The following is the interview question:
Machine coding round: (Time 1hr)
Expression is given and a string testCase, need to evaluate the testCase is valid or not for expression
Expression may contain:
letters [a-z]
'.' ('.' represents any char in [a-z])
'*' ('*' has same property as in normal RegExp)
'^' ('^' represents start of the String)
'$' ('$' represents end of String)
Sample cases:
Expression Test Case Valid
ab ab true
a*b aaaaaab true
a*b*c* abc true
a*b*c aaabccc false
^abc*b abccccb true
^abc*b abbccccb false
^abcd$ abcd true
^abc*abc$ abcabc true
^abc.abc$ abczabc true
^ab..*abc$ abyxxxxabc true
My approach:
Convert the given regular expression into concatenation(ab), alteration(a|b), (a*) kleenstar.
And add + for concatenation.
For example:
abc$ => .*+a+b+c
^ab..*abc$ => a+b+.+.*+a+b+c
Convert into postfix notation based on precedence.
(parantheses>kleen_star>concatenation>..)
(a|b)*+c => ab|*c+
Build NFA based on Thompson construction
Backtracking / traversing through NFA by maintaining a set of states.
When I started implementing it, it took me a lot more than 1 hour. I felt that the step 3 was very time consuming. I built the NFA by using postfix notation +stack and by adding new states and transitions as needed.
So, I was wondering if there is faster alternative solution this question? Or maybe a faster way to implement step 3. I found this CareerCup link where someone mentioned in the comment that it was from some programming contest. So If someone has solved this previously or has a better solution to this question, I'd be happy to know where I went wrong.
Some derivation of Levenshtein distance comes to mind - possibly not the fastest algorithm, but it should be quick to implement.
We can ignore ^ at the start and $ at the end - anywhere else is invalid.
Then we construct a 2D grid where each row represents a unit [1] in the expression and each column represents a character in the test string.
[1]: A "unit" here refers to a single character, with the exception that * shall be attached to the previous character
So for a*b*c and aaabccc, we get something like:
a a a b c c c
a*
b*
c
Each cell can have a boolean value indicating validity.
Now, for each cell, set it to valid if either of these hold:
The value in the left neighbour is valid and the row is x* or .* and the column is x (x being any character a-z)
This corresponds to a * matching one additional character.
The value in the upper-left neighbour is valid and the row is x or . and the column is x (x being any character a-z)
This corresponds to a single-character match.
The value in the top neighbour is valid and the row is x* or .*.
This corresponds to the * matching nothing.
Then check if the bottom-right-most cell is valid.
So, for the above example, we get: (V indicating valid)
a a a b c c c
a* V V V - - - -
b* - - - V - - -
c - - - - V - -
Since the bottom-right cell isn't valid, we return invalid.
Running time: O(stringLength*expressionLength).
You should notice that we're mostly exploring a fairly small part of the grid.
This solution can be improved by making it a recursive solution making use of memoization (and just calling the recursive solution for the bottom-right cell).
This will give us a best-case performance of O(1), but still a worst-case performance of O(stringLength*expressionLength).
My solution assumes the expression must match the entire string, as inferred from the result of the above example being invalid (as per the question).
If it can instead match a substring, we can modify this slightly so, if the cell is in the top row it's valid if:
The row is x* or .*.
The row is x or . and the column is x.
Given only 1 hour we can use simple way.
Split pattern into tokens: a*b.c => { a* b . c }.
If pattern doesn't start with ^ then add .* in the beginning, else remove ^.
If pattern doesn't end with $ then add .* in the end, else remove $.
Then we use recursion: going 3 way in case if we have recurring pattern (increase pattern index by 1, increase word index by 1, increase both indices by 1), going one way if it is not recurring pattern (increase both indices by 1).
