number 010011 ------> 101100 new 4th largest number |
v 010101 ------> 101010 |
so after we solve the new problem, we just need to bitwise not.
Now the main problem is how to solve the new problem. First, let A be the array, so A[m]{1<=m<=2n} only can be 1 or 0, let DP[v][q] be the amount of numbers which satisfy condition2 and condition #(1)=q in {A[2n-v+1]~A[2n]}, so the DP[2n][n] is the amount of satisfied numbers.
A[1] only can be 1 or 0, if A[1]=1, the amount of numbers is DP[2n-1][n-1], if A[1]=0, the amount of numbers is DP[2n-1][n], now we want to find the kth largest number, if k<=DP[2n-1][n-1], kth largest number's A[1] must be 1, then we can judge A[2] with DP[2n-2][n-2]; if k>DP[2n-1][n-1], kth largest number's A[1] must be 0 and k=k-DP[2n-1][n-1], then we can judge A[2] with DP[2n-2][n-1]. So with the same theory, we can judge A[j] one by one until there is no number to compare. Now we can give a example to understand (n=3, k=4)
(We use dynamic programming to determine DP matrix, the DP equation is DP[v][q]=DP[v-1][q-1]+DP[v-1][q])
Intention: we need the number in leftest row can be compared,
so we add a row on DP's left row, but it's not include by DP matrix
in the row, all the number is 1.
the number include by bracket are initialized by ourselves
the theory of initialize just follow the mean of DP matrix
DP matrix = (1) (0) (0) (0) 4<=DP[5][2]=5 --> A[1]=1
(1) (1) (0) (0) 4>DP[4][1]=3 --> A[2]=0, k=4-3=1
(1) (2) (0) (0) 1<=DP[3][1]=3 --> A[3]=1
(1) (3) 2 (0) 1<=1 --> a[4]=1
(1) (4) 5 (0) no number to compare, A[5]~A[6]=0
(1) (5) 9 5 so the number is 101100
If you have not understand clearly, you can use the code to understand
Intention：DP[2n][n] increase very fast, so the code can only work when n<=19, in the problem n<1000, so you can use big number programming, and the code can be optimize by bit operation, so the code is just a reference
/*--------------------------------------------------
Environment: X86 Ubuntu GCC
Author: Cong Yu
Blog: aimager.com
Mail: funcemail#gmail.com
Build_Date: Mon Dec 16 21:52:49 CST 2013
Function:
--------------------------------------------------*/
#include <stdio.h>
int DP[2000][1000];
// kth is the result
int kth[1000];
void Oper(int n, int k){
int i,j,h;
// temp is the compare number
// jishu is the
int temp,jishu=0;
// initialize
for(i=1;i<=2*n;i++)
DP[i-1][0]=i-1;
for(j=2;j<=n;j++)
for(i=1;i<=2*j-1;i++)
DP[i-1][j-1]=0;
for(i=1;i<=2*n;i++)
kth[i-1]=0;
// operate DP matrix with dynamic programming
for(j=2;j<=n;j++)
for(i=2*j;i<=2*n;i++)
DP[i-1][j-1]=DP[i-2][j-2]+DP[i-2][j-1];
// the main thought
if(k>DP[2*n-1][n-1])
printf("nothing\n");
else{
i=2*n;
j=n;
for(;j>=1;i--,jishu++){
if(j==1)
temp=1;
else
temp=DP[i-2][j-2];
if(k<=temp){
kth[jishu]=1;
j--;
}
else{
kth[jishu]=0;
if(j==1)
k-=1;
else
k-=DP[i-2][j-2];
}
}
for(i=1;i<=2*n;i++){
kth[i-1]=1-kth[i-1];
printf("%d",kth[i-1]);
}
printf("\n");
}
}
int main(){
int n,k;
scanf("%d",&n);
scanf("%d",&k);
Oper(n,k);
return 0;
}
Related
Let us call a number "steady" if sum of digits on odd positions is equal to sum of digits on even positions. For example 132 or 4059. Given a number N, program should output smallest/first "steady" number greater than N. For example if N = 4, answer = 11, if N = 123123, answer = 123134.
