A different implementation of digit DP problems

Revision en1, by shadow9236, 2024-03-11 14:22:09

Hello Codeforces,

I have seen multiple blogs on iterative digit DP. In this blog, I would like to share my recursive implementation of digit dp related problems.

Problem Structure

Typically, digit dp is used in problems of finding the number of values in a range $$$[l,r]$$$ which follow a certain property. Let us call this $$$f(l,r)$$$. The following method will work only when $$$f(l,r) = f(0,r) - f(0,l-1)$$$, which is true for most of the problems. We will now focus on how to calculate $$$g(n) = f(0,n)$$$.

Idea behind the algorithm

To count all numbers less than or equal to $$$n$$$ which follow a certain property, we first count all numbers with the same number of digits as $$$n$$$. Here the numerical order will be the same as lexicographical order. We can now iterate over the common prefix of the two numbers and the first character which differs after the prefix. This gives us the following template.


int g(string s){ int n = s.size(); int ans = 0; for(int i = 0; i < n; i++){ // i is the first position of difference for(int j = (i == 0); j < s[i] - '0'; j++){ // keeping digit j at position i ans += valid_strings(n-i-1); // number of valid strings of length n-i-1 } } if(valid(s)){ ans += 1; } if(n > 1){ // All numbers with less digits ans += g(string(n-1,'9')); } return ans; }

Adding memoisation to this recursive function gives polylog complexity in most cases. Note that $$$j$$$ is initialised to $$$1$$$ for $$$i == 0$$$ since our numbers can't have leading zeroes.

Some examples

Consider ARC173 A. This problem can be solved with binary search. Here we have to find the number of numbers less than or equal to $$$mid$$$ that no two consecutive digits are the same. Here I get the following function.

Solution
Tags tutorial, dynamic programming, digit dp

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en11 English shadow9236 2024-03-11 15:01:30 4 Tiny change: 'odeforces,\n Digit ' -> 'odeforces,<br>\n Digit '
en10 English shadow9236 2024-03-11 15:01:10 0 (published)
en9 English shadow9236 2024-03-11 15:00:29 100
en8 English shadow9236 2024-03-11 14:59:09 4 Tiny change: 'r>\n\n<br><br>\n[CF ' -> 'r>\n\n<br>\n[CF '
en7 English shadow9236 2024-03-11 14:58:31 52
en6 English shadow9236 2024-03-11 14:57:06 2
en5 English shadow9236 2024-03-11 14:56:06 989
en4 English shadow9236 2024-03-11 14:51:36 5 Tiny change: '\n</li>\n<\ul>\n' -> '\n</li>\n</ui>\n'
en3 English shadow9236 2024-03-11 14:50:25 742
en2 English shadow9236 2024-03-11 14:26:19 233
en1 English shadow9236 2024-03-11 14:22:09 2450 Initial revision (saved to drafts)