Sample code in C#
using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;
namespace ReTest
{
class Program
{
static void Main(string[] args)
{
Debug.Assert(IsMatch("ab", "ab") == true);
Debug.Assert(IsMatch("aaaaaab", "a*b") == true);
Debug.Assert(IsMatch("abc", "a*b*c*") == true);
Debug.Assert(IsMatch("aaabccc", "a*b*c") == true); /* original false, but it should be true */
Debug.Assert(IsMatch("abccccb", "^abc*b") == true);
Debug.Assert(IsMatch("abbccccb", "^abc*b") == false);
Debug.Assert(IsMatch("abcd", "^abcd$") == true);
Debug.Assert(IsMatch("abcabc", "^abc*abc$") == true);
Debug.Assert(IsMatch("abczabc", "^abc.abc$") == true);
Debug.Assert(IsMatch("abyxxxxabc", "^ab..*abc$") == true);
}
static bool IsMatch(string input, string pattern)
{
List<PatternToken> patternTokens = new List<PatternToken>();
for (int i = 0; i < pattern.Length; i++)
{
char token = pattern[i];
if (token == '^')
{
if (i == 0)
patternTokens.Add(new PatternToken { Token = token, Occurence = Occurence.Single });
else
throw new ArgumentException("input");
}
else if (char.IsLower(token) || token == '.')
{
if (i < pattern.Length - 1 && pattern[i + 1] == '*')
{
patternTokens.Add(new PatternToken { Token = token, Occurence = Occurence.Multiple });
i++;
}
else
patternTokens.Add(new PatternToken { Token = token, Occurence = Occurence.Single });
}
else if (token == '$')
{
if (i == pattern.Length - 1)
patternTokens.Add(new PatternToken { Token = token, Occurence = Occurence.Single });
else
throw new ArgumentException("input");
}
else
throw new ArgumentException("input");
}
PatternToken firstPatternToken = patternTokens.First();
if (firstPatternToken.Token == '^')
patternTokens.RemoveAt(0);
else
patternTokens.Insert(0, new PatternToken { Token = '.', Occurence = Occurence.Multiple });
PatternToken lastPatternToken = patternTokens.Last();
if (lastPatternToken.Token == '$')
patternTokens.RemoveAt(patternTokens.Count - 1);
else
patternTokens.Add(new PatternToken { Token = '.', Occurence = Occurence.Multiple });
return IsMatch(input, 0, patternTokens, 0);
}
static bool IsMatch(string input, int inputIndex, IList<PatternToken> pattern, int patternIndex)
{
if (inputIndex == input.Length)
{
if (patternIndex == pattern.Count || (patternIndex == pattern.Count - 1 && pattern[patternIndex].Occurence == Occurence.Multiple))
return true;
else
return false;
}
else if (inputIndex < input.Length && patternIndex < pattern.Count)
{
char c = input[inputIndex];
PatternToken patternToken = pattern[patternIndex];
if (patternToken.Token == '.' || patternToken.Token == c)
{
if (patternToken.Occurence == Occurence.Single)
return IsMatch(input, inputIndex + 1, pattern, patternIndex + 1);
else
return IsMatch(input, inputIndex, pattern, patternIndex + 1) ||
IsMatch(input, inputIndex + 1, pattern, patternIndex) ||
IsMatch(input, inputIndex + 1, pattern, patternIndex + 1);
}
else
return false;
}
else
return false;
}
class PatternToken
{
public char Token { get; set; }
public Occurence Occurence { get; set; }
public override string ToString()
{
if (Occurence == Occurence.Single)
return Token.ToString();
else
return Token.ToString() + "*";
}
}
enum Occurence
{
Single,
Multiple
}
}
}
Here is a solution in Java. Space and Time is O(n). Inline comments are provided for more clarity:
/**
* #author Santhosh Kumar
*
*/
public class ExpressionProblemSolution {
public static void main(String[] args) {
System.out.println("---------- ExpressionProblemSolution - start ---------- \n");
ExpressionProblemSolution evs = new ExpressionProblemSolution();
evs.runMatchTests();
System.out.println("\n---------- ExpressionProblemSolution - end ---------- ");
}
// simple node structure to keep expression terms
class Node {
Character ch; // char [a-z]
Character sch; // special char (^, *, $, .)
