# Shen's Developer Diary # Longest Palindromic Substring

## Problem Statement

Given a string s, return the longest palindromic substring in s.

### Example 1

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


### Example 2

Input: s = "cbbd"
Output: "bb"


Constraints:

• 1 <= s.length <= 1000

• s consist of only digits and English letters.

### Note

Palindrome refers to a string that reads the same forward and backward, the reverse of the string is the same as the string, e.g., hoh, aba, abba…

## Discussion

### Concepts

The concepts in this question are slightly more implicit, but the keyword is on palindrome itself

1. Dynamic Programming - Look at any of the palindromes, you’d find that the central substring of a palindrome is also a palindrome! e.g., in abcba, bcd and c are both palindromes. In other words, we can determine if a string is palindromic by checking if its substring is palindromic, and this is the pattern of Dynamic programming: Solving a problem by solving subproblems.

### Brute Force Attempt

To easily solve the problem, we can just examine all the substrings of s and check for the longest palindrome.

1. Iterating all substrings which will require $$O(n^2)$$ time complexity:

 longest = ""
for i in range(len(s)):
for j in range(i + 1, len(s)):
substring = s[i:j]
if (isPalindrome(substring)) and len(substring) > len(longest):
longest = substring

2. Checking if a string is palindromic:

 def isPalindrome(s):
return s == s[::-1]


Of course, to save time, you can just check half of the string, but $$O(n/2) \thickapprox O(n)$$, so yeah

The time complexity of this brute force attempt is $$O(n^3)$$, for details, please refer to here as the methods are similar.

## Optimal Approach 1 (Dynamic Programming)

In the discussion, I mentioned solving the subproblems, so let’s write it out clearly:

### Idea

A substring, s[left:right + 1] (substring from index left to right, inclusive) is a palindrome if and only if s[left + 1 : right] (substring from index left + 1 to right - 1, inclusive) is palindrome and s[left] == s[right].

Furthermore, there’ll be two base conditions:

1. s[left:right + 1] is a substring if left == right, one character strings are palindromic.

2. s[left:right + 1] is a substring if s[left] == s[right] and right - left == 1, two character strings are palindromic if both characters are the same

For other cases, we can apply the idea and build upon the base conditions.

### Examples

Here are two examples of the idea.

#### Example 1 #### Example 2 ### Recursion Approach (Brute Force)

Let’s implement this idea using the recursion approach:

def recurseIsPalindrome(s, left, right):
if left == right:
return True
elif right - left == 1:
return s[left] == s[right]
else:
return s[left] == s[right] and recurseIsPalindrome(s, left + 1, right - 1)

s = "ebababd"
longest = ""
for right in range(len(s)):
for left in range(right + 1):
if recurseIsPalindrome(s, left, right) and right - left + 1 > len(longest):
longest = s[left : right + 1]

return longest


So what about it? It’s basically reimplementing the initial brute force attempt using recursion, and it has the same time complexity of $$O(n^3)$$. However, we can build upon this by using Dynamic Programming: We store the previous results.

Notice that there have been some repetitions, e.g.,

1. left = 1 and right = 5. To check if the substring “babab” is a palindrome, we examine substrings “aba” and “b”. 2. left = 2 and right = 4. To check if the substring “bab” is a palindrome, we examine substrings “a”. Notice the repetition? I’ve examined “aba” in (1), but I’m examining “aba” again in (2).

### Tabulation Method

This is where we will be optimizing, we will cache the results through the tabulation method from Dynamic Programming.

To achieve this, for every pair of [left, right] we came across, we will cache the results for a later lookup.

One way to do it is through hashmap, but since we know the exact size, array will be a better approach to reduce space usage:

class Solution:
def longestPalindrome(self, s: str) -> str:
## The "cache"
dp = [[False] * len(s) for _ in range(len(s))]
res = ""
for right in range(len(s)):
for left in range(right + 1):
## Reimplementing recurseIsPalindrome using cache
if right == left:
dp[right][left] = True
elif right == left + 1:
dp[right][left] = s[right] == s[left]
else:
dp[right][left] = dp[right - 1][left + 1] and s[right] == s[left]
if dp[right][left] and right - left + 1 > len(res):
res = s[left:right + 1]
return res


Notice the similarity of this with the recursion approach? That’s the idea.

Here’s how the table will be filled up from the algorithm:

1. From our first condition, all single character strings (the diagonal) are palindrome, let’s fill them with True: 2. Next, we look at the second condition, two characters strings (the second diagonal), where right == left + 1, they are palindrome if s[right] == s[left]:

(somehow all are False) 3. Finally, we look at last condition, the box will be True if dp[right -1][left + 1](that's it, the top right box) is True, and s[right] == s[left]: 4. And so on…

Note that understanding the table might or might not help in understanding the algorithm, they’re just illustrations.

