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# Leet Code : Longest Palindromic Substring Java | CPP | Java Script | Python Solution

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

A string is called a palindrome string if the reverse of that string is the same as the original string.

Example 1:

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

Example 2:

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

Solution :

JavaScript :

``````var longestPalindrome = function(s) {
const n = s.length;
const f = new Array(1001).fill(new Array(1001).fill(false));
let res = s[0];
let max = -1;

if(n == 1) {
return s;
}

if(n == 2) {
return s[0] == s[1] ? s : s[0];
}

for(let i=0; i<n; i++) {
f[i][i] = true;
}

for(let i=1; i<n; i++) {
for(let j=0; j<i; j++) {
if(s[i] == s[j]) {
if(i - j == 1) {
f[i][j] = true;
} else {
f[i][j] = f[i-1][j+1];
}
if(f[i][j] && i-j+1 > max) {
max = i-j+1;
res = s.substring(j, i+1);
}
} else {
f[i][j] = false;
}
}
}

return res;
};``````

CPP :

``````char * longestPalindrome(char * s){
char dp[1001][1001] = {0};
unsigned maxlen = 0;
unsigned startidx = 0;

// Initial setup
for (int i = 0; i < strlen(s); i++)
dp[i][i]=1;

for (int i = strlen(s); i >= 0; i--)
{
for (int j = strlen(s); j > i; j--)
{
unsigned current_len = j-i;
dp[i][j] = 0;
// If the first and the last letter are the same and (the middle is a palindrome OR there is no middle)
if (s[i] == s[j] && (dp[i+1][j-1] == 1 || (current_len == 1)))
{
dp[i][j] = 1;
if (current_len > maxlen)
{
maxlen = current_len;
startidx = i;
}
}
}
}
*(s+startidx+maxlen+1) = '\0';
return (s + startidx);
}``````

Python :

``````class Solution:
def longestPalindrome(self, s: str) -> str:

"""edge case and intialization"""
if not s or len(s) == 1:
return s
LEN = len(s)
dp = [[False for c in range(LEN)] for r in range(LEN)]

"""
all single characters are palindromes.
strings starting at i and ending at i are single characters
"""
maxlength = 1
maxsubstring = s[0]
for i in range(LEN):
dp[i][i] = True

"""
2 character string -> check if they are palindromes
start at i and end at i+1
"""
for i in range(LEN - 1):
if s[i] == s[i + 1]:
dp[i][i+1] = True
maxsubstring = s[i:i+2] #splicing is non-inclusive

"""starting at strings of len 3 and up"""
for l in range(2, LEN):
# you can picture processing diagnally strings of length l + 1
for i in range(LEN - l):
#i is the start index, j is the end index.
length = l + 1
j = i + l

"""
main logic:
boundry characters should be equal
and the non-boundry substring should be a palindrome
"""
if dp[i+1][j-1] and s[i] == s[j]:
dp[i][j] = True
"""
in this section of the code
the maxlength is bound to be
greater than the previous iteration since we
are iterating by string size in the outer most loop
"""
maxlength = length
maxsubstring = s[i:j+1]
continue
return maxsubstring``````