LeetCode em substring palíndromo mais longo em Python

Question

Esta é uma pergunta de programação do LeetCode :

Dada uma string s, retorna a substring palindrômica mais longa em s.

Exemplo 1:

Entrada: s = "babad" Saída: "bab" Nota: "aba" também é uma resposta válida.

Abaixo está meu código que falha na seguinte entrada devido ao "Limite de tempo excedido":

""

class Solution(object):

    def longestPalindrome(self, s):
        """
        :type s: str
        :rtype: str
        """
        if len(s) == 0:
            return None
        if len(s) == 1:
            return s

        P = [[False]*len(s) for i in range(len(s))]

        for i in range(len(s)):
            P[i][i]   = True

        for i in range(len(s)-1):
            P[i][i+1] = (s[i]==s[i+1])

        for s_len in range(3,len(s)+1):
            for i in range(len(s)+1-s_len):
                P[i][i+s_len-1] = P[i+1][i+s_len-2] and (s[i]==s[i+s_len-1])

        ip = 0
        jp = 0
        max_len = 1

        for i in range(len(s)):
            for j in range(len(s)):
                if P[i][j] and j+1-i > max_len:
                    max_len = j+1-i
                    ip = i
                    jp = j 
                    continue

        return s[ip:jp+1]

Eu estava tentando seguir a seguinte abordagem descrita na solução do site. Alguém poderia ajudar a ver como tornar meu código mais eficiente?

---

Answer 1

Isenção de responsabilidade: não é um revisor de código

Aqui estão alguns comentários curtos:

• Você está fazendo loops duas vezes.
• Isso tornaria a força bruta.
• A força bruta geralmente falha para algumas questões médias e difíceis no LeetCode.

Solução alternativa

• Aqui, faríamos um loop através de uma vez:

class Solution:
    def longestPalindrome(self, s):
        if len(s) < 1:
            return s

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

        left, right = 0, 1
        for index in range(1, len(s)):
            if index - right > 0 and isPalindrome(index - right - 1, index + 1):
                left, right = index - right - 1, right + 2
            if index - right >= 0 and isPalindrome(index - right, index + 1):
                left, right = index - right, right + 1
        return s[left: left + right]

Sua solução

• Acabei de testar sua solução (passa marginalmente):

class Solution(object):

    def longestPalindrome(self, s):
        """
        :type s: str
        :rtype: str
        """
        if len(s) < 1:
            return s

        P = [[False] * len(s) for i in range(len(s))]

        for i in range(len(s)):
            P[i][i] = True

        for i in range(len(s) - 1):
            P[i][i + 1] = (s[i] == s[i + 1])

        for s_len in range(3, len(s) + 1):
            for i in range(len(s) + 1 - s_len):
                P[i][i + s_len - 1] = P[i + 1][i + s_len - 2] and (s[i] == s[i + s_len - 1])

        ip = 0
        jp = 0
        max_len = 1

        for i in range(len(s)):
            for j in range(len(s)):
                if P[i][j] and j + 1 - i > max_len:
                    max_len = j + 1 - i
                    ip = i
                    jp = j
                    continue

        return s[ip:jp + 1]

• Como os tempos de execução são altos, é possível que às vezes falhe.

• Acho que o LeetCode tem um limite de tempo para cada problema, talvez 10 segundos seja o limite para esse problema específico.

• Provavelmente com base na geolocalização / hora, o tempo de execução também seria diferente.

---

Um pouco mais de otimização:

• Por favor, veja esta linha for j in range(i + 1, len(s))::

class Solution(object):

    def longestPalindrome(self, s):
        """
        :type s: str
        :rtype: str
        """
        if len(s) < 1:
            return s

        P = [[False] * len(s) for _ in range(len(s))]

        for i in range(len(s)):
            P[i][i] = True

        for i in range(len(s) - 1):
            P[i][i + 1] = (s[i] == s[i + 1])

        for s_len in range(3, len(s) + 1):
            for i in range(len(s) + 1 - s_len):
                P[i][i + s_len - 1] = P[i + 1][i + s_len - 2] and (s[i] == s[i + s_len - 1])

        ip = 0
        jp = 0
        max_len = 1

        for i in range(len(s)):
            for j in range(i + 1, len(s)):
                if P[i][j] and j + 1 - i > max_len:
                    max_len = j + 1 - i
                    ip = i
                    jp = j
                    continue

        return s[ip:jp + 1]

• Reduz cerca de 1 segundo, mas ainda não é bom.

