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Solution Tricky part is to figure out how to encode the string....

solution: First take a look at a few cases: 1,2,3,4 -> 0 1,1,1,1,1,2 -> 4 1,1,1,1,2 ->...

Solution: DFS + Trie The key is how we find the matches of word from the dictionary. W...

Solution: BFS while (!queue.isEmpty()) { int curCourse = queue.poll();...

Solution BFSt minimumSemesters(int n, int[][] relations) { int[] inDegree = new int[n ...

 ```java class Solut...

 ```java class MyHas...

Solution: Array + LinkedList706. Design HashMap706. Design HashMap706. Design HashMap7...

 ```java class Solut...

solution: DFS O(N), O(N)iven the root of a binary tree, return the length of the longe...

solution: Sliding window + counter, O(N), O(K) simplest approach: sliding window + two...

solution: dp O(N^2 * k),O(NK) dp[i][j] represents the maximum length of subsequence en...

Intuitive: For each data center, can upgrade 1, 2, 3,... up to K servers, using O(N) l...

Solution Build the Tree: Use an graph to represent the undirected tree. DFS: Bob going...

Solution: N-tree pre order traversal 「else return x's oldest child who's not in curOrd...

1.Keep record of previos of 1 2. Group by 0 s // 0000000001111 if (seats[prev] != 1)...

solutiuon List<Integer> list1 = new ArrayList<>(); List<Integer> list2 = new ArrayList...

SOLUTION: DP dp[i][j] = number of ways to build the first i characters of target using...

solution DFS int leftSum = resLeft[0], leftCount = resLeft[1]; int rightSum = resRight...

Solution Use a map to track pattern character -> substring. the mapping is bi-directio...