This is the homework spec
// Returns a string where the same characters next each other in
// string n are separated by "22"
//
// Pseudocode Example:
// doubleDouble("goodbye") => "go22odbye"
// doubleDouble("yyuu") => "y22yu22u"
// doubleDouble("aaaa") => "a22a22a22a"
//
string doubleDouble(string n)
{
return ""; // This is not always correct.
}
This is part of a homework set that deals with recursion. I know how to use recursion with an int function but I'm not entirely sure how to approach this problem when a string is passed through. Is it as simple as n.length == n.length() +1 ? and then simply insert "22" ? Alongside this, how would one traverse the string? Thanks!
I would say that the base case be if n turns out to be just a blank space, or if it has a size 0, then simply return the string back, no changes made.
This seems OK to me (untested code)
string doubleDouble(string n)
{
if (n.size() < 2)
return n;
else if (n[0] == n[1])
return n[0] + ("22" + doubleDouble(n.substr(1)));
else
return n[0] + doubleDouble(n.substr(1));
}
As you can see you recurse on substrings of the original string.
Undoubtedly there are more efficient ways to do this (that minimise the string copying) but I'll leave that for you.
PS you need to enable C++11 to get all the required versions of + for strings.
string doubleDouble(string n)
{
if(n.length() == 2){
if(n[0]==n[1])
return n.insert(1,"22");
else
return n;
}
else if(n.length() == 1)
return n;
else
if(n.length() % 2 == 1 || n[n.length()/2]!=n[n.length()/2+1])
return doubleDouble(n.substr(0,n.length() / 2)) + doubleDouble(n.substr((n.length() / 2),n.length()));
else if(n[n.length()/2]==n[n.length()/2+1])
return doubleDouble(n.substr(0,n.length() / 2)) +"22"+ doubleDouble(n.substr((n.length() / 2),n.length()));
}
My solution uses 'Divide-and-conquer' paradigm to separate the string in 2 halfs again and again until it finds 1 or 2 characters only. The solution above is simpler. I tried an academic style.
Related
This Leetcode problem is about how to match a pattern string against a text string as efficiently as possible. The pattern string can consists of letters, dots, and stars, where a letter only matches itself, a dot matches any individual character, and a star matches any number of copies of the preceding character. For example, the pattern
ab*c.
would match ace and abbbbcc. I know that it's possible to solve this original problem using dynamic programming.
My question is whether it's possible to see whether two patterns match one another. For example, the pattern
bdbaa.*
can match
bdb.*daa
Is there a nice algorithm for solving this pattern-on-pattern matching problem?
Here's one approach that works in polynomial time. It's slightly heavyweight and there may be a more efficient solution, though.
The first observation that I think helps here is to reframe the problem. Rather than asking whether these patterns match each other, let's ask this equivalent question:
Given patterns P1 and P2, is there a string w where P1 and P2 each match w?
In other words, rather than trying to get the two patterns to match one another, we'll search for a string that each pattern matches.
You may have noticed that the sorts of patterns you're allowed to work with are a subset of the regular expressions. This is helpful, since there's a pretty elaborate theory of what you can do with regular expressions and their properties. So rather than taking aim at your original problem, let's solve this even more general one:
Given two regular expressions R1 and R2, is there a string w that both R1 and R2 match?
The reason for solving this more general problem is that it enables us to use the theory that's been developed around regular expressions. For example, in formal language theory we can talk about the language of a regular expression, which is the set of all strings that the regex matches. We can denote this L(R). If there's a string that's matched by two regexes R1 and R2, then that string belongs to both L(R1) and L(R2), so our question is equivalent to
Given two regexes R1 and R2, is there a string w in L(R1) ∩ L(R2)?
So far all we've done is reframe the problem we want to solve. Now let's go solve it.
The key step here is that it's possible to convert any regular expression into an NFA (a nondeterministic finite automaton) so that every string matched by the regex is accepted by the NFA and vice-versa. Even better, the resulting NFA can be constructed in polynomial time. So let's begin by constructing NFAs for each input regex.
