# LeetCode #10 - Regular Expression Matching

Hello happy people 👋! It’s a new day, and we have a new problem at hand -

## Problem Statement

Given an input string s and a pattern p, implement regular expression matching with support for . and * where:

• . Matches any single character.​​​​
• * Matches zero or more of the preceding element.

The matching should cover the entire input string (not partial).

### Constraints:

• 0 <= s.length <= 20
• 0 <= p.length <= 30
• s contains only lowercase English letters.
• p contains only lowercase English letters, ., and *.
• It is guaranteed for each appearance of the character *, there will be a previous valid character to match.

### Examples

Example 1:

Input: s = "aa", p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input: s = "aa", p = "a*"
Output: true
Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input: s = "ab", p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

Example 4:

Input: s = "aab", p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore, it matches "aab".

Example 5:

Input: s = "mississippi", p = "mis*is*p*."
Output: false

## Analysis

We are given two strings —

• s → which we need to match
• p → which has the pattern

There can be two special characters —

• * → This can match 0 or more characters right before it. For e.g. if s = aa and p = a*, then * in p can match 0 or more a (because a is right before *). Thus, we can have one a in place of *, and we are left with only one a in both s and p, which is same. Hence, s and p are a match.
• . → This can match any single character. For e.g., if s = ab and p = .*, then since, . is right before * which means * can be replaced by .. This makes p = .. which means there can be any two characters. These “any two characters” can be a and b, hence, it’s a match.

I know this is slightly complicated 😩 but as we look at some more examples, it would be easier to understand.

## Approach

To understand the approach, let’s take some examples —

s = "aa" and p = "aa", since all the character in both s and p are same, hence it’s a match.

Now, what about s = "aabb" and p = "aab*" 🤔? We know that substrings bb and b* are match because * can be replaced by one b. Since, we already know that remaining substrings aa and aa are match, hence the whole strings also a match.

What can we infer from this? Right, if we have solution of part of a problem, we can use that partial result and can go forward. Also, we can use the already calculated result without calculating it again.

Does this ring a bell 🔔? Yes, this problem satisfies the following two properties -

• Optimal Substructure — Any problem has optimal substructure property if its overall optimal solution can be constructed from the optimal solutions of its subproblems.
• Overlapping Subproblems — Any problem has overlapping sub-problems if finding its solution involves solving the same subproblem multiple times.

It is now evident that we can use good old Dynamic Programming to solve this problem. Below are the steps —

1. Create a boolean 2D dp array with sizes as boolean[][] dp = new boolean[s.length() + 1][p.length() + 1]. We are adding extra 1 to incorporate the case in case either or both of the strings are empty.
2. If both strings are empty, then it’s a match, thus, dp[0][0] = true.
3. Let’s take an example s = "aab" and p = "c*a*b" and create a DP table.

c * a * b
0 1 2 3 4 5
0 TRUE FALSE TRUE FALSE TRUE FALSE
a 1 FALSE FALSE FALSE TRUE TRUE FALSE
a 2 FALSE FALSE FALSE FALSE TRUE FALSE
b 3 FALSE FALSE FALSE FALSE FALSE TRUE

1. First column — it means p is empty and it will match to s only if s is also empty which we have stored in dp[0][0]. Thus, remaining values of dp[0][i] will be false.
2. First row — this is not so easy. It means which p matches empty s. The answer is either an empty pattern or a pattern that represents an empty string such as "a*", "x*y*", "l*m*n*" and so on. In the above example, if s = "" and p = "c*", then due to *, c can be replaced by 0 cs which gives us an empty string. Hence, dp[0][2] = true.
3. For non-empty strings, let’s say that s[i - 1] == p[j - 1] this means the (i - 1)th and (j - 1)th characters are same. This means, we have to check if the remaining strings are a match or not. If they are a match, then the current substrings will be a match, otherwise they won’t be a match i.e., dp[i][j] = dp[i - 1][j - 1]. We’re taking (i - 1)th and (j - 1)th characters to offset empty strings as we’re assuming our strings start from index 1.
4. If p[j - 1] == ".", then it means any single character can be matched. Therefore, here also, we will have to check if the remaining string is a match or not. Thus, dp[i][j] = dp[i - 1][j - 1].
5. If p[j - 1] == "*", then it means either it’s represents an empty string (0 characters), thus dp[i][j] = dp[i][j - 2] or s[i - 1] == p[j - 2] || p[j - 2] == ".", then current character of string equals the char preceding * in pattern so the result is dp[i-1][j].

💡 Try to evaluate the table by yourself to make it more clear.

### Time Complexity

Since we are dealing with each character of both s and p the time complexity will be O(m × n) where m and n are the lengths of s and p respectively.

