Skip to main content

Leetcode 647. Palindromic Substrings. Python (O(n^2))

647Palindromic Substrings 


Given a string s, return the number of palindromic substrings in it.

A string is a palindrome when it reads the same backward as forward.

substring is a contiguous sequence of characters within the string.

 

Example 1:

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

Example 2:

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

 

Constraints:

  • 1 <= s.length <= 1000
  • s consists of lowercase English letters.


Solution :


class Solution:
    def countSubstrings(self, s: str) -> int:
        ans = 0
        
        for i in range(len(s)):
            ans += self.pcount(s, i, i)
            ans += self.pcount(s, i, i+1)

        return ans
    
    def pcount(self, s, l, r):
        ans = 0
        while l >= 0 and r < len(s) and s[l] == s[r]:
            ans += 1
            l -= 1
            r += 1
        return ans



Explanation :







Comments

Popular posts from this blog

Leetcode 424. Longest Repeating Character Replacement. Python (Sliding Window)

  424 .  Longest Repeating Character Replacement You are given a string  s  and an integer  k . You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most  k  times. Return  the length of the longest substring containing the same letter you can get after performing the above operations .   Example 1: Input: s = "ABAB", k = 2 Output: 4 Explanation: Replace the two 'A's with two 'B's or vice versa. Example 2: Input: s = "AABABBA", k = 1 Output: 4 Explanation: Replace the one 'A' in the middle with 'B' and form "AABBBBA". The substring "BBBB" has the longest repeating letters, which is 4.   Constraints: 1 <= s.length <= 10 5 s  consists of only uppercase English letters. 0 <= k <= s.length Solution :  class Solution: def characterReplacement(self, s: str, k: int) -> int: hm = {} ans = 0 ...

Leetcode 322. Coin Change. Python (Greedy? vs DP?)

322 .  Coin Change You are given an integer array  coins  representing coins of different denominations and an integer  amount  representing a total amount of money. Return  the fewest number of coins that you need to make up that amount . If that amount of money cannot be made up by any combination of the coins, return  -1 . You may assume that you have an infinite number of each kind of coin.   Example 1: Input: coins = [1,2,5], amount = 11 Output: 3 Explanation: 11 = 5 + 5 + 1 Example 2: Input: coins = [2], amount = 3 Output: -1 Example 3: Input: coins = [1], amount = 0 Output: 0   Constraints: 1 <= coins.length <= 12 1 <= coins[i] <= 2 31  - 1 0 <= amount <= 10 4   Solution : class Solution:     def coinChange(self, coins: List[int], amount: int) -> int:         dp = [amount + 1] * (amount + 1) #[0...7]         dp[0] = 0        ...

May-6 2020 Challenge

  6.   Majority Element Given an array of size  n , find the majority element. The majority element is the element that appears  more than   ⌊ n/2 ⌋  times. You may assume that the array is non-empty and the majority element always exist in the array. Example 1: Input: [3,2,3] Output: 3 Example 2: Input: [2,2,1,1,1,2,2] Output: 2 Solution in Java  class Solution {     public int majorityElement(int[] num) {         int m = num[0], cnt= 1;     for (int i = 1; i < num.length; i++) {         if (cnt == 0) {             m= num[i];             cnt = 1;         } else if (num[i] == m) {             cnt++;         } else              cnt--;    }      return m;...