Now that we have those NFAs, we want to answer this question: is there a string that both NFAs accept? And fortunately, there's a quick way to answer this. There's a common construction on NFAs called the product construction that, given two NFAs N1 and N2, constructs a new NFA N' that accepts all the strings accepted by both N1 and N2 and no other strings. Again, this construction runs in polynomial time.
Once we have N', we're basically done! All we have to do is run a breadth-first or depth-first search through the states of N' to see if we find an accepting state. If so, great! That means there's a string accepted by N', which means that there's a string accepted by both N1 and N2, which means that there's a string matched by both R1 and R2, so the answer to the original question is "yes!" And conversely, if we can't reach an accepting state, then the answer is "no, it's not possible."
I'm certain that there's a way to do all of this implicitly by doing some sort of implicit BFS over the automaton N' without actually constructing it, and it should be possible to do this in something like time O(n2). If I have some more time, I'll revisit this answer and expand on how to do that.
I have worked on my idea of DP and came out with the below implementation of the above problem. Please feel free to edit the code in case someone finds any test cases failed. From my side, I tried few test cases and passed all of them, which I will be mentioning below as well.
Please note that I have extended the idea which is used to solve the regex pattern matching with a string using DP. To refer to that idea, please refer to the LeetCode link provided in the OP and look out for discussion part. They have given the explanation for regex matching and the string.
The idea is to create a dynamic memoization table, entries of which will follow the below rules:
If pattern1[i] == pattern2[j], dp[i][j] = dp[i-1][j-1]
If pattern1[i] == '.' or pattern2[j] == '.', then dp[i][j] = dp[i-1][j-1]
The trick lies here: If pattern1[i] = '*', then if dp[i-2][j] exists, then
dp[i][j] = dp[i-2][j] || dp[i][j-1] else dp[i][j] = dp[i][j-1].
If pattern2[j] == '*', then if pattern1[i] == pattern2[j-1], then
dp[i][j] = dp[i][j-2] || dp[i-1][j]
else dp[i][j] = dp[i][j-2]
pattern1 goes row-wise and pattern2 goes column-wise. Also, please note that this code should also work for normal regex pattern matching with any given string. I have verified it by running it on LeetCode and it passed all the available test cases there!
Below is the complete working implementation of the above logic:
boolean matchRegex(String pattern1, String pattern2){
boolean dp[][] = new boolean[pattern1.length()+1][pattern2.length()+1];
dp[0][0] = true;
//fill up for the starting row
for(int j=1;j<=pattern2.length();j++){
if(pattern2.charAt(j-1) == '*')
dp[0][j] = dp[0][j-2];
}
//fill up for the starting column
for(int j=1;j<=pattern1.length();j++){
if(pattern1.charAt(j-1) == '*')
dp[j][0] = dp[j-2][0];
}
//fill for rest table
for(int i=1;i<=pattern1.length();i++){
for(int j=1;j<=pattern2.length();j++){
//if second character of pattern1 is *, it will be equal to
//value in top row of current cell
if(pattern1.charAt(i-1) == '*'){
dp[i][j] = dp[i-2][j] || dp[i][j-1];
}
else if(pattern1.charAt(i-1)!='*' && pattern2.charAt(j-1)!='*'
&& (pattern1.charAt(i-1) == pattern2.charAt(j-1)
|| pattern1.charAt(i-1)=='.' || pattern2.charAt(j-1)=='.'))
dp[i][j] = dp[i-1][j-1];
else if(pattern2.charAt(j-1) == '*'){
boolean temp = false;
if(pattern2.charAt(j-2) == pattern1.charAt(i-1)
|| pattern1.charAt(i-1)=='.'
|| pattern1.charAt(i-1)=='*'
|| pattern2.charAt(j-2)=='.')
temp = dp[i-1][j];
dp[i][j] = dp[i][j-2] || temp;
}
}
}
//comment this portion if you don't want to see entire dp table
for(int i=0;i<=pattern1.length();i++){
for(int j=0;j<=pattern2.length();j++)
System.out.print(dp[i][j]+" ");
System.out.println("");
}
return dp[pattern1.length()][pattern2.length()];
}
Driver method:
System.out.println(e.matchRegex("bdbaa.*", "bdb.*daa"));
Input1: bdbaa.* and bdb.*daa
Output1: true
Input2: .*acd and .*bce
Output2: false
Input3: acd.* and .*bce
Output3: true
Time complexity: O(mn) where m and n are lengths of two regex patterns given. Same will be the space complexity.