But the constraint is that N can be very large. Number of digits in N can be 100. And time limit is 1 second.
My approach was to take in N as a string store each digit in array of int type and add 1 using long arithmetic, than test if the number is steady or not, if Yes output it, if No add 1 again and test if it is steady. Do this until you get the answer.
It works on many tests, but when the difference between oddSum and EvenSum is very large like in 9090909090 program exceeds time limit. I could not come up with other algorithm. Intuitively I think there might be some pattern in swapping several last digits with each other and if necessary add or subtract something to them, but I don't know. I prefer a good HINT instead of answer, because I want to do it myself.
Use the algorithm that you would use. It goes like this:
Input: 9090909090
Input: 9090909090 Odd:0 Even:45
Input: 909090909? Odd:0 Even:45
Clearly no digit will work, we can make the odd at most 9
Input: 90909090?? Odd:0 Even:36
Clearly no digit will work, we removed a 9 and there is no larger digit (we have to make the number larger)
Input: 9090909??? Odd:0 Even:36
Clearly no digit will work. Even is bigger than odd, we can only raise odd to 18
Input: 909090???? Odd:0 Even:27
Clearly no digit will work, we removed a 9
Input: 90909????? Odd:0 Even:27
Perhaps a 9 will work.
Input: 909099???? Odd:9 Even:27
Zero is the smallest number that might work
Input: 9090990??? Odd:9 Even:27
We need 18 more and only have two digits, so 9 is the smallest number that can work
Input: 90909909?? Odd:18 Even:27
Zero is the smallest number that can work.
Input: 909099090? Odd:18 Even:27
9 is the only number that can work
Input: 9090990909 Odd:27 Even:27
Success
Do you see the method? Remove digits while a solution is impossible then add them back until you have the solution. At first, remove digits until a solution is possible. Only a number than the one you removed can be used. Then add numbers back using the smallest one possible at each stage until you have the solution.
You can try Digit DP technique .
Your parameter can be recur(pos,oddsum,evensum,str)
your state transitions will be like this :
bool ans=0
for(int i=0;i<10;i++)
{
ans|=recur(pos+1,oddsum+(pos%2?i:0),evensum+(pos%2?i:0),str+(i+'0')
if(ans) return 1;
}
Base case :
if(pos>=n) return oddsum==evensum;
Memorization: You only need to save pos,oddsum,evensum in your DP array. So your DP array will be DP[100][100*10][100*10]. This is 10^8 and will cause MLE, you have to prune some memory.
As oddsum+evensum<9*100 , we can have only one parameter SUM and add / subtract when odd/even . So our new recursion will look like this : recur(pos,sum,str)
state transitions will be like this :
bool ans=0
for(int i=0;i<10;i++)
{
ans|=recur(pos+1,SUM+(pos%2?i:-i),str+(i+'0')
if(ans) return 1;
}
Base case :
if(pos>=n) return SUM==0;
Memorization: now our Dp array will be 2d having [pos][sum] . we can say DP[100][10*100]
Find the parity with the smaller sum. Starting from the smallest digit of that parity, increase digits of that parity to the min of 9 and the remaining increase needed.
This gets you a larger steady number, but it may be too big.
E.g., 107 gets us 187, but 110 would do.
Next, repeatedly decrement the value of the nonzero digit in the largest position of each parity in our steady number where doing so doesn't reduce us below our target.
187,176,165,154,143,132,121,110
This last step as written is linear in the number of decrements. That's fast enough since there are at most 9*digits of them, but it can be optimized.
I want to develop a way to be able to represent all combinations of b bits with k bits set (equal to 1). It needs to be a way that given an index, can get quickly the binary sequence related, and the other way around too. For instance, the tradicional approach which I thought would be to generate the numbers in order, like:
For b=4 and k=2:
0- 0011
1- 0101
2- 0110
3- 1001
4-1010
5-1100
If I am given the sequence '1010', I want to be able to quickly generate the number 4 as a response, and if I give the number 4, I want to be able to quickly generate the sequence '1010'. However I can't figure out a way to do these things without having to generate all the sequences that come before (or after).
