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| 1 | +/// The average complexity of this algorithm will be `O(n * 2^n)`, |
| 2 | +/// where `n` is the length of the string. This is because we spend |
| 3 | +/// `O(n^2)` time to fill the `dp` array and `O(n * 2^n)` time to iterate |
| 4 | +/// over all partitions. So the total complexity would be `O(n^2 + n * 2^n)`. |
| 5 | +/// But here we can neglect `O(n^2)` because it is of lower priority. |
| 6 | +/// So the total complexity will be `O(n * 2^n)`. |
| 7 | +/// The space complexity will be `O(n^2)`, as we store all partitions |
| 8 | +/// in a list of lists (`dp`). |
| 9 | + |
| 10 | +class Solution { |
| 11 | + List<List<String>> partition(String s) { |
| 12 | + // Create an empty list for the result |
| 13 | + List<List<String>> result = []; |
| 14 | + // Create a two-dimensional array for storing boolean values of palindromes |
| 15 | + int n = s.length; |
| 16 | + List<List<bool>> dp = List.generate(n, (_) => List.filled(n, false)); |
| 17 | + // Fill the dp array in O(n^2) time |
| 18 | + fillDP(s, dp); |
| 19 | + // Create a stack for storing the current partition and index |
| 20 | + List<Pair> stack = []; |
| 21 | + // Add an empty list and zero index to the stack |
| 22 | + stack.add(Pair([], 0)); |
| 23 | + // While the stack is not empty, continue the loop |
| 24 | + while (stack.isNotEmpty) { |
| 25 | + // Pop the last element from the stack |
| 26 | + Pair top = stack.removeLast(); |
| 27 | + // Get the current list and index from the element |
| 28 | + List<String> current = top.list; |
| 29 | + int start = top.index; |
| 30 | + // If the index is equal to the length of the string, add the current list to the result |
| 31 | + if (start == n) { |
| 32 | + result.add(List.from(current)); |
| 33 | + } else { |
| 34 | + // Otherwise, iterate over all substrings from the index to the end of the string |
| 35 | + for (int i = start; i < n; i++) { |
| 36 | + // If the substring is a palindrome according to the dp array, add it to a copy of the current list |
| 37 | + // and add the copy of the list and the next index to the stack |
| 38 | + if (dp[start][i]) { |
| 39 | + List<String> copy = List.from(current); |
| 40 | + copy.add(s.substring(start, i + 1)); |
| 41 | + stack.add(Pair(copy, i + 1)); |
| 42 | + } |
| 43 | + } |
| 44 | + } |
| 45 | + } |
| 46 | + |
| 47 | + return result; |
| 48 | + } |
| 49 | + |
| 50 | + void fillDP(String s, List<List<bool>> dp) { |
| 51 | + int n = s.length; |
| 52 | + // Iterate over all substring lengths from 1 to n |
| 53 | + for (int len = 1; len <= n; len++) { |
| 54 | + // Iterate over all starting indices of substrings |
| 55 | + for (int i = 0; i <= n - len; i++) { |
| 56 | + // Compute the ending index of the substring |
| 57 | + int j = i + len - 1; |
| 58 | + // Apply the rule for filling dp[i][j] |
| 59 | + if (s[i] == s[j] && (j - i < 2 || dp[i + 1][j - 1])) { |
| 60 | + dp[i][j] = true; |
| 61 | + } else { |
| 62 | + dp[i][j] = false; |
| 63 | + } |
| 64 | + } |
| 65 | + } |
| 66 | + } |
| 67 | +} |
| 68 | + |
| 69 | +class Pair { |
| 70 | + List<String> list; |
| 71 | + int index; |
| 72 | + Pair(this.list, this.index); |
| 73 | +} |
| 74 | + |
| 75 | +void main(List<String> args) { |
| 76 | + print(Solution().partition("aab")); // [["a","a","b"],["aa","b"]] |
| 77 | + print(Solution().partition("a")); // [["a"]] |
| 78 | +} |
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