You can use a dynamic approach tailored to this subset of a Thompson NFA style regex implementing only . and *:
You can do that either with dynamic programming (here in Ruby):
def is_match(s, p)
return true if s==p
len_s, len_p=s.length, p.length
dp=Array.new(len_s+1) { |row| [false] * (len_p+1) }
dp[0][0]=true
(2..len_p).each { |j| dp[0][j]=dp[0][j-2] && p[j-1]=='*' }
(1..len_s).each do |i|
(1..len_p).each do |j|
if p[j-1]=='*'
a=dp[i][j - 2]
b=[s[i - 1], '.'].include?(p[j-2])
c=dp[i - 1][j]
dp[i][j]= a || (b && c)
else
a=dp[i - 1][j - 1]
b=['.', s[i - 1]].include?(p[j - 1])
dp[i][j]=a && b
end
end
end
dp[len_s][len_p]
end
# 139 ms on Leetcode
Or recursively:
def is_match(s,p,memo={["",""]=>true})
if p=="" && s!="" then return false end
if s=="" && p!="" then return p.scan(/.(.)/).uniq==[['*']] && p.length.even? end
if memo[[s,p]]!=nil then return memo[[s,p]] end
ch, exp, prev=s[-1],p[-1], p.length<2 ? 0 : p[-2]
a=(exp=='*' && (
([ch,'.'].include?(prev) && is_match(s[0...-1], p, memo) ||
is_match(s, p[0...-2], memo))))
b=([ch,'.'].include?(exp) && is_match(s[0...-1], p[0...-1], memo))
memo[[s,p]]=(a || b)
end
# 92 ms on Leetcode
In each case:
The operative starting point in the string and pattern is at the second character looking for * and matches one character back for as long as s matches the character in p prior to the *
The meta character . is being used as a fill in for the actual character. This allows any character in s to match . in p
You can solve this with backtracking too, not very efficiently (because the match of the same substrings may be recalculated many times, which could be improved by introducing a lookup table where all non-matching pairs of strings are saved and the calculation only happens when they cannot be found in the lookup table), but seems to work (js, the algorithm assumes the simple regex are valid, which means not beginning with * and no two adjacent * [try it yourself]):
function canBeEmpty(s) {
if (s.length % 2 == 1)
return false;
for (let i = 1; i < s.length; i += 2)
if (s[i] != "*")
return false;
return true;
}
function match(a, b) {
if (a.length == 0 || b.length == 0)
return canBeEmpty(a) && canBeEmpty(b);
let x = 0, y = 0;
// process characters up to the next star
while ((x + 1 == a.length || a[x + 1] != "*") &&
(y + 1 == b.length || b[y + 1] != "*")) {
if (a[x] != b[y] && a[x] != "." && b[y] != ".")
return false;
x++; y++;
if (x == a.length || y == b.length)
return canBeEmpty(a.substr(x)) && canBeEmpty(b.substr(y));
}
if (x + 1 < a.length && y + 1 < b.length && a[x + 1] == "*" && b[y + 1] == "*")
// star coming in both strings
return match(a.substr(x + 2), b.substr(y)) || // try skip in a
match(a.substr(x), b.substr(y + 2)); // try skip in b
else if (x + 1 < a.length && a[x + 1] == "*") // star coming in a, but not in b
return match(a.substr(x + 2), b.substr(y)) || // try skip * in a
((a[x] == "." || b[y] == "." || a[x] == b[y]) && // if chars matching
match(a.substr(x), b.substr(y + 1))); // try skip char in b
else // star coming in b, but not in a
return match(a.substr(x), b.substr(y + 2)) || // try skip * in b
((a[x] == "." || b[y] == "." || a[x] == b[y]) && // if chars matching
match(a.substr(x + 1), b.substr(y))); // try skip char in a
}
For a little optimization you could normalize the strings first:
function normalize(s) {
while (/([^*])\*\1([^*]|$)/.test(s) || /([^*])\*\1\*/.test(s)) {
s = s.replace(/([^*])\*\1([^*]|$)/, "$1$1*$2"); // move stars right
s = s.replace(/([^*])\*\1\*/, "$1*"); // reduce
}
return s;
}
// example: normalize("aa*aa*aa*bb*b*cc*cd*dd") => "aaaa*bb*ccc*ddd*"
There is a further reduction of the input possible: x*.* and .*x* can both be replaced by .*, so to get the maximal reduction you would have to try to move as many stars as possible next to .* (so moving some stars to the left can be better than moving all to the right).
