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Palindrome Partitioning - Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s. Example 1: Input: s = "aab" Output: [["a","a","b"],["aa","b"]] Example 2: Input: s = "a" Output: [["a"]] Constraints: * 1 <= s.length <= 16 * s contains only lowercase English letters.
Palindrome Partitioning II - Given a string s, partition s such that every substring of the partition is a palindrome. Return the minimum cuts needed for a palindrome partitioning of s. Example 1: Input: s = "aab" Output: 1 Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut.
Apr 9, 2016 · Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s. Example 1: Input: s = "aab". Output: [["a","a","b"],["aa","b"]] Example 2: Input: s = "a". Output: [["a"]] Constraints:
In-depth solution and explanation for LeetCode 131. Palindrome Partitioning in Python, Java, C++ and more. Intuitions, example walk through, and complexity analysis.
class Solution: def partition (self, s: str)-> List [List [str]]: ans = [] def isPalindrome (s: str)-> bool: return s == s [::-1] def dfs (s: str, j: int, path: List [str], ans: List [List [str]])-> None: if j == len (s): ans. append (path) return for i in range (j, len (s)): if isPalindrome (s [j: i + 1]): dfs (s, i + 1, path + [s [j: i + 1 ...
Palindrome Partitioning IV - Given a string s, return true if it is possible to split the string s into three non-empty palindromic substrings. Otherwise, return false. A string is said to be palindrome if it the same string when reversed.
Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s. Example 1: Input: s = "aab". Output: [["a","a","b"],["aa","b"]] Example 2: Input: s = "a". Output: [["a"]] Constraints:
