DP Other Interval DP¶
Table of Contents¶
- 5. Longest Palindromic Substring (Medium)
- 647. Palindromic Substrings (Medium)
- 3040. Maximum Number of Operations With the Same Score II (Medium)
- 375. Guess Number Higher or Lower II (Medium)
- 1130. Minimum Cost Tree From Leaf Values (Medium)
- 96. Unique Binary Search Trees (Medium)
- 1770. Maximum Score from Performing Multiplication Operations (Hard)
- 1547. Minimum Cost to Cut a Stick (Hard)
- 1039. Minimum Score Triangulation of Polygon (Medium)
- 1000. Minimum Cost to Merge Stones (Hard)
- 2019. The Score of Students Solving Math Expression (Hard)
- 3277. Maximum XOR Score Subarray Queries (Hard)
- 87. Scramble String (Hard)
- 312. Burst Balloons (Hard)
- 664. Strange Printer (Hard)
- 546. Remove Boxes (Hard)
- 471. Encode String with Shortest Length (Hard) 👑
- 3018. Maximum Number of Removal Queries That Can Be Processed I (Hard) 👑
5. Longest Palindromic Substring¶
class longestPalindrome:
"""Return the longest palindromic substring in s."""
@staticmethod
def dp(s: str) -> str:
n = len(s)
if n <= 1:
return s
start, maxLen = 0, 1
# Init
dp = [[0] * n for _ in range(n)]
for i in range(n):
dp[i][i] = 1
# update
for j in range(1, n):
for i in range(j):
if s[i] == s[j]:
if j - i <= 2:
dp[i][j] = 1
else:
dp[i][j] = dp[i + 1][j - 1]
if dp[i][j] and j - i + 1 > maxLen:
maxLen = j - i + 1
start = i
return s[start : start + maxLen]
@staticmethod
def expand_around_center(s: str) -> str:
def expand(left, right):
while left >= 0 and right < len(s) and s[left] == s[right]:
left -= 1
right += 1
return right - left - 1
if len(s) <= 1:
return s
start, end = 0, 0
for i in range(len(s)):
odd = expand(i, i)
even = expand(i, i + 1)
maxLen = max(odd, even)
if maxLen > end - start:
start = i - (maxLen - 1) // 2
end = i + maxLen // 2
return s[start : end + 1]
if __name__ == "__main__":
s = "babad"
assert longestPalindrome.dp(s) in ["bab", "aba"]
assert longestPalindrome.expand_around_center(s) in ["bab", "aba"]
647. Palindromic Substrings¶
"""
- Return the number of palindromic substrings in `s`.
- Bottom-up DP table
| dp | a | b | b | a | e |
| :---: | :-: | :-: | :-: | :-: | :-: |
| **a** | 1 | 0 | 0 | 1 | 0 |
| **b** | 0 | 1 | 1 | 0 | 0 |
| **b** | 0 | 0 | 1 | 0 | 0 |
| **a** | 0 | 0 | 0 | 1 | 0 |
| **e** | 0 | 0 | 0 | 0 | 1 |
"""
def countSubstrings(s: str) -> int:
n = len(s)
dp = [[0] * n for _ in range(n)]
res = 0
for i in range(n - 1, -1, -1):
for j in range(i, n):
if s[i] == s[j]:
if j - i <= 1:
dp[i][j] = 1
res += 1
elif dp[i + 1][j - 1]:
dp[i][j] = 1
res += 1
return res
print(countSubstrings("abbae")) # 7