Node next;
Node(Character ch1, Character sch1) {
ch = ch1;
sch = sch1;
}
Node add(Character ch1, Character sch1) {
this.next = new Node(ch1, sch1);
return this.next;
}
Node next() {
return this.next;
}
public String toString() {
return "[ch=" + ch + ", sch=" + sch + "]";
}
}
private boolean letters(char ch) {
return (ch >= 'a' && ch <= 'z');
}
private boolean specialChars(char ch) {
return (ch == '.' || ch == '^' || ch == '*' || ch == '$');
}
private void validate(String expression) {
// if expression has invalid chars throw runtime exception
if (expression == null) {
throw new RuntimeException(
"Expression can't be null, but it can be empty");
}
char[] expr = expression.toCharArray();
for (int i = 0; i < expr.length; i++) {
if (!letters(expr[i]) && !specialChars(expr[i])) {
throw new RuntimeException(
"Expression contains invalid char at position=" + i
+ ", invalid_char=" + expr[i]
+ " (allowed chars are 'a-z', *, . ^, * and $)");
}
}
}
// Parse the expression and split them into terms and add to list
// the list is FSM (Finite State Machine). The list is used during
// the process step to iterate through the machine states based
// on the input string
//
// expression = a*b*c has 3 terms -> [a*] [b*] [c]
// expression = ^ab.*c$ has 4 terms -> [^a] [b] [.*] [c$]
//
// Timing : O(n) n -> expression length
// Space : O(n) n -> expression length decides the no.of terms stored in the list
private Node preprocess(String expression) {
debug("preprocess - start [" + expression + "]");
validate(expression);
Node root = new Node(' ', ' '); // root node with empty values
Node current = root;
char[] expr = expression.toCharArray();
int i = 0, n = expr.length;
while (i < n) {
debug("i=" + i);
if (expr[i] == '^') { // it is prefix operator, so it always linked
// to the char after that
if (i + 1 < n) {
if (i == 0) { // ^ indicates start of the expression, so it
// must be first in the expr string
current = current.add(expr[i + 1], expr[i]);
i += 2;
continue;
} else {
throw new RuntimeException(
"Special char ^ should be present only at the first position of the expression (position="
+ i + ", char=" + expr[i] + ")");
}
} else {
throw new RuntimeException(
"Expression missing after ^ (position=" + i
+ ", char=" + expr[i] + ")");
}
} else if (letters(expr[i]) || expr[i] == '.') { // [a-z] or .
if (i + 1 < n) {
char nextCh = expr[i + 1];
if (nextCh == '$' && i + 1 != n - 1) { // if $, then it must
// be at the last
// position of the
// expression
throw new RuntimeException(
"Special char $ should be present only at the last position of the expression (position="
+ (i + 1)
+ ", char="
+ expr[i + 1]
+ ")");
}
if (nextCh == '$' || nextCh == '*') { // a* or b$
current = current.add(expr[i], nextCh);
i += 2;
continue;
} else {
current = current.add(expr[i], expr[i] == '.' ? expr[i]
: null);
i++;
continue;
}
} else { // a or b
current = current.add(expr[i], null);
i++;
continue;
}
} else {
throw new RuntimeException("Invalid char - (position=" + (i)
+ ", char=" + expr[i] + ")");
}
}
debug("preprocess - end");
return root;
}
// Traverse over the terms in the list and iterate and match the input string
// The terms list is the FSM (Finite State Machine); the end of list indicates
// end state. That is, input is valid and matching the expression
//
// Timing : O(n) for pre-processing + O(n) for processing = 2O(n) = ~O(n) where n -> expression length
// Timing : O(2n) ~ O(n)
// Space : O(n) where n -> expression length decides the no.of terms stored in the list
public boolean process(String expression, String testString) {
Node root = preprocess(expression);
print(root);
Node current = root.next();
if (root == null || current == null)
return false;
int i = 0;
int n = testString.length();
debug("input-string-length=" + n);
char[] test = testString.toCharArray();
// while (i < n && current != null) {
while (current != null) {
debug("process: i=" + i);
debug("process: ch=" + current.ch + ", sch=" + current.sch);
if (current.sch == null) { // no special char just [a-z] case
if (test[i] != current.ch) { // test char and current state char
// should match
return false;
} else {
i++;
current = current.next();
continue;
}
} else if (current.sch == '^') { // process start char
if (i == 0 && test[i] == current.ch) {
i++;
current = current.next();
continue;
} else {
return false;
}
} else if (current.sch == '$') { // process end char
if (i == n - 1 && test[i] == current.ch) {
i++;
current = current.next();
continue;