### Space Complexity

We need a DP array for our intermediate operations of dimensions m × n, hence the space complexity will also be O(m × n).

Let's look at the code now ⬇.

## Code

### Java

public class RegularExpressionMatching {

public boolean isMatch(String s, String p) {
int rows = s.length();
int columns = p.length();
/// Base conditions
if (rows == 0 && columns == 0) {
return true;
}
if (columns == 0) {
return false;
}
// DP array
boolean[][] dp = new boolean[rows + 1][columns + 1];
// Empty string and empty pattern are a match
dp[0][0] = true;
// Deals with patterns with *
for (int i = 2; i < columns + 1; i++) {
if (p.charAt(i - 1) == '*') {
dp[0][i] = dp[0][i - 2];
}
}
// For remaining characters
for (int i = 1; i < rows + 1; i++) {
for (int j = 1; j < columns + 1; j++) {
if (s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == '.') {
dp[i][j] = dp[i - 1][j - 1];
} else if (j > 1 && p.charAt(j - 1) == '*') {
dp[i][j] = dp[i][j - 2];
if (p.charAt(j - 2) == '.' || p.charAt(j - 2) == s.charAt(i - 1)) {
dp[i][j] = dp[i][j] | dp[i - 1][j];
}
}
}
}
return dp[rows][columns];
}
}

### Python

def isMatch(s: str, p: str) -> bool:
rows, columns = (len(s), len(p))
# Base conditions
if rows == 0 and columns == 0:
return True
if columns == 0:
return False
# DP array
dp = [[False for j in range(columns + 1)] for i in range(rows + 1)]
# Since empty string and empty pattern are a match
dp[0][0] = True
# Deals with patterns containing *
for i in range(2, columns + 1):
if p[i - 1] == '*':
dp[0][i] = dp[0][i - 2]
# For remaining characters
for i in range(1, rows + 1):
for j in range(1, columns + 1):
if s[i - 1] == p[j - 1] or p[j - 1] == '.':
dp[i][j] = dp[i - 1][j - 1]
elif j > 1 and p[j - 1] == '*':
dp[i][j] = dp[i][j - 2]
if p[j - 2] == '.' or p[j - 2] == s[i - 1]:
dp[i][j] = dp[i][j] or dp[i - 1][j]
return dp[rows][columns]

### JavaScript

var isMatch = function (s, p) {
const rows = s.length;
const columns = p.length;
/// Base conditions
if (rows == 0 && columns == 0) {
return true;
}
if (columns == 0) {
return false;
}
// DP array
const dp = Array.from({ length: s.length + 1 }, () => [false]);
// Empty string and empty pattern are a match
dp[0][0] = true;
// Deals with patterns with *
for (let i = 1; i < columns + 1; i++) {
if (p[i - 1] === '*') {
dp[0][i] = dp[0][i - 2];
}
else {
dp[0][i] = false;
};
}
// For remaining characters
for (let i = 1; i < rows + 1; i++) {
for (let j = 1; j < columns + 1; j++) {
if (p[j - 1] === '*') {
if (p[j - 2] === s[i - 1] || p[j - 2] === '.') {
dp[i][j] = dp[i][j - 2] || dp[i - 1][j];
} else {
dp[i][j] = dp[i][j - 2];
}
} else if (p[j - 1] === s[i - 1] || p[j - 1] === '.') {
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = false;
}
}
}
return dp[rows][columns];
};

### Kotlin

fun isMatch(s: String, p: String): Boolean {
val rows = s.length
val columns = p.length
/// Base conditions
if (rows == 0 && columns == 0) {
return true
}
if (columns == 0) {
return false
}
// DP array
val dp = Array(rows + 1) { BooleanArray(columns + 1) }
// Empty string and empty pattern are a match
dp[0][0] = true
// Deals with patterns with *
for (i in 2 until columns + 1) {
if (p[i - 1] == '*') {
dp[0][i] = dp[0][i - 2]
}
}
// For remaining characters
for (i in 1 until rows + 1) {
for (j in 1 until columns + 1) {
if (s[i - 1] == p[j - 1] || p[j - 1] == '.') {
dp[i][j] = dp[i - 1][j - 1]
} else if (j > 1 && p[j - 1] == '*') {
dp[i][j] = dp[i][j - 2]
if (p[j - 2] == '.' || p[j - 2] == s[i - 1]) {
dp[i][j] = dp[i][j] or dp[i - 1][j]
}
}
}
}
return dp[rows][columns]
}`

## Conclusion

Congratulations 👏! We have solved another hard problem from LeetCode.

I hope you enjoyed this post. Feel free to share your thoughts on this.

You can find the complete source code on my GitHub repository. If you like what you learn, feel free to fork 🔪 and star ⭐ it.

Till next time… Happy coding 😄 and Namaste 🙏!