Recursive String Transformations - c++

Question

EDIT: I've made the main change of using iterators to keep track of successive positions in the bit and character strings and pass the latter by const ref. Now, when I copy the sample inputs onto themselves multiple times to test the clock, everything finishes within 10 seconds for really long bit and character strings and even up to 50 lines of sample input. But, still when I submit, CodeEval says the process was aborted after 10 seconds. As I mention, they don't share their input so now that "extensions" of the sample input work, I'm not sure how to proceed. Any thoughts on an additional improvement to increase my recursive performance would be greatly appreciated.
NOTE: Memoization was a good suggestion but I could not figure out how to implement it in this case since I'm not sure how to store the bit-to-char correlation in a static look-up table. The only thing I thought of was to convert the bit values to their corresponding integer but that risks integer overflow for long bit strings and seems like it would take too long to compute. Further suggestions for memoization here would be greatly appreciated as well.
This is actually one of the moderate CodeEval challenges. They don't share the sample input or output for moderate challenges but the output "fail error" simply says "aborted after 10 seconds," so my code is getting hung up somewhere.
The assignment is simple enough. You take a filepath as the single command-line argument. Each line of the file will contain a sequence of 0s and 1s and a sequence of As and Bs, separated by a white space. You are to determine whether the binary sequence can be transformed into the letter sequence according to the following two rules:
1) Each 0 can be converted to any non-empty sequence of As (e.g, 'A', 'AA', 'AAA', etc.)
2) Each 1 can be converted to any non-empty sequences of As OR Bs (e.g., 'A', 'AA', etc., or 'B', 'BB', etc) (but not a mixture of the letters)
The constraints are to process up to 50 lines from the file and that the length of the binary sequence is in [1,150] and that of the letter sequence is in [1,1000].
The most obvious starting algorithm is to do this recursively. What I came up with was for each bit, collapse the entire next allowed group of characters first, test the shortened bit and character strings. If it fails, add back one character from the killed character group at a time and call again.
Here is my complete code. I removed cmd-line argument error checking for brevity.
#include <iostream>
#include <fstream>
#include <string>
#include <iterator>
using namespace std;
//typedefs
typedef string::const_iterator str_it;
//declarations
//use const ref and iterators to save time on copying and erasing
bool TransformLine(const string & bits, str_it bits_front, const string & chars, str_it chars_front);
int main(int argc, char* argv[])
{
//check there are at least two command line arguments: binary executable and file name
//ignore additional arguments
if(argc < 2)
{
cout << "Invalid command line argument. No input file name provided." << "\n"
<< "Goodybe...";
return -1;
}
//create input stream and open file
ifstream in;
in.open(argv[1], ios::in);
while(!in.is_open())
{
char* name;
cout << "Invalid file name. Please enter file name: ";
cin >> name;
in.open(name, ios::in);
}
//variables
string line_bits, line_chars;
//reserve space up to constraints to reduce resizing time later
line_bits.reserve(150);
line_chars.reserve(1000);
int line = 0;
//loop over lines (<=50 by constraint, ignore the rest)
while((in >> line_bits >> line_chars) && (line < 50))
{
line++;
//impose bit and char constraints
if(line_bits.length() > 150 ||
line_chars.length() > 1000)
continue; //skip this line
(TransformLine(line_bits, line_bits.begin(), line_chars, line_chars.begin()) == true) ? (cout << "Yes\n") : (cout << "No\n");
}
//close file
in.close();
return 0;
}
bool TransformLine(const string & bits, str_it bits_front, const string & chars, str_it chars_front)
{
//using iterators so store current length as local const
//can make these const because they're not altered here
int bits_length = distance(bits_front, bits.end());
int chars_length = distance(chars_front, chars.end());
//check success rule
if(bits_length == 0 && chars_length == 0)
return true;
//Check fail rules:
//1. next bit is 0 but next char is B
//2. bits length is zero (but char is not, by previous if)
//3. char length is zero (but bits length is not, by previous if)
if((*bits_front == '0' && *chars_front == 'B') ||
bits_length == 0 ||
chars_length == 0)
return false;
//we now know that chars_length != 0 => chars_front != chars.end()
//kill a bit and then call recursively with each possible reduction of front char group
bits_length = distance(++bits_front, bits.end());
//current char group tracker
const char curr_char_type = *chars_front; //use const so compiler can optimize
int curr_pos = distance(chars.begin(), chars_front); //position of current front in char string
//since chars are 0-indexed, the following is also length of current char group
//start searching from curr_pos and length is relative to curr_pos so subtract it!!!
int curr_group_length = chars.find_first_not_of(curr_char_type, curr_pos)-curr_pos;
//make sure this isn't the last group!
if(curr_group_length < 0 || curr_group_length > chars_length)
curr_group_length = chars_length; //distance to end is precisely distance(chars_front, chars.end()) = chars_length
//kill the curr_char_group
//if curr_group_length = char_length then this will make chars_front = chars.end()
//and this will mean that chars_length will be 0 on next recurssive call.
chars_front += curr_group_length;
curr_pos = distance(chars.begin(), chars_front);
//call recursively, adding back a char from the current group until 1 less than starting point
int added_back = 0;
while(added_back < curr_group_length)
{
if(TransformLine(bits, bits_front, chars, chars_front))
return true;
//insert back one char from the current group
else
{
added_back++;
chars_front--; //represents adding back one character from the group
}
}
//if here then all recursive checks failed so initial must fail
return false;
}
They give the following test cases, which my code solves correctly:
Sample input:
1| 1010 AAAAABBBBAAAA
2| 00 AAAAAA
3| 01001110 AAAABAAABBBBBBAAAAAAA
4| 1100110 BBAABABBA
Correct output:
1| Yes
2| Yes
3| Yes
4| No
Since a transformation is possible if and only if copies of it are, I tried just copying each binary and letter sequences onto itself various times and seeing how the clock goes. Even for very long bit and character strings and many lines it has finished in under 10 seconds.
My question is: since CodeEval is still saying it is running longer than 10 seconds but they don't share their input, does anyone have any further suggestions to improve the performance of this recursion? Or maybe a totally different approach?
Thank you in advance for your help!

