Grokking Dynamic Programming Patterns for Coding Interviews
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Count of Palindromic Substrings

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A

Anonymous

· 2 years ago

class Solution:
  def findCPS(self, st):
    
    def is_palindrom(s, i, j):
      while i <= j :
        if s[i] != s[j]:
          return False
        i += 1
        j -= 1
      return True

    count = 0
    for i in range(0, len(st)):
      for j in range(i, len(st)):
        if is_palindrom(st, i, j):
            count += 1
    
    return count
A

Anonymous

· 4 years ago

Can you add the brute force, swapping the max for count does not work on the leetcode problem

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