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Formatted question description: https://leetcode.ca/all/1332.html
1332. Remove Palindromic Subsequences (Easy)
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.
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: String is already palindrome
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".
Example 4:
Input: s = "" Output: 0
Constraints:
0 <= s.length <= 1000
s
only consists of letters 'a' and 'b'
Related Topics:
String
Solution 1.
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class Solution { public int removePalindromeSub(String s) { if (s == null || s.length() == 0) return 0; int low = 0, high = s.length() - 1; while (low < high) { if (s.charAt(low) == s.charAt(high)) { low++; high--; } else return 2; } return 1; } } ############ class Solution { public int removePalindromeSub(String s) { if (s.length() == 0) { return 0; } if (new StringBuilder(s).reverse().toString().equals(s)) { return 1; } return 2; } }
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// OJ: https://leetcode.com/problems/remove-palindromic-subsequences/ // Time: O(N) // Space: O(1) class Solution { public: int removePalindromeSub(string s) { if (s.empty()) return 0; bool isPalindrome = true; for (int i = 0, j = s.size() - 1; i < j && isPalindrome; ++i, --j) { isPalindrome = s[i] == s[j]; } return isPalindrome ? 1 : 2; } };
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class Solution: def removePalindromeSub(self, s: str) -> int: if not s: return 0 if s[::-1] == s: return 1 return 2
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func removePalindromeSub(s string) int { if len(s) == 0 { return 0 } if s == reverse(s) { return 1 } return 2 } func reverse(s string) string { r := []byte(s) for i, j := 0, len(r)-1; i < j; i, j = i+1, j-1 { r[i], r[j] = r[j], r[i] } return string(r) }
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function removePalindromeSub(s: string): number { if (s.length == 0) return 0; if (s == s.split('').reverse().join('')) return 1; return 2; }
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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 } }