It is not necessary to generate the sequences in that order, you could do 0-1001, 1-0110, 2-0011 and so on, but there has to be no repetition between 0 and the (combination of b choose k) - 1 and all sequences have to be represented.
How would you approach this? Is there a better algorithm than the one I'm using?
pkpnd's suggestion is on the right track, essentially process one digit at a time and if it's a 1, count the number of options that exist below it via standard combinatorics.
nCr() can be replaced by a table precomputation requiring O(n^2) storage/time. There may be another property you can exploit to reduce the number of nCr's you need to store by leveraging the absorption property along with the standard recursive formula.
Even with 1000's of bits, that table shouldn't be intractably large. Storing the answer also shouldn't be too bad, as 2^1000 is ~300 digits. If you meant hundreds of thousands, then that would be a different question. :)
import math
def nCr(n,r):
return math.factorial(n) // math.factorial(r) // math.factorial(n-r)
def get_index(value):
b = len(value)
k = sum(c == '1' for c in value)
count = 0
for digit in value:
b -= 1
if digit == '1':
if b >= k:
count += nCr(b, k)
k -= 1
return count
print(get_index('0011')) # 0
print(get_index('0101')) # 1
print(get_index('0110')) # 2
print(get_index('1001')) # 3
print(get_index('1010')) # 4
print(get_index('1100')) # 5
Nice question, btw.
Consider we have a sacks of gold and thief wants to get the maximum gold. Thief can take the gold to get maximum by,
1) Taking the Gold from contiguous sacks.
2) Thief should take the same amount of gold from all sacks.
N Sacks 1 <= N <= 1000
M quantity of Gold 0 <= M <= 100
Sample Input1:
3 0 5 4 4 4
Output:
16
Explanation:
4 is the minimum amount he can take from the sacks 3 to 6 to get the maximum value of 16.
Sample Input2:
2 4 3 2 1
Output:
8
Explanation:
2 is the minimum amount he can take from the sacks 1 to 4 to get the maximum value of 8.
I approached the problem using subtracting the values from array and taking the transition point from negative to positive, but this doesn't solves the problem.
EDIT: code provided by OP to find the index:
int temp[6];
for(i=1;i<6;i++){
for(j=i-1; j>=0;j--) {
temp[j] = a[j] - a[i];
}
}
for(i=0;i<6;i++){
if(temp[i]>=0) {
index =i;
break;
}
}
The best amount of gold (TBAG) taken from every sack is equal to weight of some sack. Let's put indexes of candidates in a stack in order.
When we meet heavier weight (than stack contains), it definitely continues "good sequence", so we just add its index to the stack.
When we meet lighter weight (than stack top), it breaks some "good sequences" and we can remove heavier candidates from the stack - they will not have chance to be TBAG later. Remove stack top until lighter weight is met, calculate potentially stolen sum during this process.
Note that stack always contains indexes of strictly increasing sequence of weights, so we don't need to consider items before index at the stack top (intermediate AG) in calculation of stolen sum (they will be considered later with another AG value).
for idx in Range(Sacks):
while (not Stack.Empty) and (Sacks[Stack.Peek] >= Sacks[idx]): //smaller sack is met
AG = Sacks[Stack.Pop]
if Stack.Empty then
firstidx = 0
else
firstidx = Stack.Peek + 1
//range_length * smallest_weight_in_range
BestSUM = MaxValue(BestSUM, AG * (idx - firstidx))
Stack.Push(idx)
now check the rest:
repeat while loop without >= condition
Every item is pushed and popped once, so linear time and space complexity.
P.S. I feel that I've ever seen this problem in another formulation...
I see two differents approaches for the moment :
Naive approach: For each pair of indices (i,j) in the array, compute the minimum value m(i,j) of the array in the interval (i,j) and then compute score(i,j) = |j-i+1|*m(i,j). Take then the maximum score over all the pairs (i,j).
-> Complexity of O(n^3).
Less naive approach:
Compute the set of values of the array
For each value, compute the maximum score it can get. For that, you just have to iterate once over all the values of the array. For example, when your sample input is [3 0 5 4 4 4] and the current value you are looking is 3, then it will give you a score of 12. (You'll first find a value of 3 thanks to the first index, and then a score of 12 due to indices from 2 to 5).