} else {
return false;
}
} else if (current.sch == '*') { // process repeat char
if (letters(current.ch)) { // like a* or b*
while (i < n && test[i] == current.ch)
i++; // move i till end of repeat char
current = current.next();
continue;
} else if (current.ch == '.') { // like .*
Node nextNode = current.next();
print(nextNode);
if (nextNode != null) {
Character nextChar = nextNode.ch;
Character nextSChar = nextNode.sch;
// a.*z = az or (you need to check the next state in the
// list)
if (test[i] == nextChar) { // test [i] == 'z'
i++;
current = current.next();
continue;
} else {
// a.*z = abz or
// a.*z = abbz
char tch = test[i]; // get 'b'
while (i + 1 < n && test[++i] == tch)
; // move i till end of repeat char
current = current.next();
continue;
}
}
} else { // like $* or ^*
debug("process: return false-1");
return false;
}
} else if (current.sch == '.') { // process any char
if (!letters(test[i])) {
return false;
}
i++;
current = current.next();
continue;
}
}
if (i == n && current == null) {
// string position is out of bound
// list is at end ie. exhausted both expression and input
// FSM reached the end state, hence the input is valid and matches the given expression
return true;
} else {
return false;
}
}
public void debug(Object str) {
boolean debug = false;
if (debug) {
System.out.println("[debug] " + str);
}
}
private void print(Node node) {
StringBuilder sb = new StringBuilder();
while (node != null) {
sb.append(node + " ");
node = node.next();
}
sb.append("\n");
debug(sb.toString());
}
public boolean match(String expr, String input) {
boolean result = process(expr, input);
System.out.printf("\n%-20s %-20s %-20s\n", expr, input, result);
return result;
}
public void runMatchTests() {
match("ab", "ab");
match("a*b", "aaaaaab");
match("a*b*c*", "abc");
match("a*b*c", "aaabccc");
match("^abc*b", "abccccb");
match("^abc*b", "abccccbb");
match("^abcd$", "abcd");
match("^abc*abc$", "abcabc");
match("^abc.abc$", "abczabc");
match("^ab..*abc$", "abyxxxxabc");
match("a*b*", ""); // handles empty input string
match("xyza*b*", "xyz");
}}
int regex_validate(char *reg, char *test) {
char *ptr = reg;
while (*test) {
switch(*ptr) {
case '.':
{
test++; ptr++; continue;
break;
}
case '*':
{
if (*(ptr-1) == *test) {
test++; continue;
}
else if (*(ptr-1) == '.' && (*test == *(test-1))) {
test++; continue;
}
else {
ptr++; continue;
}
break;
}
case '^':
{
ptr++;
while ( ptr && test && *ptr == *test) {
ptr++; test++;
}
if (!ptr && !test)
return 1;
if (ptr && test && (*ptr == '$' || *ptr == '*' || *ptr == '.')) {
continue;
}
else {
return 0;
}
break;
}
case '$':
{
if (*test)
return 0;
break;
}
default:
{
printf("default case.\n");
if (*ptr != *test) {
return 0;
}
test++; ptr++; continue;
}
break;
}
}
return 1;
}
int main () {
printf("regex=%d\n", regex_validate("ab", "ab"));
printf("regex=%d\n", regex_validate("a*b", "aaaaaab"));
printf("regex=%d\n", regex_validate("^abc.abc$", "abcdabc"));
printf("regex=%d\n", regex_validate("^abc*abc$", "abcabc"));
printf("regex=%d\n", regex_validate("^abc*b", "abccccb"));
printf("regex=%d\n", regex_validate("^abc*b", "abbccccb"));
return 0;
}
My friend give this wild card(*) matching algorithm . Here is the code .
//This function compares text strings, one of which can have wildcards ('*').
//
BOOL GeneralTextCompare(
char * pTameText, // A string without wildcards
char * pWildText, // A (potentially) corresponding string with wildcards
BOOL bCaseSensitive = FALSE, // By default, match on 'X' vs 'x'
char cAltTerminator = '\0' // For function names, for example, you can stop at the first '('
)
{
BOOL bMatch = TRUE;
char * pAfterLastWild = NULL; // The location after the last '*', if we’ve encountered one
char * pAfterLastTame = NULL; // The location in the tame string, from which we started after last wildcard
char t, w;
// Walk the text strings one character at a time.
while (1)
{
t = *pTameText;
w = *pWildText;
// How do you match a unique text string?
if (!t || t == cAltTerminator)
{
// Easy: unique up on it!
if (!w || w == cAltTerminator)
{
break; // "x" matches "x"
}
else if (w == '*')
{
pWildText++;
continue; // "x*" matches "x" or "xy"
}
else if (pAfterLastTame)
{
if (!(*pAfterLastTame) || *pAfterLastTame == cAltTerminator)
{
bMatch = FALSE;
break;
}
pTameText = pAfterLastTame++;
pWildText = pAfterLastWild;
continue;
}
bMatch = FALSE;
break; // "x" doesn't match "xy"
}
else
{
if (!bCaseSensitive)
{
// Lowercase the characters to be compared.
