Dmitrii Volyx

Given a string s, return all the palindromic permutations (without duplicates) of it. You may return the answer in any order. If s has no palindromic permutation, return an empty list. Example 1: Input: s = "aabb" Output: ["abba","baab"] Example 2: Input: s = "abc" Output: [] Constraints: 1 <= s.length <= 16 s consists of only lowercase English letters. Solution class Solution { public List<String> generatePalindromes(String s) { Set<String> res = new HashSet<>(); int[] freq = new int[256]; int count = 0; for (char c : s.toCharArray()) { freq[c]++; } if (!canPermute(freq)) { return Collections.emptyList(); } char mid = ' '; int k = 0; char[] st = new char[s.length() / 2]; for (int i = 0; i < freq.length; i++) { if (freq[i] % 2 == 1) { mid = (char) i; } for (int j = 0; j < freq[i] / 2; j++) { st[k++] = (char) i; } } permute(0, st, mid, res); return new ArrayList<>(res); } void permute(int i, char[] st, char mid, Set<String> res) { if (i == st.length) { String m = ((mid == ' ') ? "": "" + mid); String val = new String(st) + m + new StringBuilder(new String(st)).reverse().toString(); res.add(val); } else { for (int j = i; j < st.length; j++) { swap(i, j, st); permute(i + 1, st, mid, res); swap(i, j, st); } } } boolean canPermute(int[] freq) { int count = 0; for (int f: freq) { count += f % 2; } return count <= 1; } void swap(int i, int j, char[] st) { char t = st[i]; st[i] = st[j]; st[j] = t; } }

Given a string s which consists of lowercase or uppercase letters, return the length of the longest palindrome that can be built with those letters. Letters are case sensitive, for example, “Aa” is not considered a palindrome here. Example 1: Input: s = "abccccdd" Output: 7 Explanation: One longest palindrome that can be built is "dccaccd", whose length is 7. Example 2: Input: s = "a" Output: 1 Example 3: Input: s = "bb" Output: 2 Constraints: ...

Given a string s, return true if a permutation of the string could form a palindrome. Example 1: Input: s = "code" Output: false Example 2: Input: s = "aab" Output: true Example 3: Input: s = "carerac" Output: true Constraints: 1 <= s.length <= 5000 s consists of only lowercase English letters. Solution class Solution { public boolean canPermutePalindrome(String s) { int[] freq = new int[256]; for (char c: s.toCharArray()) { freq[c]++; } int count = 0; for (int f: freq) { count += f % 2; } return count <= 1; } }

You are given an m x n grid rooms initialized with these three possible values. -1 A wall or an obstacle. 0 A gate. INF Infinity means an empty room. We use the value 231 - 1 = 2147483647 to represent INF as you may assume that the distance to a gate is less than 2147483647. Fill each empty room with the distance to its nearest gate. If it is impossible to reach a gate, it should be filled with INF. ...

Given an integer array nums, return the number of longest increasing subsequences. Notice that the sequence has to be strictly increasing. Example 1: Input: nums = [1,3,5,4,7] Output: 2 Explanation: The two longest increasing subsequences are [1, 3, 4, 7] and [1, 3, 5, 7]. Example 2: Input: nums = [2,2,2,2,2] Output: 5 Explanation: The length of longest continuous increasing subsequence is 1, and there are 5 subsequences' length is 1, so output 5. Constraints: ...

A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> … -> sk such that: Every adjacent pair of words differs by a single letter. Every si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList. sk == endWord Given two words, beginWord and endWord, and a dictionary wordList, return all the shortest transformation sequences from beginWord to endWord, or an empty list if no such sequence exists. Each sequence should be returned as a list of the words [beginWord, s1, s2, …, sk]. ...

Given a pattern and a string s, return true if s matches the pattern. A string s matches a pattern if there is some bijective mapping of single characters to strings such that if each character in pattern is replaced by the string it maps to, then the resulting string is s. A bijective mapping means that no two characters map to the same string, and no character maps to two different strings. ...

You are given an array routes representing bus routes where routes[i] is a bus route that the ith bus repeats forever. For example, if routes[0] = [1, 5, 7], this means that the 0th bus travels in the sequence 1 -> 5 -> 7 -> 1 -> 5 -> 7 -> 1 -> … forever. You will start at the bus stop source (You are not on any bus initially), and you want to go to the bus stop target. You can travel between bus stops by buses only. ...

We can shift a string by shifting each of its letters to its successive letter. For example, “abc” can be shifted to be “bcd”. We can keep shifting the string to form a sequence. For example, we can keep shifting “abc” to form the sequence: “abc” -> “bcd” -> … -> “xyz”. Given an array of strings strings, group all strings[i] that belong to the same shifting sequence. You may return the answer in any order. ...

Given two strings s and t of lengths m and n respectively, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return the empty string “”. The testcases will be generated such that the answer is unique. A substring is a contiguous sequence of characters within the string. Example 1: Input: s = "ADOBECODEBANC", t = "ABC" Output: "BANC" Explanation: The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t. Example 2: Input: s = "a", t = "a" Output: "a" Explanation: The entire string s is the minimum window. Example 3: Input: s = "a", t = "aa" Output: "" Explanation: Both 'a's from t must be included in the window. Since the largest window of s only has one 'a', return empty string. Constraints: ...