Take the maximum over all values found at step 2.
-> Complexity is here O(n*m), since you have to do at most m times the step 2, and the step 2 can be done in O(n).
Maybe there is a better complexity, but I don't have a clue yet.
The problem statement is the following:
Given a number x, find the smallest Sparse number which greater than or equal to x
A number is Sparse if there are no two adjacent 1s in its binary representation. For example 5 (binary representation: 101) is sparse, but 6 (binary representation: 110) is not sparse.
I'm taking the problem from this post where the most efficient solution is listed as having a running time of O(logn):
1) Find binary of the given number and store it in a
boolean array.
2) Initialize last_finalized bit position as 0.
2) Start traversing the binary from least significant bit.
a) If we get two adjacent 1's such that next (or third)
bit is not 1, then
(i) Make all bits after this 1 to last finalized
bit (including last finalized) as 0.
(ii) Update last finalized bit as next bit.
What isn't clear in the post is what is meant by "finalized bit." It seems that the algorithm starts out by inserting the binary representation of a number into a std::vector using a while loop in which it ANDS the input (which is a number x) with 1 and then pushes that back into the vector but, at least from the provided description, its not clear why this is done. Is there a clearer explanation (or even approach) to an efficient solution for this problem?
EDIT:
// Start from second bit (next to LSB)
for (int i=1; i<n-1; i++)
{
// If current bit and its previous bit are 1, but next
// bit is not 1.
if (bin[i] == 1 && bin[i-1] == 1 && bin[i+1] != 1)
{
// Make the next bit 1
bin[i+1] = 1;
// Make all bits before current bit as 0 to make
// sure that we get the smallest next number
for (int j=i; j>=last_final; j--)
bin[j] = 0;
// Store position of the bit set so that this bit
// and bits before it are not changed next time.
last_final = i+1;
}
}
If you see any sequence "011" in the binary representation of your number, then change the '0' to a '1' and set every bit after it to '0' (since that gives the minimum).
The algorithm suggests starting from the right (the least significant bit), but if you start from the left, find the leftmost sequence "011" and do as above, you get the solution one half of the time. The other half is when the next bit to the left of this sequence is a '1'. When you change the '0' to a '1', you create a new "011" sequence that needs to be treated the same way.
The "last finalized bit" is the leftmost '0' bit that sees only '0' bits to its right. This is because all of those '0's won't change in the next steps.
So here are some observation to solve this question:-
Convert the number in its binary format now if the last digits is 0 then we can append 1 and 0 both at the end but if the last digit is 1 then we can only append 0 at the end.
So naive approach is to do a simple iteration and check for every number but we can optimize this approach so for that if we look closely to some example
let say n=5 -> 101 next sparse is 5 (101)
let say n=14 -> 1110 next sparse is 16 (10000)
let say n=39 ->100111 next sparse is 40 (101000)
let say n=438 -> 110110110 next sparse is 512 (1000000000)
To optimize naive approach the idea here is to use the BIT-MANIPULATION
the concept that if we AND a bit sequence with a shifted version of itself, we’re effectively removing the trailing 1 from every sequence of consecutive 1s.
for n=5
0101 (5)
& 1010 (5<<1)
---------
0000
so as you get the value of n&(n<<1) to be zero means the number you have does not have any consecutive 1's in it ( because if it is not zero then there must be a sequence of consecutive 1's in our number) so this will be answer
for n=14
01110 (14)
& 11100 (14<<1)
----------------
01100
so the value is not zero then just increment our number by 1 so our new number is 15
now again perform same things
01111 (15)
& 11110 (15<<1)
------------------------------
01110
again our number is not zero then increment number by 1 and perform same for n = 16
010000 (16)
& 100000 (16<<1)
------------------------
000000
so now our number become zero so we have now encounter a number which does not contains any consecutive 1's so our answer is 16.
So in the similar manner you can check for other number too.
Hope you get the idea if so then upvote. Happy Coding!
int nextSparse(int n) {
// code here
while(true)
{
if(n&(n<<1))
n++;
else
return n;
}
}
Time Complexity will be O(logn).
