Interview: Machine coding / regex (Better alternative to my solution) - regex

Question

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.

Answer 1

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.

Answer 2

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
}
}
}

Answer 3

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");
}}

Answer 4

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;
}