## Description

Given a string, your task is to count how many palindromic substrings in this string.

The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.

Example 1:

Input: "abc"
Output: 3
Explanation: Three palindromic strings: "a", "b", "c".


Example 2:

Input: "aaa"
Output: 6
Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".


Note:

1. The input string length won’t exceed 1000.

## Solutions

### 1. Brute Force

第一次写出了脑子里有泡的 BF：

# Time: O(n^3)
# Space: O(1)
class Solution:
def countSubstrings(self, s: str) -> int:
n = len(s)
res = []
for i in range(n):
for j in range(i, n):
if self.is_palindromic(s[i:j+1]):
res.append(s[i:j+1])
return len(res)

def is_palindromic(self, s: str) -> bool:
l, r = 0, len(s) - 1
while l <= r:
if s[l] != s[r]:
return False
l += 1
r -= 1
return True
# Time Limit Exceeded
# 128/130 cases passed (N/A)

# Time: O(n^2)
# Space: O(1)
class Solution:
def countSubstrings(self, s: str) -> int:
n = len(s)
res = 0
for i in range(n):
res += self.count(s, i, i) + self.count(s, i, i+1)
return res

def count(self, s: str, i: int, j:int) -> bool:
n = len(s)
cnt = 0
while i >= 0 and j < n:
if s[i] == s[j]:
cnt += 1
else:
break
i -= 1
j += 1
return cnt

# 130/130 cases passed (116 ms)
# Your runtime beats 86.22 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.6 MB)


### 2. DP

# Time: O(n^2)
# Space: O(n^2)
class Solution:
def countSubstrings(self, s: str) -> int:
n = len(s)
cnt = 0
dp = [[False for _ in range(n)] for _ in range(n)]

for i in range(n-1, -1, -1):
for j in range(i, n):
# j - i < 3: AXA
dp[i][j] = (s[i] == s[j]) and (j - i < 3 or dp[i+1][j-1])
if dp[i][j]:
cnt += 1
return cnt

# 130/130 cases passed (352 ms)
# Your runtime beats 29.95 % of python3 submissions
# Your memory usage beats 34.61 % of python3 submissions (21.4 MB)