Question

1332. Remove Palindromic Subsequences

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Description

You are given a string s consisting only of letters 'a' and 'b'. In a single step you can remove one palindromic subsequence from s.

Return _the minimum number of steps to make the given string empty_.

A string is a subsequence of a given string if it is generated by deleting some characters of a given string without changing its order. Note that a subsequence does not necessarily need to be contiguous.

A string is called palindrome if is one that reads the same backward as well as forward.

 

Example 1:

Input: s = "ababa"
Output: 1
Explanation: s is already a palindrome, so its entirety can be removed in a single step.

Example 2:

Input: s = "abb"
Output: 2
Explanation: "abb" -> "bb" -> "". 
Remove palindromic subsequence "a" then "bb".

Example 3:

Input: s = "baabb"
Output: 2
Explanation: "baabb" -> "b" -> "". 
Remove palindromic subsequence "baab" then "b".

 

Constraints:

• 1 <= s.length <= 1000
• s[i] is either 'a' or 'b'.

Solutions

Solution 1

class Solution:
  def removePalindromeSub(self, s: str) -> int:
    return 1 if s[::-1] == s else 2

class Solution {
  public int removePalindromeSub(String s) {
    for (int i = 0, j = s.length() - 1; i < j; ++i, --j) {
      if (s.charAt(i) != s.charAt(j)) {
        return 2;
      }
    }
    return 1;
  }
}

class Solution {
public:
  int removePalindromeSub(string s) {
    for (int i = 0, j = s.size() - 1; i < j; ++i, --j) {
      if (s[i] != s[j]) {
        return 2;
      }
    }
    return 1;
  }
};

func removePalindromeSub(s string) int {
  for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 {
    if s[i] != s[j] {
      return 2
    }
  }
  return 1
}

function removePalindromeSub(s: string): number {
  for (let i = 0, j = s.length - 1; i < j; ++i, --j) {
    if (s[i] !== s[j]) {
      return 2;
    }
  }
  return 1;
}

impl Solution {
  pub fn remove_palindrome_sub(s: String) -> i32 {
    let mut l = 0;
    let mut r = s.len() - 1;
    let s: Vec<char> = s.chars().collect();
    while l < r {
      if s[l] != s[r] {
        return 2;
      }
      l += 1;
      r -= 1;
    }
    1
  }
}