Can anyone help me with some algorithm for this problem?
We have a big number (19 digits) and, in a loop, we subtract one of the digits of that number from the number itself.
We continue to do this until the number reaches zero. We want to calculate the minimum number of subtraction that makes a given number reach zero.
The algorithm must respond fast, for a 19 digits number (10^19), within two seconds. As an example, providing input of 36 will give 7:
1. 36 - 6 = 30
2. 30 - 3 = 27
3. 27 - 7 = 20
4. 20 - 2 = 18
5. 18 - 8 = 10
6. 10 - 1 = 9
7. 9 - 9 = 0
Thank you.
The minimum number of subtractions to reach zero makes this, I suspect, a very thorny problem, one that will require a great deal of backtracking potential solutions, making it possibly too expensive for your time limitations.
But the first thing you should do is a sanity check. Since the largest digit is a 9, a 19-digit number will require about 1018 subtractions to reach zero. Code up a simple program to continuously subtract 9 from 1019 until it becomes less than ten. If you can't do that within the two seconds, you're in trouble.
By way of example, the following program (a):
#include <stdio.h>
int main (int argc, char *argv[]) {
unsigned long long x = strtoull(argv[1], NULL, 10);
x /= 1000000000;
while (x > 9)
x -= 9;
return x;
}
when run with the argument 10000000000000000000 (1019), takes a second and a half clock time (and CPU time since it's all calculation) even at gcc insane optimisation level of -O3:
real 0m1.531s
user 0m1.528s
sys 0m0.000s
And that's with the one-billion divisor just before the while loop, meaning the full number of iterations would take about 48 years.
So a brute force method isn't going to help here, what you need is some serious mathematical analysis which probably means you should post a similar question over at https://math.stackexchange.com/ and let the math geniuses have a shot.
(a) If you're wondering why I'm getting the value from the user rather than using a constant of 10000000000000000000ULL, it's to prevent gcc from calculating it at compile time and turning it into something like:
mov $1, %eax
Ditto for the return x which will prevent it noticing I don't use the final value of x and hence optimise the loop out of existence altogether.
I don't have a solution that can solve 19 digit numbers in 2 seconds. Not even close. But I did implement a couple of algorithms (including a dynamic programming algorithm that solves for the optimum), and gained some insight that I believe is interesting.
Greedy Algorithm
As a baseline, I implemented a greedy algorithm that simply picks the largest digit in each step:
uint64_t countGreedy(uint64_t inputVal) {
uint64_t remVal = inputVal;
uint64_t nStep = 0;
while (remVal > 0) {
uint64_t digitVal = remVal;
uint_fast8_t maxDigit = 0;
while (digitVal > 0) {
uint64_t nextDigitVal = digitVal / 10;
uint_fast8_t digit = digitVal - nextDigitVal * 10;
if (digit > maxDigit) {
maxDigit = digit;
}
digitVal = nextDigitVal;
}
remVal -= maxDigit;
++nStep;
}
return nStep;
}
Dynamic Programming Algorithm
The idea for this is that we can calculate the optimum incrementally. For a given value, we pick a digit, which adds one step to the optimum number of steps for the value with the digit subtracted.
With the target function (optimum number of steps) for a given value named optSteps(val), and the digits of the value named d_i, the following relationship holds:
optSteps(val) = 1 + min(optSteps(val - d_i))
This can be implemented with a dynamic programming algorithm. Since d_i is at most 9, we only need the previous 9 values to build on. In my implementation, I keep a circular buffer of 10 values:
static uint64_t countDynamic(uint64_t inputVal) {
uint64_t minSteps[10] = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
uint_fast8_t digit0 = 0;
for (uint64_t val = 10; val <= inputVal; ++val) {
digit0 = val % 10;
uint64_t digitVal = val;
uint64_t minPrevStep = 0;
bool prevStepSet = false;
while (digitVal > 0) {
uint64_t nextDigitVal = digitVal / 10;
uint_fast8_t digit = digitVal - nextDigitVal * 10;
if (digit > 0) {
uint64_t prevStep = 0;
if (digit > digit0) {
prevStep = minSteps[10 + digit0 - digit];
} else {
prevStep = minSteps[digit0 - digit];
}
if (!prevStepSet || prevStep < minPrevStep) {
minPrevStep = prevStep;
prevStepSet = true;
}
}
digitVal = nextDigitVal;
}
minSteps[digit0] = minPrevStep + 1;
}
return minSteps[digit0];
}
Comparison of Results
This may be considered a surprise: I ran both algorithms on all values up to 1,000,000. The results are absolutely identical. This suggests that the greedy algorithm actually calculates the optimum.
