1-bit and 2-bit Characters

-
Sometimes a recursive approach to a problem might seem like a bad idea "because of the stack overhead and potential explosion of permutations and overly complicated logic"... yada, yada, yada.

Sure, sometimes such arguments are right on the target, but sometimes (many times) computer science problems can mirror life problems quite accurately, where the answer to the problem can be a boring "it depends".

Take a look at this problem by LeetCode:

https://leetcode.com/problems/1-bit-and-2-bit-characters/description/

 We have two special characters. The first character can be represented by one bit 0. The second character can be represented by two bits (10 or 11).
Now given a string represented by several bits. Return whether the last character must be a one-bit character or not. The given string will always end with a zero.
Example 1:
Input: 
bits = [1, 0, 0]
Output: True
Explanation: 
The only way to decode it is two-bit character and one-bit character. So the last character is one-bit character.
Example 2:
Input: 
bits = [1, 1, 1, 0]
Output: False
Explanation: 
The only way to decode it is two-bit character and two-bit character. So the last character is NOT one-bit character.

The problem statement is practically giving you the recipe on how to solve it.
I make the argument here that in such a case a basic recursive solution will do the trick quite well. It can be stated in four simple lines (literally) and since this is an fast-forward-moving tail recursion, it will be executing very fast.

Definitely it can be done interactively but I think the recursive solution here really does a great job in following the rules stated in the problem description in a very mathematical manner.

Hope you enjoy! Marcelo


public class Solution
{
public bool IsOneBitCharacter(int[] bits)
{
return IsOneBitCharacterInternal(bits, 0);
}

 private bool IsOneBitCharacterInternal(int[] bits, int index)
{
if (index >= bits.Length) return false;
if (index == bits.Length - 1) return bits[index] == 0;
if (bits[index] == 1) return IsOneBitCharacterInternal(bits, index + 2);
return IsOneBitCharacterInternal(bits, index + 1);
}
}

Comments

  1. TarasNovember 11, 2017 at 8:54 PM

    Very nice, Marcelo! Interestingly even though I also think that your recursive solution models the problem very well, the first thought that came to my mind when I saw this problem is Huffman encoding and since this problem is actually using it, I decided to approach it from the point of view of an Huffman encoding decoder - if current element is 1 than we know we can skip 2 bits, and if current element is zero, we can skip only a single element. The code is here:
    class Solution {
    public:
    bool isOneBitCharacter(const vector& bits) {
    int i = 0;
    for (; i < bits.size() - 1; i += bits[i] + 1) {}
    return i == bits.size() - 1;
    }
    };

    Have a splendid rest of the weekend!
    Cheers,
    Taras

    ReplyDelete
    Replies
    1. TarasNovember 11, 2017 at 8:57 PM

      also, here is a slightly less ugly version:
      class Solution {
      public:
      bool isOneBitCharacter(const vector& bits) {
      int i = 0;
      while (i < bits.size() - 1) i += bits[i] + 1;
      return i == bits.size() - 1;
      }
      };

      https://ideone.com/Y9bhbb

      Delete
    2. Marcelo M De BarrosNovember 11, 2017 at 11:09 PM

      Oh I love the bits[i]+1, encapsulates the core of the problem so well! Brilliant!!!

      Delete
    3. TarasNovember 11, 2017 at 11:14 PM

      Ha, I'm glad that you liked it - it was mostly just to remove an extra if statement, since they hurt performance, but as a side effect they also made the solution a bit shorter. In your implementation you could also replaced
      if (bits[index] == 1) return IsOneBitCharacterInternal(bits, index + 2);
      return IsOneBitCharacterInternal(bits, index + 1);

      with
      return IsOneBitCharacterInternal(bits, index + bits[index] + 1);

      which in theory should make your solution even faster and more elegant :)

      Delete

Post a Comment

Popular posts from this blog

Changing the root of a binary tree

-
Image
If you work with graphics and eventually you want to pivot a binary tree using a different node as the root, this might be interesting to you. Here is the problem:  Change the Root of a Binary Tree - LeetCode 1666. Change the Root of a Binary Tree Medium 6 14 Add to List Share Given the  root  of a binary tree and a  leaf  node, reroot the tree so that the  leaf  is the new root. You can reroot the tree with the following steps for each node  cur  on the path  starting from the  leaf  up to the  root ​​​  excluding the root : If  cur  has a left child, then that child becomes  cur 's right child. Note that it is guaranteed that  cur  will have at most one child. cur 's original parent becomes  cur 's left child. Return  the new root  of the rerooted tree. Note:  Ensure that your solution sets the  Node.parent  pointers correctly after rerooting or you will receive "Wrong Answer".   Example 1: Input: root = [3,5,1,6,2,0,8,null,null,7,4], leaf = 7 Output: [7,2,nul
Read more

ProjectEuler Problem 719 (some hints, but no spoilers)

-
Image
Last time that I solved a ProjectEuler (PE) problem was in 2015... This is their latest problem as of today:  https://projecteuler.net/problem=719 Number Splitting        Problem 719 We define an  S S -number to be a natural number,  n n , that is a perfect square and its square root can be obtained by splitting the decimal representation of  n n  into 2 or more numbers then adding the numbers. For example, 81 is an  S S -number because  81 − − √ = 8 + 1 81 = 8 + 1 . 6724 is an  S S -number:  6724 − − − − √ = 6 + 72 + 4 6724 = 6 + 72 + 4 . 8281 is an  S S -number:  8281 − − − − √ = 8 + 2 + 81 = 82 + 8 + 1 8281 = 8 + 2 + 81 = 82 + 8 + 1 . 9801 is an  S S -number:  9801 − − − − √ = 98 + 0 + 1 9801 = 98 + 0 + 1 . Further we define  T ( N ) T ( N )  to be the sum of all  S S  numbers  n ≤ N n ≤ N . You are given  T ( 10 4 ) = 41333 T ( 10 4 ) = 41333 . Find  T ( 10 12 ) The rules of PE prevent me from disclosing the solution (they are very picky about it, rightfully so), but I can chat abo
Read more

The Power Sum, a recursive problem by HackerRank

-
Image
Another one by HackerRank:  https://www.hackerrank.com/challenges/the-power-sum : Find the number of ways that a given integer,  , can be expressed as the sum of the   power of unique, natural numbers. Input Format The first line contains an integer  .  The second line contains an integer  . Constraints Output Format Output a single integer, the answer to the problem explained above. Sample Input 0 10 2 Sample Output 0 1 Explanation 0 If   and  , we need to find the number of ways that   can be represented as the sum of squares of unique numbers. This is the only way in which   can be expressed as the sum of unique squares. Sample Input 1 100 2 Sample Output 1 3 Explanation 1 Sample Input 2 100 3 Sample Output 2 1 Explanation 2  can be expressed as the sum of the cubes of  .  . There is no other way to express   as th
Read more