IIUC, you are asking: "Can a regex pattern match another regex pattern?"
Yes, it can. Specifically, . matches "any character" which of course includes . and *. So if you have a string like this:
bdbaa.*
How could you match it? Well, you could match it like this:
bdbaa..
Or like this:
b.*
Or like:
.*ba*.*
I am asking the user to input an expression which will be evaluated in postfix notation. The beginning of the expression is the variable name where the answer of the evaluated expression will be stored. Ex: A 4 5 * 6 + 2 * 1 – 6 / 4 2 + 3 * * = where A is the variable name and the equal sign means the answer to the expression will be stored in the variable A. The OUT A statement means that the number stored in the variable A will be printed out.
What I need help with is that when I input the second expression, I do not get the right answer. For example, my first expression A 4 5 * 6 + 2 * 1 – 6 / 4 2 + 3 * * = will evaluate to 153 and then when I input my second expression B A 10 * 35.50 + =, it has to evaluate to 1565.5, but it doesn't. It evaluates to 35.5. I cannot figure out why I am getting the wrong answer. Also, I need help with the OUT statement.
else if (isalpha(expr1[i]))
{
stackIt.push(mapVars1[expr1[i]]);
}
Will place the variable, or zero if the variable has not been set, onto the stack.
else if (isalpha(expr1[i]))
{
map<char, double>::iterator found = mapVars1.find(expr1[i]);
if (found != mapVars1.end())
{
stackIt.push(found->second);
}
else
{
// error message and exit loop
}
}
Is probably better.
Other suggestions:
Compilers are pretty sharp these days, but you may get a bit out of char cur = expr1[i]; and then using cur (or suitably descriptive variable name) in place of the remaining expr1[i]s in the loop.
Consider using isdigit instead of expr1[i] >= '0' && expr1[i] <= '9'
Test your code for expressions with multiple spaces in a row or a space after an operator. It looks like you will re-add the last number you parsed.
Test for input like 123a456. You might not like the result.
If spaces after each token in the expression are specified in the expression protocol, placing your input string into a stringstream will allow you to remove a great deal of your parsing code.
stringstream in(expr1);
string token;
while (in >> token)
{
if (token == "+" || token == "-'" || ...)
{
// operator code
}
else if (token == "=")
{
// equals code
}
else if (mapVars1.find(token) != mapVars1.end())
{
// push variable
}
else if (token.length() > 0)
{
char * endp;
double val = strtod(token.c_str(), &endp);
if (*endp == '\0')
{
// push val
}
}
}
To use previous symbol names in subsequent expressions add this to the if statements in your parsing loop:
else if (expr1[i] >= 'A' && expr1[i] <= 'Z')
{
stackIt.push(mapVars1[expr[i]]);
}
Also you need to pass mapVars by reference to accumulate its contents across Eval calls:
void Eval(string expr1, map<char, double> & mapVars1)
For the output (or any) other command I would recommend parsing the command token that's at the front of the string first. Then call different evaluators based on the command string. You are trying to check for OUT right now after you have already tried to evaluate the string as an arithmetic assignment command. You need to make that choice first.
I am currently finishing up an assignment I have to complete for my OOP class and I am struggling with 1 part in particular. Keep in mind I am still a beginner. The question is as followed:
If the string contains 13 characters, all of characters are digits and the check digit is modulo 10, this function returns true; false otherwise.
This is in regards to a EAN. I basically have to separate every second digit from the rest digits. for example 9780003194876 I need to do calculations with 7,0,0,1,4,7. I have no clue about doing this.