if (t >= 'A' && t <= 'Z')
{
t += ('a' - 'A');
}
if (w >= 'A' && w <= 'Z')
{
w += ('a' - 'A');
}
}
// How do you match a tame text string?
if (t != w)
{
// The tame way: unique up on it!
if (w == '*')
{
pAfterLastWild = ++pWildText;
pAfterLastTame = pTameText;
w = *pWildText;
if (!w || w == cAltTerminator)
{
break; // "*" matches "x"
}
continue; // "*y" matches "xy"
}
else if (pAfterLastWild)
{
if (pAfterLastWild != pWildText)
{
pWildText = pAfterLastWild;
w = *pWildText;
if (!bCaseSensitive && w >= 'A' && w <= 'Z')
{
w += ('a' - 'A');
}
if (t == w)
{
pWildText++;
}
}
pTameText++;
continue; // "*sip*" matches "mississippi"
}
else
{
bMatch = FALSE;
break; // "x" doesn't match "y"
}
}
}
pTameText++;
pWildText++;
}
return bMatch;
}
This algo works as follow (according to me)
mississippi *sip*
mississippi sip*
ississippi sip*
ssissippi sip*
sissippi ip*
sissippi sip* pAfterLastWild is used to restore the location
issippi ip*
ssippi p*
ssippi sip* again pAfterLastWild is used here.
sippi ip*
sippi sip* here also.
ippi ip*
ppi p*
pi *
i *
I am not able to figure out why pAfterLastTame is needed and what does this piece of code is doing here as i am not able to find use of it .
else if (pAfterLastTame)
{
if (!(*pAfterLastTame) || *pAfterLastTame == cAltTerminator)
{
bMatch = FALSE;
break;
}
pTameText = pAfterLastTame++;
pWildText = pAfterLastWild;
continue;
}
This algo is pretty fast as number of comparisons are equal to size of tameString (correct me i am wrong) .
Does any one know more efficient algorithm than this ??
I've been trying to implement shunting yard algorithm. The code gets an input from the user and after it is evaluated by another function (which is done already), it will be converted to postfix notation and then passed to be computed. The code below is just for the algorithm itself. ie is the vector of tokens made from user's input.
The code makes sense to me but it doesn't compile, yet can't figure out where is not quite right.
double eval_infix_expr(vector<Token> ie, map<string,double> sym_tab)
{
vector<Token> postfix_expr;
static bool IsOperator(const string& token) {
return token == "+" ||
token == "-" ||
token == "*" ||
token == "/" ||
token == "%";
}
static int PrecedenceOf(const string& token) {
if (token == "+" || token == "-") return 0;
if (token == "*" || token == "/" || token == "%") return 1;
throw runtime_error("Unknown operator: " + token);
}
bool expectingOperator = false;
for (size_t i = 0; i < ie.size(); ++i) {
if (IsOperator(ie[i])) {
if (!expectingOperator)
throw runtime_error("Unexpected operator: " + ie[i]);
while (!sym_tab.empty() && IsOperator(sym_tab.top()) &&
PrecedenceOf(sym_tab.top()) >= PrecedenceOf(ie[i])) {
postfix_expr.push_back(sym_tab.top()); sym_tab.pop();
}
sym_tab.push(ie[i]);
expectingOperator = false;
}
else if (ie[i] == "(") {
if (expectingOperator)
throw runtime_error("Expected operator, found (.");
sym_tab.push(ie[i]);
}
else if (ie[i] == ")") {
if (!expectingOperator)
throw runtime_error("Expected value, found ).");
while (!sym_tab.empty() && sym_tab.top() != "(") {
postfix_expr.push_back(sym_tab.top()); sym_tab.pop();
}
if (sym_tab.empty())
throw runtime_error("Imbalanced parentheses.");
sym_tab.pop();
expectingOperator = true;
}
else {
if (expectingOperator)
throw runtime_error("Expecting operator, found " + ie[i]);
postfix_expr.push_back(ie[i]);
expectingOperator = true;
}
}
if (!expectingOperator)
throw runtime_error("Expected value, didn't find one.");
while (!sym_tab.empty()) {
if (sym_tab.top() == "(")
throw runtime_error("Imbalanced parentheses.");
postfix_expr.push_back(sym_tab.top()); sym_tab.pop();
}
}
postfix_Evaluator pe(postfix_expr);
return pe.eval();
}
}
I have a homework and its just bugging for the last 2 days, I've been doing exactly what the pseudo code and still haven't got it right yet.
