**Number of Palindromic Substrings - Amazon Top Interview Questions**

### Problem Statement :

Given a lowercase alphabet string s, return the number of palindromic substrings in s. Constraints 1 ≤ n ≤ 1,000 where n is the length of s Example 1 Input s = "tacocat" Output 10 Explanation The palindromic substrings are: "t" "a" "c" "o" "c" "a" "t" "coc" "acoca" "tacocat"

### Solution :

` ````
Solution in C++ :
int solve(string s) {
int count = 0;
for (int i = 0; i < s.size(); i++) {
int j = i - 1, k = i;
while (j >= 0 && k < s.size() && s[j] == s[k]) {
count++;
j--;
k++;
}
j = i, k = i;
while (j >= 0 && k < s.size() && s[j] == s[k]) {
count++;
j--;
k++;
}
}
return count;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(String s) {
int res = 0;
for (int i = 0; i < s.length(); i++) {
int u = expand(s, i, i + 1); // even
int v = expand(s, i, i); // odd
res += u / 2;
res += v / 2;
}
res += s.length();
return res;
}
public int expand(String s, int l, int r) {
while (l >= 0 && r < s.length() && s.charAt(l) == s.charAt(r)) {
l--;
r++;
}
return r - l - 1;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, s):
def isP(s):
return s == s[::-1]
@lru_cache(None)
def dp(s):
count = 0
if len(s) == 1:
return 1
for j in range(1, len(s) + 1):
if isP(s[:j]):
count += 1
return count + dp(s[1:])
return dp(s)
```

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