Any help would be greatly appreciated!
bool isValid(const char* str){
if (atoi(str) == 13){
}
return false;
}
You can start with a for loop which increments itself by 2 for each execution:
for (int i = 1, len = strlen(str); i < len; i += 2)
{
int digit = str[i] - '0';
// do something with digit
}
The above is just an example though...
Since the question was tagged as C++ (Not C, so I suggest other answerers to not solve this using C libraries, please. Let us getting OP's C++ knoweledge in the right way since the beggining), and is an OOP class I'm going to solve this with the C++ way: Use the std::string class:
bool is_valid( const std::string& str )
{
if( str.size() == 13 )
{
for( std::size_t i = 0 ; i < 13 ; i += 2 )
{
int digit = str[i] - '0';
//Do what you wan't with the digit
}
}
else
return false;
}
First, if it's EAN, you have to process every digit, not just
every other one. In fact, all you need to do is a weighted sum
of the digits; for EAN-13, the weigths alternate between 1 and
3, starting with three. The simplest solution is probably to
put them in a table (i.e. int weigt[] = { 1, 3, 1, 3... };,
and iterate over the string (in this case, using an index rather
than iterators, since you want to be able to index into
weight as well), converting each digit into a numerical value
(str[i] - '0', if isdigit(static_cast<unsigned char>(str[i])
is true; if it's false, you haven't got a digi.), then
multiplying it by the running total. When you're finished, if
the total, modulo 10, is 0, it's correct. Otherwise, it isn't.
You certainly don't want to use atoi, since you don't want the
numerical value of the string; you want to treat each digit
separately.
Just for the record, professionally, I'd write something like:
bool
isValidEAN13( std::string const& value )
{
return value.size() == 13
&& std::find_if(
value.begin(),
value.end(),
[]( unsigned char ch ){ return !isdigit( ch ); } )
== value.end()
&& calculateEAN13( value ) == value.back() - '0';
}
where calculateEAN13 does the actual calculations (and can be
used for both generation and checking). I suspect that this
goes beyond the goal of the assignment, however, and that all
your teacher is looking for is the calculateEAN13 function,
with the last check (which is why I'm not giving it in full).
I am trying to solve problem PLD on SPOJ, but I'm getting a WA on the 9th testcase.
My Approach:
I am implementing Manacher's Algorithm and I believe that if something wrong is there, then it can be wrong in this code.
if((k%2==0)&&(p[i]>=k)&&(temp[i]=='#'))
count++;
if((k%2==1)&&(p[i]>=k)&&(temp[i]!='#'))
count++;
But according to my approach if character is #, then the maximum length of palindromic string centered at it can be even only, so if p[i] >= k, then I am increasing count if we are finding a palindromic string of even length.
Similarly for characters [considering input character i.e other than #] centered at i-th location but for odd length strings.
#include<stdio.h>
#include<string.h>
char a[30002],temp[60010];
int p[60010];
int min(int a,int b)
{
if(a<b)
return a;
return b;
}
int main()
{
//freopen("input.txt","r+",stdin);
//freopen("a.txt","w+",stdout);
int k,len,z;
scanf("%d",&k);
getchar();
gets(a);
len=strlen(a);
//Coverting String
temp[0]='$';
temp[1]='#';
z=2;
for(int i=1;i<=len;i++)
{
temp[z++]=a[i-1];
temp[z++]='#';
}
len=z;
int r=0,c=0,check=0,idash,t,count=0;
for(int i=1;i<len;i++)
{
check=0;
idash=c-(i-c);
p[i]=r>i?min(r-i,p[idash]):0;
t=p[i];
while(temp[i+p[i]+1]==temp[i-1-p[i]])
p[i]++;
if(r<i+p[i])
{
c=i;
r=i+p[i];
}
if((k%2==0)&&(p[i]>=k)&&(temp[i]=='#'))
count++;
if((k%2==1)&&(p[i]>=k)&&(temp[i]!='#'))
count++;
}
printf("%d",count);
//getchar();
//getchar();
return 0;
}
You may want to take advantage of C++ short-circuit evaluation of logical expressions.
For example, rearrange the order so you check for '#' first:
if ((temp[i] == '#') && (k % 2 == 0) && (p[i] >= k))
In the above rearrangement, if the character is not '#', none of the other expressions are evaluated.