For example, if I put in "mike]" or "mike]123", my program will crash, due to the stack is empty...From what I observe, the program will crash when:
- The stack is empty
- And there is a close parenthesis
PS: with the help of us2012, I can fix the crash problem. However, the result is not right.
Instead of printing out "invalid", it outputs "valid"
:(
Here is the pseudo code from my professor:
def parse_parenthesis(str):
stack = create a new empty stack of OpenParen objects
for i from 0 to str.size() - 1:
if str[i] is an open parenthesis
stack.push(new OpenParen(str[i]))
else if str[i] is not a close parenthesis:
# str[i] is not a parenthesis of any kind, so ignore it
continue
# otherwise str[i] must be a close parenthesis, try to
# match it with the most recent open paren, on the top
# of the stack
else if stack is empty
return false;
else if stack.peek() is of the same type as str[i]:
# close properly
stack.pop()
else
return false;
if stack is not empty
return false;
else
return true
and here is what I have so far:
.cpp file
bool ParenMatching(const string& s, unique_ptr<string>& output)
{
unique_ptr<OpenParen> stack(new OpenParen);
bool validOpen, validClose, valid;
bool match; //haveCloseParen;
/*string unMatch = "Unmatch";
string unExpected = "Unexpected close paren";
string noError = "No Error";*/
for (size_t i = 0; i < s.length(); i++)
{
// check if its open parenthesis
validOpen = stack->IsOpenParen(s[i]);
// check if its close parenthesis
validClose = stack->IsCloseParen(s[i]);
// if there is open paren, push into the stack
if(validOpen)
stack->PushObj(s[i]);
else if(!validClose)
{
continue;
}
else if(stack->GetObj().IsEmpty())
valid = false;
else if(match = IsMatchParen(s[i], stack))
stack->PopObj();
else
valid = false;
}
if(!stack->GetObj().IsEmpty())
valid = false;
else
valid = true;
return valid;
}
bool IsMatchParen(const char c, const unique_ptr<OpenParen>& stack)
{
bool valid;
if(c == ')' && stack->PeekObj() == '(')
valid = true;
else if (c == ']' && stack->PeekObj() == '[')
valid = true;
else if (c == '}' && stack->PeekObj() == '{')
valid = true;
else if (c == '>' && stack->PeekObj() == '<')
valid = true;
else
valid = false;
return valid;
}
OpenParen.cpp
// Check if its open paren
bool OpenParen::IsOpenParen(const char c)
{
bool isOpen;
if(c == '(' || c == '[' || c == '{' || c == '<')
isOpen = true;
else
isOpen = false;
return isOpen;
}
// check if its close paren
bool OpenParen::IsCloseParen(const char c)
{
bool isClose;
if(c == ')' || c == ']' || c == '}' || c == '>')
isClose = true;
else
isClose = false;
return isClose;
}
gcc 4.7.3: g++ -Wall -Wextra -std=c++0x parens.cpp
#include <iostream>
#include <stack>
#include <string>
#include <vector>
bool isOpen(char c) {
return c == '(' || c == '[' || c == '{' || c == '<'; }
bool isClose(char c) {
return c == ')' || c == ']' || c == '}' || c == '>'; }
bool isMatch(char c1, char c2) {
return (c1 == '(' && c2 == ')')
|| (c1 == '[' && c2 == ']')
|| (c1 == '{' && c2 == '}')
|| (c1 == '<' && c2 == '>'); }
bool parse(const std::string& s) {
std::stack<std::string::value_type> stk;
for (std::string::size_type i = 0; i < s.size(); ++i) {
if (isOpen(s[i])) { stk.push(s[i]); }
else if (isClose(s[i])) {
if (!stk.empty() && isMatch(stk.top(), s[i])) { stk.pop(); }
else { return false; } } }
return stk.empty(); }
int main() {
std::vector<std::string> ptests = {
"", "()", "()()", "(())", "a(a)a" };
std::vector<std::string> ftests = {
"(", ")", ")(", ")()(", "))((" };
for (const auto& t : ptests) {
if (!parse(t)) { std::cout << "fail: " << t << std::endl; } }
for (const auto& t : ftests) {
if (parse(t)) { std::cout << "fail: " << t << std::endl; } }
}
One important thing you should keep in mind about C++ : Multiple else ifs do not live at the same level. That's because else if is not a single entity, it's an else that belongs to the preceding statement and an ifthat begins a new statement, so
if (cond1)
a();
else if (cond 2)
b();
else if (cond 3)
c();
else
d();
is actually
if (cond1)
a();
else {
if (cond 2)
b();
else {
if (cond 3)
c();
else
d();
}
}
Therefore, your check whether the stack is empty needs to be before the check whether the current close parens matches the top of the stack. Otherwise, your program will try to examine the top of the stack when it's empty, and that results in a crash.
