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).

Q6.)

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

Try palindrome in this way your code execution will be fast

Beat my code execution speed of 4seconds and get some super quick questions

Sometimes 1second also

134 ms means a good size application

11 minutes 3 codes 2 times rejection

Longest Palindromic Substring ?

Input: s = "babad"
Output: "bab"
Explanation: "aba" is also a valid answer.

Common mistake

Some people will be tempted to come up with a quick solution, which is unfortunately flawed (however can be corrected easily):

Reverse SSS and become S′S'S
′
. Find the longest common substring between SSS and S′S'S
′
, which must also be the longest palindromic substring.

This seemed to work, let’s see some examples below.

For example, SSS = "caba", S′S'S
′
= "abac".

The longest common substring between SSS and S′S'S
′
is "aba", which is the answer.

Let’s try another example: SSS = "abacdfgdcaba", S′S'S
′
= "abacdgfdcaba".

The longest common substring between SSS and S′S'S
′
is "abacd". Clearly, this is not a valid palindrome.

Algorithm

We could see that the longest common substring method fails when there exists a reversed copy of a non-palindromic substring in some other part of SSS. To rectify this, each time we find a longest common substring candidate, we check if the substring’s indices are the same as the reversed substring’s original indices. If it is, then we attempt to update the longest palindrome found so far; if not, we skip this and find the next candidate.

This gives us an O(n2)O(n^2)O(n
2
) Dynamic Programming solution which uses O(n2)O(n^2)O(n
2
) space (could be improved to use O(n)O(n)O(n) space). Please read more about Longest Common Substring here.

Approach 2: Brute Force
The obvious brute force solution is to pick all possible starting and ending positions for a substring, and verify if it is a palindrome.

Complexity Analysis

Time complexity : O(n3)O(n^3)O(n
3
). Assume that nnn is the length of the input string, there are a total of (n2)=n(n−1)2\binom{n}{2} = \frac{n(n-1)}{2}(
2
n
​
)=
2
n(n−1)
​
such substrings (excluding the trivial solution where a character itself is a palindrome). Since verifying each substring takes O(n)O(n)O(n) time, the run time complexity is O(n3)O(n^3)O(n
3
).

Space complexity : O(1)O(1)O(1).

Approach 3: Dynamic Programming
To improve over the brute force solution, we first observe how we can avoid unnecessary re-computation while validating palindromes. Consider the case "ababa". If we already knew that "bab" is a palindrome, it is obvious that "ababa" must be a palindrome since the two left and right end letters are the same.

We define P(i,j)P(i,j)P(i,j) as following:

P(i,j)={true,if the substring Si…Sj is a palindromefalse,otherwise. P(i,j) = \begin{cases} \text{true,} &\quad\text{if the substring } S_i \dots S_j \text{ is a palindrome}\\ \text{false,} &\quad\text{otherwise.} \ \end{cases}
P(i,j)={
true,
false,
​

if the substring S
i
​
…S
j
​
is a palindrome
otherwise.
​

Therefore,

P(i,j)=(P(i+1,j−1) and Si==Sj)P(i, j) = ( P(i+1, j-1) \text{ and } S_i == S_j )
P(i,j)=(P(i+1,j−1) and S
i
​
==S
j
​
)
The base cases are:

P(i,i)=trueP(i, i) = true
P(i,i)=true
P(i,i+1)=(Si==Si+1)P(i, i+1) = ( S_i == S_{i+1} )
P(i,i+1)=(S
i
​
==S
i+1
​
)
This yields a straight forward DP solution, which we first initialize the one and two letters palindromes, and work our way up finding all three letters palindromes, and so on...

Complexity Analysis

Time complexity : O(n2)O(n^2)O(n
2
). This gives us a runtime complexity of O(n2)O(n^2)O(n
2
).

Space complexity : O(n2)O(n^2)O(n
2
). It uses O(n2)O(n^2)O(n
2
) space to store the table.

Additional Exercise

Could you improve the above space complexity further and how?

Approach 4: Expand Around Center
In fact, we could solve it in O(n2)O(n^2)O(n
2
) time using only constant space.

We observe that a palindrome mirrors around its center. Therefore, a palindrome can be expanded from its center, and there are only 2n−12n - 12n−1 such centers.

You might be asking why there are 2n−12n - 12n−1 but not nnn centers? The reason is the center of a palindrome can be in between two letters. Such palindromes have even number of letters (such as "abba") and its center are between the two 'b's.

Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays.

The overall run time complexity should be O(log (m+n)).

Beat the top 20% ❤️ to be no.1

The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P A H N
A P L S I I G
Y I R
And then read line by line: "PAHNAPLSIIGYIR"

Write the code that will take a string and make this conversion given a number of rows:

string convert(string s, int numRows);


Example 1:

Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"
Example 2:

Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:
P I N
A L S I G
Y A H R
P I
Example 3:

Input: s = "A", numRows = 1
Output: "A"


Constraints:

1 <= s.length <= 1000
s consists of English letters (lower-case and upper-case), ',' and '.'.
1 <= numRows <= 1000

Solve it n get 30 for successful testcases passing of 60 testcases