### Analysis

#### Time Complexity

As shown in the table, we are only filling up the bottom left portion of the table, hence it will be $$O(n^2 / 2) \thickapprox O(n^2)$$

Of course, the easier way to see is that the for loop is only nested for one level

#### Space Complexity

We have initialized a list / array of size $$n \times n$$ , hence the space complexity is $$O(n^2)$$ as well.

### Improving Space Complexity

It’s common in dynamic programming to improve the space complexity two using only two rows.

Notice from the diagrams that I’ll only be referring to data from the previous row? Thus, we will only store the data of current and previous rows.

To do this, we will alternate between two rows, we will use row 1 if it is in an odd iteration, and row 0 if it is in an even iteration.

class Solution:
def longestPalindrome(self, s: str) -> str:
dp = [[False] * len(s) for _ in range(2)]

res = ""
for right in range(len(s)):
cur = right % 2 == 0
for left in range(right + 1):
if right == left:
dp[cur][left] = True
elif right == left + 1:
dp[cur][left] = s[right] == s[left]
else:
dp[cur][left] = dp[not cur][left + 1] and s[right] == s[left]
if dp[cur][left] and right - left + 1 > len(res):
res = s[left:right + 1]
return res


Basically, not cur refers to the previous row while cur is the current row.

That’s it, by using a simple trick, we could easily trim down our space complexity from $$O(n^2)$$ to $$O(n)$$, and here’s the performance after optimization: ## Optimal Approach 2

On the other hand, the second approach is more trivial and straightforward. The idea is that instead of shrinking from outside in, we expand inside out.

### Idea

How does it work? We will treat every character in a string as a center of a palindrome, and find the largest palindrome from there.

To put it simply, rather than checking from two ends, we check from the center, and get the longest palindrome from this center. However, our center might be two characters as well. In this case, we will assume the center to be in between two characters. ### Implementation

The ideas can be implemented through the following steps:

1. Iterate through every character in the string. 2. Expand from the point:

1. Every character is a center 2. The point between the current character and the next character is a center 3. Compare and get the longest substring from 2(a) and 2(b) However, it’ll be difficult to represent the concept of center between the characters, hence, left & right pointers will be used. They will be implemented slightly differently:

1. For scenario 2(a) - left = center = right 2. For scenario 2(b) - left = right of center and right = left of center You must be wondering if I wrote it wrongly, no, in fact I didn’t.

This is to ensure that the initial string of palindrome, s[left: right + 1] == “”, and not “dd” as the initial pair of characters may not always be palindromic, e.g., “db”.

And finally, here’s the codes

def expandFromCenter(s, left, right):
leftP, rightP = left, right
# Making sure that it's still within the boundary
while leftP > 0 and rightP < len(s) - 1 and s[leftP - 1] == s[rightP + 1]:
leftP -= 1
rightP += 1
return rightP - leftP

class Solution:
def longestPalindrome(self, s: str) -> str:
res = ""
for center in range(len(s)):
onePointCenter = expandFromCenter(s, center, center)
twoPointCenter = expandFromCenter(s, center + 1, center)
best = max(onePointCenter, twoPointCenter)
if best + 1 > len(res):
isTwoPoint = best % 2 == 1
leftEnd = center - best // 2
rightEnd = center + best // 2 + isTwoPoint
res = s[leftEnd:rightEnd + 1]
return res


I believe the only doubt here is on the isTwoPoint.

So let’s look back at both scenarios, and notice that scenario (1) will always result in even best while scenario (2) will result in odd best. (best = length - 1 = right - left) Hence, for scenario (1), the leftEnd and rightEnd is simply $$center \pm best / 2$$.

(as best / 2 is the distance from the center to the end)

However, it’s a bit tricky for scenario (2). Note the distance from center to left and the right end is best / 2 + 1 since best is odd, and we are doing floor division. Besides, from the codes, our center is always the left character of the center point. Thus, the leftEnd remains center - best / 2.

(without the need of + 1, as it's already on the left side)

### Analysis

#### Time Complexity

expandFromCenter operation is O(n), and iterating the string is $$O(n)$$. Since we do expandFromCenter in every iteration, the resulting time complexity is $$O(n^2)$$.

#### Space Complexity

As we only store pointers, it can be said that space complexity is $$O(1)$$.

Of course, the improvement is provided on the assumption that the result res is not taken into account, otherwise, it will remain the same as O(n) as res can have the same length as s.

Oh and here's the performance on Leetcode using this approach: ## Last Words

This problem represents a typical dynamic programming style problem. Hence, for those who have not learned about DP, solution 1 might be more challenging. Nonetheless, solution 2 should be friendly to most.

Lastly, there’s also a more advanced and very specific $$O(n)$$ algorithm that solves this particular problem.

If you’re interested, you can check out Wikipedia and this amazing article on it. (though I felt like the codes provided by the article is incorrect)