Palindrome Partitioning Leetcode Solution
Difficulty: Medium
Topics: String, Dynamic Programming, Backtracking
Given a string s, partition s such that every substring[^1] of the partition is a palindrome[^2]. 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 <= 16scontains only lowercase English letters.
[^1]: Substring A substring is a contiguous non-empty sequence of characters within a string.
[^2]: Palindrome A palindrome is a string that reads the same forward and backward.
Solution:
We can use a combination of backtracking and dynamic programming. The goal is to explore all possible partitions and check if each partition is a palindrome.
Steps to Solve:
- Backtracking:
- We’ll explore every possible partitioning of the string. For each possible starting index, we try to partition the string at every possible end index.
- If a substring is a palindrome, we recursively partition the remaining string.
- Palindrome Check:
- We need a helper function to check if a given substring is a palindrome. This can be done by comparing the substring with its reverse.
- Collect Results:
- When we reach the end of the string, we’ve found a valid partitioning, so we add it to the result list.
Let’s implement this solution in PHP: 131. Palindrome Partitioning
<?php
// Example usage:
$s1 = "aab";
$s2 = "a";
print_r(partition($s1)); // Output: [["a","a","b"],["aa","b"]]
print_r(partition($s2)); // Output: [["a"]]
?>Explanation:
partition($s): This function initializes the backtracking process.backtrack($s, $start, &$current, &$result):- It explores each substring starting at index
start. - If a palindrome is found, it’s added to the current partition (
$current), and the function is called recursively for the next part of the string. - Once all partitions from the current start point are explored, the last substring is removed from
$currentto backtrack and try other possibilities.
- It explores each substring starting at index
isPalindrome($s): This helper function checks if a string is a palindrome by comparing characters from the start and end.
Time Complexity:
- The time complexity is exponential in the worst case, as we explore all possible partitions (backtracking).
- The palindrome check is (O(n)) for each substring.
This solution works efficiently for the given constraint where the maximum length of the string is 16.
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