I don't have a formal proof that this is indeed true for all possible values. It intuitively kind of makes sense to me. If in any given step, you choose a smaller digit than the maximum, you compromise the immediate progress with the goal of getting into a more favorable situation that allows you to catch up and pass the greedy approach. But in all the scenarios I thought about, the situation after taking a sub-optimal step just does not get significantly more favorable. It might make the next step bigger, but that is at most enough to get even again.
Complexity
While both algorithms look linear in the size of the value, they also loop over all digits in the value. Since the number of digits corresponds to log(n), I believe the complexity is O(n * log(n)).
I think it's possible to make it linear by keeping counts of the frequency of each digit, and modifying them incrementally. But I doubt it would actually be faster. It requires more logic, and turns a loop over all digits in the value (which is in the range of 2-19 for the values we are looking at) into a fixed loop over 10 possible digits.
Runtimes
Not surprisingly, the greedy algorithm is faster to calculate a single value. For example, for value 1,000,000,000, the runtimes on my MacBook Pro are:
greedy: 3 seconds
dynamic: 36 seconds
On the other hand, the dynamic programming approach is obviously much faster at calculating all the values, since its incremental approach needs them as intermediate results anyway. For calculating all values from 10 to 1,000,000:
greedy: 19 minutes
dynamic: 0.03 seconds
As already shown in the runtimes above, the greedy algorithm gets about as high as 9 digit input values within the targeted runtime of 2 seconds. The implementations aren't really tuned, and it's certainly possible to squeeze out some more time, but it would be fractional improvements.
Ideas
As already explored in another answer, there's no chance of getting the result for 19 digit numbers in 2 seconds by subtracting digits one by one. Since we subtract at most 9 in each step, completing this for a value of 10^19 needs more than 10^18 steps. We mostly use computers that perform in the rough range of 10^9 operations/second, which suggests that it would take about 10^9 seconds.
Therefore, we need something that can take shortcuts. I can think of scenarios where that's possible, but haven't been able to generalize it to a full strategy so far.
For example, if your current value is 9999, you know that you can subtract 9 until you reach 9000. So you can calculate that you will make 112 steps ((9999 - 9000) / 9 + 1) where you subtract 9, which can be done in a few operations.
As said in comments already, and agreeing with #paxdiablo’s other answer, I’m not sure if there is an algorithm to find the ideal solution without some backtracking; and the size of the number and the time constraint might be tough as well.
A general consideration though: You might want to find a way to decide between always subtracting the highest digit (which will decrease your current number by the largest possible amount, obviously), and by looking at your current digits and subtracting which of those will give you the largest “new” digit.
Say, your current number only consists of digits between 0 and 5 – then you might be tempted to subtract the 5 to decrease your number by the highest possible value, and continue with the next step. If the last digit of your current number is 3 however, then you might want to subtract 4 instead – since that will give you 9 as new digit at the end of the number, instead of “only” 8 you would be getting if you subtracted 5.
Whereas if you have a 2 and two 9 in your digits already, and the last digit is a 1 – then you might want to subtract the 9 anyway, since you will be left with the second 9 in the result (at least in most cases; in some edge cases it might get obliterated from the result as well), so subtracting the 2 instead would not have the advantage of giving you a “high” 9 that you would otherwise not have in the next step, and would have the disadvantage of not lowering your number by as high an amount as subtracting the 9 would …
But every digit you subtract will not only affect the next step directly, but the following steps indirectly – so again, I doubt there is a way to always chose the ideal digit for the current step without any backtracking or similar measures.