You may want to extract (p[i] >= k) to an outside if statement since it is common to both:
if (p[i] >= k)
{
if ((temp[i] == '#') && (k % 2 == 0)) ++count;
if ((temp[i] != '#') && (k % 2 == 1)) ++count;
}
The above modification will result in only one evaluation of the expression (p[i] >= k).
Also examine your for loop to see if there are statements or expressions that don't change or are repeated. If a statement or expression doesn't change inside the loop, it is called an invariant, and can be moved before or after the loop.
Statements or expressions that are duplicated (such as array index calculations) can be evaluated and stored in a temporary variable. Although good compilers may do this (depending on the optimization level), in your performance requirements, you may want to help out the compiler.
Another suggestion is to replace p[i] with a pointer to the location or a reference to the location. Again, this is to help out the compiler when the optimization is not set optimally:
int& p_slot_i = p[i]; // This syntax needs checking
// or
int * p_slot_i = &p[i];
//...
t = *p_slot_i;
while(temp[i + *p_slot_i + 1] == temp[i - 1 - *p_slot_i)
{
*p_slot_i++;
}
Lastly, Elimination of spaces, blank lines and curly braces DOES NOT AFFECT PROGRAM PERFORMANCE. A program that is one line or spaced across multiply lines will have the exact assembly translation and the exact performance. So please, add spaces, blank lines and curly braces to improve readability.
Edit 1: performance of min()
You may want to declare you min() function as inline to suggest to the compiler you want the function pasted where it is called, rather than calling the function. Function calls slow down a programs execution.
Please tell me if I am understanding the the substr member function correctly?
result = result.substr(0, pos) + result.substr(pos + 1);
It takes the string from pos, 0 until (but not including), remove[i]
and then + result.substr(pos + 1); concatenates the rest of the string, except but not including the string / char in remove?
string removeLetters2(string text, string remove)
{
int pos;
string result = text;
for (int i = 0; i < remove.length(); i++)
{
while (true)
{
pos = result.find(remove[i]);
if (pos == string::npos)
{
break;
}
else
{
result = result.substr(0, pos) +
result.substr(pos + 1);
}
}
}
return result;
}
In short, you are asking if
result = result.substr(0, pos) +
result.substr(pos + 1);
removes the character at position pos, right?
Short Answer:
Yes.
Longer Answer:
The two-argument call takes the start index and the length (the one argument call goes to the end of string).
It helps to imagine the string like this:
F o o / B a r
0 1 2 3 4 5 6 <- indices
Now remove /:
F o o / B a r
0 1 2 3 4 5 6 <- indices
1 2 3 | <- 1st length
| 1 2 3 <- 2nd length
result = result.substr(0, 3) <- from index 0 with length 3
+ result.substr(4); <- from index 4 to end
As a programmer, always be aware of the difference between distance/index and length.
Better: If index is known:
Your code creates two new, temporary strings, which are then concatenated into a third temporary string, which is then copied to result.
It would be better to ask string to erase (wink wink) in place:
result.erase(pos,1);
// or by iterator
string::iterator it = ....;
result.erase(it,it+1);
This leaves more optimization freedom to the string implementer, who may choose to just move all characters after pos by one to the left. This could, in a specialized scenario, be implemented with a single assignment, a single loop, and within the loop with the x86 swap instruction.
Better: If characters to be deleted are known:
Or, but I am not sure if this gives better performance, but it may give better code, the algorithm remove_if:
#include <algorithm>
// this would remove all slashes, question marks and dots
....
std::string foobar = "ab/d?...";
std::remove_if (foobar.begin(), foobar.end(), [](char c) {
return c=='/' || c=='?' || '.';
});
remove_if accepts any function object.
If there is just one character, it gets easier:
// this would remove all slashes
std::remove (foobar.begin(), foobar.end(), '/');
Although the answer to your question is "yes", there is a better way to go about what you are trying to do. Use string::erase, like this:
result.erase(pos, 1);
This API is designed for removal of characters from the string; it achieves the same result much more efficiently.
Yes, this function removes all letters in remove from text.
since you seem to delete more than one type of character have a look at remove_if from <algorithm> with a special predicate too, although the response of dasblinkenlignt is the good one