• Tenho certeza de que existem mais maneiras de otimizar.

• Espere um pouco! Existem bons revisores Python aqui. Provavelmente ajudaria você.

---

Com alguns comentários:

class Solution:
    def longestPalindrome(self, s):
        if len(s) < 1:
            return s

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

        # We set the left pointer on the first index
        # We set the right pointer on the second index
        # That's the minimum true palindrome
        left, right = 0, 1

        # We visit the alphabets from the second index forward once
        for index in range(1, len(s)):
            # Here we move the right pointer twice and once checking for palindromeness
            # We boundary check using index - right, to remain positive
            if index - right > 0 and isPalindrome(index - right - 1, index + 1):
                print(f"Step {index - 1}: Left pointer is at {index - right - 1} and Right pointer is at {index + 1}")
                print(f"Palindromeness start: {index - right - 1} - Palindromeness end: {index + 1}")
                print(f"Window length: {right}")
                print(f"Before: Left is {left} and Right is {left + right}")
                left, right = index - right - 1, right + 2
                print(f"After: Left is {left} and Right is {left + right}")
                print(f"String: {s[left: left + right]}")
                print('#' * 50)
            if index - right >= 0 and isPalindrome(index - right, index + 1):
                print(f"Step {index - 1}: Left pointer is at {index - right} and Right pointer is at {index + 1}")
                print(f"Palindromeness start: {index - right - 1} - Palindromeness end: {index + 1}")
                print(f"Window length: {right + 1}")
                print(f"Before: Left is {left} and Right is {left + right}")
                left, right = index - right, right + 1
                print(f"After: Left is {left} and Right is {left + right}")
                print(f"String: {s[left: left + right]}")
                print('#' * 50)
        return s[left: left + right]


Solution().longestPalindrome("glwhcebdjbdroiurzfxxrbhzibilmcfasshhtyngwrsnbdpzgjphujzuawbebyhvxfhtoozcitaqibvvowyluvdbvoqikgojxcefzpdgahujuxpiclrrmalncdrotsgkpnfyujgvmhydrzdpiudkfchtklsaprptkzhwxsgafsvkahkbsighlyhjvbburdfjdfvjbaiivqxdqwivsjzztzkzygcsyxlvvwlckbsmvwjvrhvqfewjxgefeowfhrcturolvfgxilqdqvitbcebuooclugypurlsbdfquzsqngbscqwlrdpxeahricvtfqpnrfwbyjvahrtosovsbzhxtutyfjwjbpkfujeoueykmbcjtluuxvmffwgqjgrtsxtdimsescgahnudmsmyfijtfrcbkibbypenxnpiozzrnljazjgrftitldcueswqitrcvjzvlhionutppppzxoepvtzhkzjetpfqsuirdcyqfjsqhdewswldawhdyijhpqtrwgyfmmyhhkrafisicstqxokdmynnnqxaekzcgygsuzfiguujyxowqdfylesbzhnpznayzlinerzdqjrylyfzndgqokovabhzuskwozuxcsmyclvfwkbimhkdmjacesnvorrrvdwcgfewchbsyzrkktsjxgyybgwbvktvxyurufsrdufcunnfswqddukqrxyrueienhccpeuqbkbumlpxnudmwqdkzvsqsozkifpznwapxaxdclxjxuciyulsbxvwdoiolgxkhlrytiwrpvtjdwsssahupoyyjveedgqsthefdyxvjweaimadykubntfqcpbjyqbtnunuxzyytxfedrycsdhkfymaykeubowvkszzwmbbjezrphqildkmllskfawmcohdqalgccffxursvbyikjoglnillapcbcjuhaxukfhalcslemluvornmijbeawxzokgnlzugxkshrpojrwaasgfmjvkghpdyxt")

Impressões:

Step 18: Left pointer is at 18 and Right pointer is at 20
Palindromeness start: 17 - Palindromeness end: 20
Window length: 2
Before: Left is 0 and Right is 1
After: Left is 18 and Right is 20
String: xx
##################################################
Step 25: Left pointer is at 24 and Right pointer is at 27
Palindromeness start: 23 - Palindromeness end: 27
Window length: 3
Before: Left is 18 and Right is 20
After: Left is 24 and Right is 27
String: ibi
##################################################
Step 462: Left pointer is at 460 and Right pointer is at 464
Palindromeness start: 459 - Palindromeness end: 464
Window length: 4
Before: Left is 24 and Right is 27
After: Left is 460 and Right is 464
String: pppp
##################################################

Happy Coding! (ˆ_ˆ)