Answer 1

Here's what I found:
Pass by constant reference
Strings and other large data structures should be passed by constant reference.
This allows the compiler to pass a pointer to the original object, rather than making a copy of the data structure.
Call functions once, save result
You are calling bits.length() twice. You should call it once and save the result in a constant variable. This allows you to check the status again without calling the function.
Function calls are expensive for time critical programs.
Use constant variables
If you are not going to modify a variable after assignment, use the const in the declaration:
const char curr_char_type = chars[0];
The const allows compilers to perform higher order optimization and provides safety checks.
Change data structures
Since you are perform inserts maybe in the middle of a string, you should use a different data structure for the characters. The std::string data type may need to reallocate after an insertion AND move the letters further down. Insertion is faster with a std::list<char> because a linked list only swaps pointers. There may be a trade off because a linked list needs to dynamically allocate memory for each character.
Reserve space in your strings
When you create the destination strings, you should use a constructor that preallocates or reserves room for the largest size string. This will prevent the std::string from reallocating. Reallocations are expensive.
Don't erase
Do you really need to erase characters in the string?
By using starting and ending indices, you overwrite existing letters without have to erase the entire string.
Partial erasures are expensive. Complete erasures are not.
For more assistance, post to Code Review at StackExchange.

Answer 2

This is a classic recursion problem. However, a naive implementation of the recursion would lead to an exponential number of re-evaluations of a previously computed function value. Using a simpler example for illustration, compare the runtime of the following two functions for a reasonably large N. Lets not worry about the int overflowing.
int RecursiveFib(int N)
{
if(N<=1)
return 1;
return RecursiveFib(N-1) + RecursiveFib(N-2);
}
int IterativeFib(int N)
{
if(N<=1)
return 1;
int a_0 = 1, a_1 = 1;
for(int i=2;i<=N;i++)
{
int temp = a_1;
a_1 += a_0;
a_0 = temp;
}
return a_1;
}
You would need to follow a similar approach here. There are two common ways of approaching the problem - dynamic programming and memoization. Memoization is the easiest way of modifying your approach. Below is a memoized fibonacci implementation to illustrate how your implementation can be speeded up.
int MemoFib(int N)
{
static vector<int> memo(N, -1);
if(N<=1)
return 1;
int& res = memo[N];
if(res!=-1)
return res;
return res = MemoFib(N-1) + MemoFib(N-2);
}

Answer 3

Your failure message is "Aborted after 10 seconds" -- implying that the program was working fine as far as it went, but it took too long. This is understandable, given that your recursive program takes exponentially more time for longer input strings -- it works fine for the short (2-8 digit) strings, but will take a huge amount of time for 100+ digit strings (which the test allows for). To see how your running time goes up, you should construct yourself some longer test inputs and see how long they take to run. Try things like
0000000011111111 AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAABBBBBBBBBBAAAAAAAA
00000000111111110000000011111111 AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAABBBBBBBBBBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAABBBBBBBBBBAAAAAAAA
and longer. You need to be able to handle up to 150 digits and 1000 letters.

Answer 4

At CodeEval, you can submit a "solution" that just outputs what the input is, and do that to gather their test set. They may have variations so you may wish to submit it a few times to gather more samples. Some of them are too difficult to solve manually though... the ones you can solve manually will also run very quickly at CodeEval too, even with inefficient solutions, so there's that to consider.
Anyway, I did this same problem at CodeEval (using VB of all things), and my solution recursively looked for the "next index" of both A and B depending on what the "current" index is for where I was in a translation (after checking stoppage conditions first thing in the recursive method). I did not use memoization but that might've helped speed it up even more.
PS, I have not run your code, but it does seem curious that the recursive method contains a while loop within which the recursive method is called... since it's already recursive and should therefore encompass every scenario, is that while() loop necessary?