'Palindrome Partitioning' soln

Author: joewnga

leetcode/palindrome_partitioning.rs

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+/// 
+/// Problem: 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.
+/// 
+
+// # Solution
+// Time complexity: O(N * 2^N) where N is the length of the string
+// Space complexity: O(N^2)
+
+
+impl Solution {
+    pub fn partition(s: String) -> Vec<Vec<String>> {
+        let mut result = Vec::new();
+       let mut current_partition = Vec::new();
+       let s_chars: Vec<char> = s.chars().collect();
+       
+       Self::backtrack(&s_chars, 0, &mut current_partition, &mut result);
+       
+       result
+   }
+   
+   fn backtrack(s: &[char], start: usize, current: &mut Vec<String>, result: &mut Vec<Vec<String>>) {
+       if start >= s.len() {
+           result.push(current.clone());
+           return;
+       }
+       
+       for end in start..s.len() {
+           if Self::is_palindrome(s, start, end) {
+               let substring: String = s[start..=end].iter().collect();
+               current.push(substring);
+               
+               Self::backtrack(s, end + 1, current, result);
+               
+               current.pop();
+           }
+       }
+   }
+   
+   fn is_palindrome(s: &[char], start: usize, end: usize) -> bool {
+       let mut i = start;
+       let mut j = end;
+       
+       while i < j {
+           if s[i] != s[j] {
+               return false;
+           }
+           i += 1;
+           j -= 1;
+       }
+       
+       true
+    }
+}