Also, setting valid = false is not the right thing to do when you find a condition that indicates a non-match. The loop will still continue and can reset valid to true in a later iteration. You need to immediately return false, as you can already see in your pseudocode.
I want to create a kind of parser of the form:
#include <iostream>
#include <string>
#include <sstream>
#include <cctype>
using namespace std;
bool isValid(istringstream& is)
{
char ch;
is.get(ch); //I know get(ch) is a good start but this is as for as I got :)
.......
....
}
int main()
{
string s;
while(getline(cin,s))
{
istringstream is(s);
cout<<(isValid(is)? "Expression OK" : "Not OK")<<endl;
}
}
A boolean function that returns TRUE if the sequence of char is of the form "5" or "(5+3)" or "((5+3)+6)" or "(((4+2)+1)+6)" ...etc and FALSE for any other case
Basically, an expression will be considered as valid if it is either a single digit or of the form "open parenthesis-single digit-plus sign-single digit-close parenthesis"
Valid Expression = single digit
and
Valid Expression = (Valid Expression + Valid Expression)
Given that there is no limit to the size of the above form (number of opening and closing parenthesis..etc.) I'd like to do that using recursion
Being the newbie that I am.. Thank you for any helpful input!
To do a recursive solution you're gonna want to read the string into a buffer first, then do something like this:
int expression(char* str) {
if (*str == '(') {
int e1 = expression(str + 1);
if (e1 == -1 || *(str + 1 + e) != '+') {
return -1;
}
int e2 = expression(str + 1 + e + 1);
if (e2 == -1 || *(str + 1 + e + 1 + e2) != ')') {
return -1;
}
return 1 + e1 + 1 + e2 + 1;
}
if (*str >= '0' || *str <= '9') {
return 1;
}
return -1;
}
bool isvalid(char* str) {
int e1 = expression(str);
if (e1 < 0) {
return false;
}
if (e1 == strlen(str)) {
return true;
}
if (*(str + e1) != '+') {
return false;
}
int e2 = expression(str + e1 + 1);
if (e2 < 0) {
return false;
}
return (e1 + 1 + e2 == strlen(str));
}
Basically, the expression function returns the length of the valid expression at it's argument. If it's argument begins with a parenthesis, it gets the length of the expression after that, verifies the plus after that, then verifies the closing parenthesis after the next expression. If the argument begins with a number, return 1. If something is messed up, return -1. Then using that function we can figure out whether or not the string is valid by some sums and the length of the string.
I haven't tested the function at all, but the only case this might fail in that I can think of would be excessive parenthesis: ((5)) for example.
An alternative to recursion could be some sort of lexical parsing such as this:
enum {
ExpectingLeftExpression,
ExpectingRightExpression,
ExpectingPlus,
ExpectingEnd,
} ParseState;
// returns true if str is valid
bool check(char* str) {
ParseState state = ExpectingLeftExpression;
do {
switch (state) {
case ExpectingLeftExpression:
if (*str == '(') {
} else if (*str >= '0' && *str <= '9') {
state = ExpectingPlus;
} else {
printf("Error: Expected left hand expression.");
return false;
}
break;
case ExpectingPlus:
if (*str == '+') {
state = ExpectingRightExpression;
} else {
printf("Error: Expected plus.");
return false;
}
break;
case ExpectingRightExpression:
if (*str == '(') {
state = ExpectingLeftExpression;
} else if (*str >= '0' && *str <= '9') {
state = ExpectingEnd;
} else {
printf("Error: Expected right hand expression.");
return false;
}
break;
}
} while (*(++str));
return true;
}
That function's not complete at all, but you should be able to see where it's going. I think the recursion works better in this case anyways.